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\newtheorem{example}{\"Ornek}
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\begin{document}
\title{Do\u grusal Cebir}
\subtitle{Lineer Cebire Giri\c s dersi \"ozeti}
\author{David Pierce}
\date{13 Ocak 2020}
\publishers{Matematik B\"ol\"um\"u, MSGS\"U\\
%\mbox{}\\
\url{dpierce@msgsu.edu.tr}\\
\url{mat.msgsu.edu.tr/~dpierce/}\\
\url{polytropy.com}}
\uppertitleback{MAT 114 dersinin resmi i\c ceri\u gi,
\begin{quote}\relscale{0.9}
Matrisler, sat{\i}r i\c slemleri, sat{\i}r-denklik. 
Homojen ve homojen olmayan lineer denklem sistemleri. 
Vekt\"or uzaylar{\i}, alt uzaylar, toplam ve direkt toplam. 
Lineer ba\u g{\i}ml{\i}l{\i}k, baz [taban], boyut kavramlar{\i}. 
Verilen lineer ba\u g{\i}ms{\i}z k\"umeyi bir baza tamamlama, b\"ol\"um uzaylar{\i}, baz de\u gi\c stirme, ge\c ci\c s matrisleri. 
Lineer d\"on\"u\c s\"umler, lineer d\"on\"u\c s\"umlerin matris g\"osterimleri.
\end{quote}}
\lowertitleback{\tableofcontents}

\maketitle

\chapter{Somut}


\section{Do\u grusal denklem sistemleri}

Do\u grusal (veya lineer) cebirde temel etkinli\u gimiz,
\begin{equation}\label{eqn:lin-sys}
\left\{
\begin{array}{*4{c@{\:}}c@{\;}c@{\;}c}
a_{11}x_1&+&\cdots&+&a_{1n}x_n&=&b_1\\
\hdotsfor7\\
a_{m1}x_1&+&\cdots&+&a_{mn}x_n&=&b_m
\end{array}\right.
\end{equation}
bi\c cimli
do\u grusal denklem sistemlerini \c c\"ozmektir.
\"Ornek \numaraya{ex:1} bak\i n.
Genel \"ornek \denklemde{eqn:lin-sys}
$a_{ij}$ katsay\i lar\i\
bir \emph{cisimden} geliyor.
Cisimlerin genel tan\i m\i\ \sayfada{def:field}d\i r;
\c simdilik 
a\c sa\u g\i daki \"ornekleri bilmek yeter.

\begin{example}\label{ex:fields}
  Olu\c sturur
\begin{compactitem}\label{fields}
\item
ger\c cel say\i lar, $\R$ cismini;
\item
kesirli say\i lar, $\Q$ cismini;
\item
karma\c s\i k say\i lar, $\C$ cismini;
\item
asal bir $p$ mod\"ul\"une g\"ore tamsay\i lar, 
$\Z_p$ (veya $\F_p$) cismini.
\end{compactitem}
\end{example}

Her cismin $0$ olmayan $x$ eleman\i n\i n
$\nicefrac1x$ (veya $x\inv$)
tersi vard\i r.
$\Z_p$ cisminde $p=0$, 
dolay\i s\i yla $\nicefrac 1p$ tan\i mlanmaz.

Katsay\i lar\i\ bir $F$ cisminden gelen
do\u grusal bir sistem, $F$ \textbf{\"uzerindedir.}
$F$'nin elemanlar\i na \textbf{skalerler} denir.
Bu notlarda
ba\c ska bir \c sey s\"oylenmezse,
$F$'nin $\R$ oldu\u gu varsay\i labilir.

Sistem \denklemin{eqn:lin-sys} \c c\"oz\"umlerinin her biri,
bir
\begin{equation*}
  (x_1,\dots,x_n)
\qquad\text{veya}\qquad
\begin{bmatrix}
x_1\\\vdots\\x_n
\end{bmatrix}
\end{equation*}
\textbf{$n$-bile\c senlisidir.}
Bunun daha genel bir ad\i,
\textbf{vekt\"ord\"ur.}
Bile\c senleri bir $F$ cisminden gelen $n$-bile\c slenliler,
\begin{equation*}\label{F^n}
F^n
\end{equation*}
k\"umesini olu\c sturur.
Bu k\"umenin elemanlar\i,
a\c sa\u g\i daki kurallar \denkleme{eqn:vec} g\"ore
toplanabilir ve skalerlerle \c carp\i labilir:
\begin{align}\label{eqn:vec}
\begin{bmatrix}
x_1\\\vdots\\x_n
\end{bmatrix}
+
\begin{bmatrix}
y_1\\\vdots\\y_n
\end{bmatrix}
&=
\begin{bmatrix}
x_1+y_1\\\vdots\\x_n+y_n
\end{bmatrix}
,
&t\cdot
\begin{bmatrix}
x_1\\\vdots\\x_n
\end{bmatrix}
&=
\begin{bmatrix}
t\cdot x_1\\\vdots\\t\cdot x_n
\end{bmatrix}
.	
\end{align}

\begin{example}\label{ex:1}
\begin{equation*}
\left\{\begin{array}{*8{c@{\:}}c@{\;}c@{\;}c}
x_1&+&6x_2& &   &-&5x_4& &   &=& 3\\
   & &    & &x_3&+&4x_4& &   &=&-1\\
	 & &    & &   & &    & &x_5&=& 2
	\end{array}\right.
\end{equation*}
%sisteminde $x_2$ ve $x_4$, \textbf{serbest de\u gi\c skendir,} ve 
sisteminin \c c\"oz\"um\"u,
\begin{equation*}
\begin{bmatrix}
x_1\\x_2\\x_3\\x_4\\x_5
\end{bmatrix}
=
\left[
\begin{array}{*4{c@{\:}}c}
 3&-&6x_2&+&5x_4\\
  & & x_2& &    \\
-1& &    &-&4x_4\\
  & &    & & x_4\\
 2& &    & &
\end{array}\right]
=
\begin{bmatrix}
 3\\
 0\\
-1\\
 0\\
 2
\end{bmatrix}
+x_2
\begin{bmatrix}
-6\\1\\0\\0\\0
\end{bmatrix}
+x_4
\begin{bmatrix}
5\\0\\-4\\1\\0
\end{bmatrix}
.
\end{equation*}
\end{example}

Sistem \denkleme{eqn:lin-sys} kar\c s\i l\i k gelen
\begin{equation}\label{eqn:hom}
\left\{
\begin{array}{*4{c@{\:}}c@{\;}c@{\;}c}
a_{11}x_1&+&\cdots&+&a_{1n}x_n&=&0\\
\hdotsfor7\\
a_{m1}x_1&+&\cdots&+&a_{mn}x_n&=&0
\end{array}\right.
\end{equation}
\textbf{homojen sistemi} vard\i r.

\begin{example}\label{ex:2}
\"Ornek \numarada{ex:1} verilmi\c s sistemin
kar\c s\i l\i k gelen homojen sistemi
\begin{equation*}
\left\{\begin{array}{*8{c@{\:}}c@{\;}c@{\;}c}
x_1&+&6x_2& &   &-&5x_4& &   &=&0\\
   & &    & &x_3&+&4x_4& &   &=&0\\
	 & &    & &   & &    & &x_5&=&0
	\end{array}\right.
\end{equation*}
olur, ve bu sistemin \c c\"oz\"um\"u
\begin{equation*}
\begin{bmatrix}
x_1\\x_2\\x_3\\x_4\\x_5
\end{bmatrix}
=x_2
\begin{bmatrix}
-6\\1\\0\\0\\0
\end{bmatrix}
+x_4
\begin{bmatrix}
5\\0\\-4\\1\\0
\end{bmatrix}
.
\end{equation*}
\end{example}

Sistem \denklemde{eqn:lin-sys},
\c c\"oz\"umlerini de\u gi\c stirmeden
\begin{compactenum}[i)]
\item
bir denklemin bir kat\i\, ba\c ska bir denkleme eklenebilir;
\item
iki denklemin yerleri de\u gi\c stirilebilir;
\item
bir denklem, \emph{s\i f\i r olmayan} bir skalerle \c carp\i labilir.
\end{compactenum}
Bu i\c slemlerle herhangi do\u grusal sistem \c c\"oz\"ulebilir.

\begin{example}\label{ex:3}
  Tablo \numarada{tab:sis}ki
  %A\c sa\u g\i daki
  sistemlerin \c c\"oz\"um k\"umeleri,
birbiriyle ayn\i d\i r.
\begin{table}
  \caption{Denk sistemler}
  \label{tab:sis}
  \begin{gather}\label{eqn:def}
    \left\{\begin{array}{*8{c@{\:}}c@{\;}c@{\;}c}
    & &     & &-2x_3&-& 8x_4& &    &=& 2\\
    -x_1&-& 6x_2& &     &+& 5x_4&+&2x_5&=& 1\\
    -2x_1&-&12x_2& &     &+&10x_4&+&2x_5&=&-2\\
    x_1&+& 6x_2&+&  x_3&-&  x_4&-& x_5&=& 0
    \end{array}\right\}\\\notag
    \left\{\begin{array}{*8{c@{\:}}c@{\;}c@{\;}c}
    x_1&+& 6x_2&+&  x_3&-&  x_4&-& x_5&=& 0\\
    -x_1&-& 6x_2& &     &+& 5x_4&+&2x_5&=& 1\\
    -2x_1&-&12x_2& &     &+&10x_4&+&2x_5&=&-2\\
    & &     & &-2x_3&-& 8x_4& &    &=& 2
    \end{array}\right.\\\notag
    \left\{\begin{array}{*8{c@{\:}}c@{\;}c@{\;}c}
    x_1&+& 6x_2&+&  x_3&-&  x_4&-& x_5&=& 0\\
    & &     & &  x_3&+& 4x_4&+& x_5&=& 1\\
    & &     & & 2x_3&+& 8x_4& &    &=&-2\\
    & &     & &-2x_3&-& 8x_4& &    &=&2
    \end{array}\right.\\\notag
    \left\{\begin{array}{*8{c@{\:}}c@{\;}c@{\;}c}
    x_1&+& 6x_2&+& x_3&-&  x_4&-&  x_5&=& 0\\
    & &     & & x_3&+& 4x_4&+&  x_5&=& 1\\
    & &     & &    & &     & &-2x_5&=&-4\\
    & &     & &    & &     & & 2x_5&=& 4
    \end{array}\right.\\\label{eqn:mod-2}
    \left\{\begin{array}{*8{c@{\:}}c@{\;}c@{\;}c}
    x_1&+& 6x_2&+& x_3&-&  x_4&-&  x_5&=& 0\\
    & &     & & x_3&+& 4x_4&+&  x_5&=& 1\\
    & &     & &    & &     & &-2x_5&=&-4\\
    \end{array}\right\}\\\notag
    \left\{\begin{array}{*8{c@{\:}}c@{\;}c@{\;}c}
    x_1&+& 6x_2&+& x_3&-&  x_4&-&  x_5&=& 0\\
    & &     & & x_3&+& 4x_4&+&  x_5&=& 1\\
    & &     & &    & &     & &  x_5&=& 2\\
    \end{array}\right.\\\notag
    \left\{\begin{array}{*8{c@{\:}}c@{\;}c@{\;}c}
    x_1&+& 6x_2&+& x_3&-&  x_4& &     &=& 2\\
    & &     & & x_3&+& 4x_4& &     &=&-1\\
    & &     & &    & &     & &  x_5&=& 2\\
    \end{array}\right.\\\notag
    \left\{\begin{array}{*8{c@{\:}}c@{\;}c@{\;}c}
    x_1&+& 6x_2& &    &-& 5x_4& &     &=& 3\\
    & &     & & x_3&+& 4x_4& &     &=&-1\\
    & &     & &    & &     & &  x_5&=& 2\\
    \end{array}\right.
  \end{gather}
\end{table}
Oradaki son sistemin \c c\"oz\"um\"unu \"Ornek \numarada{ex:1} bulduk.
Hesaplamar\i m\i zda
$2\neq0$ varsayd\i k.
E\u ger skaler olarak $2=0$ ise
(\"orne\u gin verilen sistem $\Z_2$ \"uzerinde ise),
o zaman sistem \eqref{eqn:mod-2},
\begin{equation*}
			\left\{\begin{array}{*8{c@{\:}}c@{\;}c@{\;}c}
     x_1& & & & x_3&+& x_4& &   x_5&=& 0\\
        & & & & x_3& &    &+&   x_5&=& 1\\
  \end{array}\right.
\end{equation*}
olur, ve bundan
\begin{equation*}
			\left\{\begin{array}{*8{c@{\:}}c@{\;}c@{\;}c}
     x_1& & & &    &+& x_4& &      &=& 1\\
        & & & & x_3& &    &+&   x_5&=& 1\\
  \end{array}\right.
\end{equation*}
sistemini elde ederiz, ve bunun \c c\"oz\"um\"u
\begin{equation*}
\begin{bmatrix}
x_1\\x_2\\x_3\\x_4\\x_5
\end{bmatrix}
=
\begin{bmatrix}
 1\\
 0\\
 1\\
 0\\
 0
\end{bmatrix}
+x_2
\begin{bmatrix}
0\\1\\0\\0\\0
\end{bmatrix}
+x_4
\begin{bmatrix}
1\\0\\0\\1\\0
\end{bmatrix}
+x_5
\begin{bmatrix}
0\\0\\1\\0\\1
\end{bmatrix}
.
\end{equation*}
\end{example}

Lineer bir sistem \c c\"ozmek i\c cin,
sistemin de\u gi\c skenlerini her ad\i mda yazmak zorunda de\u giliz.
Sistem \denklemin{eqn:lin-sys}
\textbf{geni\c sletilmi\c s katsay\i lar matrisi,}
\begin{equation*}
\begin{bmatrix}
a_{11}&\cdots&a_{1n}&b_1\\
\hdotsfor4\\
a_{m1}&\cdots&a_{mn}&b_m
\end{bmatrix}
\quad\text{veya}\quad
\begin{pmatrix}
a_{11}&\cdots&a_{1n}&b_1\\
\hdotsfor4\\
a_{m1}&\cdots&a_{mn}&b_m
\end{pmatrix}
\end{equation*}
olur.
Bir sistemin geni\c sletilmi\c s katsay\i lar matrisinde,
sistemin \c c\"oz\"umlerini de\u gi\c stirmeden,
a\c sa\u g\i daki
\textbf{elemanter sat\i r i\c slemlerini} kullan\i labilir:
\begin{compactenum}[i)]
\item
bir sat\i r\i n bir kat\i\, ba\c ska bir sat\i ra eklenebilir;
\item
iki sat\i r\i n yerleri de\u gi\c stirilebilir;
\item
bir sat\i r, \emph{s\i f\i r olmayan} bir skalerle \c carp\i labilir.
\end{compactenum}
Bir matrisin
\begin{compactenum}[i)]
\item
$i$'ninci sat\i r\i n\i n $t$ kat\i n\i n $j$'ninci sat\i ra eklenmesi
\begin{equation*}
tR_i+R_j,
\end{equation*}
\item
$i$'ninci ve $j$'ninci sat\i rlar\i n\i n yerlerinin de\u gi\c stirilmesi
\begin{equation*}
R_i\leftrightarrow R_j,
\end{equation*}
\item
$i$'ninci sat\i r\i n\i n $t$ ile \c carp\i lmas\i
\begin{equation*}
tR_i
\end{equation*}
\end{compactenum}
ile g\"osterilebilir.
(Sat\i r\i n \.Ingilizcesi \emph{Row}'dur.)

\begin{example}\label{ex:4}
\"Ornek \numarada{ex:3} kulland\i\u g\i m\i z
i\c slemler
Tablo \numarada{tab:sis-mat}
gibi g\"osterilebilir.
\begin{table}
  \caption{Matrisin indirgemesi}
  \label{tab:sis-mat}  
  \begin{gather}\notag
    \begin{bmatrix}
      0&  0&-2&-8& 0& 2\\
      -1& -6& 0& 5& 2& 1\\
      -2&-12& 0&10& 2&-2\\
      1&  6& 1&-1&-1& 0
    \end{bmatrix}
    \xrightarrow{R_1\leftrightarrow R_4}\\\notag
    \begin{bmatrix}
      1&  6& 1&-1&-1& 0\\	
      -1& -6& 0& 5& 2& 1\\
      -2&-12& 0&10& 2&-2\\
      0&  0&-2&-8& 0& 2
    \end{bmatrix}
    \xrightarrow[2R_1+R_3]{R_1+R_2}\\\notag
    \begin{bmatrix}
      1&6& 1&-1&-1& 0\\
      0&0& 1& 4& 1& 1\\
      0&0& 2& 8& 0&-2\\
      0&0&-2&-8& 0& 2
    \end{bmatrix}
    \xrightarrow[2R_2+R_4]{-2R_2+R_3}\\\label{eqn:bas}
    \begin{bmatrix}
      1&6&1&-1&-1& 0\\
      0&0&1& 4& 1& 1\\
      0&0&0& 0&-2&-4\\
      0&0&0& 0& 2& 4
    \end{bmatrix}
    \xrightarrow{R_3+R_4}\\\notag
    \begin{bmatrix}
      1&6&1&-1&-1& 0\\
      0&0&1& 4& 1& 1\\
      0&0&0& 0&-2&-4\\
      0&0&0& 0& 0& 0
    \end{bmatrix}
    \xrightarrow{-\frac12R_3}\\\notag
    \begin{bmatrix}
      1&6&1&-1&-1& 0\\
      0&0&1& 4& 1& 1\\
      0&0&0& 0& 1& 2\\
      0&0&0& 0& 0& 0
    \end{bmatrix}
    \xrightarrow[R_3+R_1]{-R_3+R_2}\\\notag
    \begin{bmatrix}
      1&6&1&-1&0& 2\\
      0&0&1& 4&0&-1\\
      0&0&0& 0&1& 2\\
      0&0&0& 0&0& 0
    \end{bmatrix}
    \xrightarrow{-R_2+R_1}
    \begin{bmatrix}
      1&6&0&-5&0&3 \\
      0&0&1& 4&0&-1\\
      0&0&0& 0&1& 2\\
      0&0&0& 0&0& 0
    \end{bmatrix}
  \end{gather}
\end{table}

\end{example}

Sistem \denklemin{eqn:lin-sys} geni\c sletilmi\c s katsay\i lar matrisi
\begin{equation*}
\left[\begin{array}{c|c}A&\bm b\end{array}\right]
\end{equation*}
bi\c ciminde yaz\i labilir.
Burada
\begin{align}\label{eqn:Ab}
	A&=
	\begin{bmatrix}
a_{11}&\cdots&a_{1n}\\
\vdots&      &\vdots\\
a_{m1}&\cdots&a_{mn}
\end{bmatrix},&
\bm b&=
\begin{bmatrix}
b_1\\\vdots\\b_n\end{bmatrix},
\end{align}
ve $A$, sistemin \textbf{katsay\i lar matrisidir.}
El yaz\i s\i yla $\bm b$, $\vec b$ olarak yaz\i labilir.
$A$ matrisi \textbf{$m\times n$'liktir}\label{lik} 
ve
\begin{equation*}
[a_{ij}]^{1\leq i\leq m}_{1\leq j\leq n}
\qquad
\text{veya k\i saca}
\qquad
[a_{ij}]^i_j
\end{equation*} 
olarak yaz\i labilir;
burada $a_{ij}$, $A$'n\i n \textbf{$(i,j)$ girdisidir.}
Bir $i$ i\c cin,
e\u ger $\{j\colon a_{ij}\neq0\}$ k\"umesi bo\c s de\u gilse,
bunun en k\"u\c c\"uk eleman\i\ $j_i$ olsun;
di\u ger durumda $j_i$ tan\i mlanmaz.
O zaman $A$ matrisinin $(i,j_i)$ girdisi,
$A$'n\i n $i$'ninci sat\i r\i n\i n
\textbf{ba\c s eleman\i d\i r.}
E\u ger bir $\ell$ i\c cin
\begin{equation*}
j_1<\dots<j_{\ell},
\end{equation*}
ama
$\ell$'nin $m$'den kesin k\"u\c c\"uk oldu\u gu durumda
 $j_{\ell+1}$, \dots, $j_m$ tan\i mlanmazsa,
o zaman $A$,
\textbf{basamakl\i} bir matristir.
E\u ger \"ustelik
\begin{equation*}
\left.
\begin{array}{cc}
\text{$k\neq i$ durumlar\i nda }&0\\
\text{$k=i$ durumunda }&1
\end{array}\right\}
=a_{kj_i}
\end{equation*}
ise, o zaman $A$,
\textbf{sat\i rca indirgenmi\c s basamakl\i} bir matristir.

\begin{example}\label{ex:5}
  \"Ornek \numarada{ex:3}
  matris \eqref{eqn:bas}
  \begin{comment}
    $\left[\begin{smallmatrix}
        1&6&1&-1&-1& 0\\
        0&0&1& 4& 1& 1\\
        0&0&0& 0&-2&-4\\
        0&0&0& 0& 0& 0
      \end{smallmatrix}\right]$ matrisi
  \end{comment}
ve sonralar\i\
	basamakl\i d\i r,
	ama \"onceleri basamakl\i\ de\u gildir.
	Sadece son matris
	\begin{comment}
          $\left[\begin{smallmatrix}
              1&6&0&-5&0&3 \\
              0&0&1& 4&0&-1\\
              0&0&0& 0&1& 2\\
              0&0&0& 0&0& 0
            \end{smallmatrix}\right]$ matrisi
        \end{comment}
        sat\i rca indirgenmi\c s basamakl\i d\i r.
\end{example}

Herhangi do\u grusal bir sistem \c c\"ozmek i\c cin:
\begin{compactenum}
\item
Geni\c sletilmi\c s katsay\i lar matrisini yaz\i n.
\item\label{item:G}
Elemanter sat\i r i\c slemleri ile,
soldan sa\u ga \c cal\i\c sarak,
matrisi basamakl\i\ bir matrise indirgeyin.
\item\label{item:GJ}
\begin{compactenum}
\item
Bu matrisin bir sat\i r\i n\i n ba\c s eleman\i,
matrisin son s\"utunda ise,
sistemin \c c\"oz\"um\"u yoktur.
\item\sloppy
Aksi durumda,
elemanter sat\i r i\c slemlerle,
sa\u gdan sola \c cal\i\c sarak,
matrisi sat\i rca indirgenmi\c s basamakl\i\ bir matrise indirgeyin.
Sistemin \c c\"oz\"um\"u, \"Ornek \numarada{ex:1}ki gibi
elde edilebilir.
\end{compactenum}
\end{compactenum}
Burada
\begin{compactitem}
  \item
ad\i m \ref{item:G},
\textbf{Gauss yok etme} y\"ontemidir;
\item
ad\i m \ref{item:GJ},
\textbf{Gauss--Jordan indirgeme} y\"ontemidir.
\end{compactitem}

\begin{exercise}
Basamakl\i\ bir matris yaz\i n.
Bu matristen, baz\i\ elemanter sat\i r i\c slemlerle,
ba\c ska bir matris elde edin.
Geni\c sletilmi\c s katsay\i lar matrisi bu matris olan
do\u grusal denklem sistemini yaz\i n.
Nas\i l elde etti\u ginizi unutarak
sistemi \c c\"oz\"un
veya arkada\c slar\i na \c c\"ozd\"ur\"un.
Ayn\i\ y\"ontemle hocan\i z
s\i z\i n i\c cin problemler elde edebilir.
\end{exercise}

Tekrar $A$ ve $\bm b$, 
e\c sitlikler \eqref{eqn:Ab} ile tan\i mlanm\i\c s matrisleri olsun,
ve $A$'n\i n $j$'ninci s\"utunu $\bm a_j$ olsun.
O zaman
\begin{equation}\label{eqn:A=}
A=
\left[\begin{array}{c|c|c}
\bm a_1&\cdots&\bm a_n
\end{array}\right],
\end{equation}
ve sistem \eqref{eqn:lin-sys}
\begin{equation*}
x_1\bm a_1+\dots+x_n\bm a_n=\bm b
\end{equation*}
bi\c ciminde yaz\i labilir.
Bu denklemin sol \"uyesi
bir \textbf{matris \c carp\i m\i d\i r:}
e\u ger
\begin{equation*}
  \begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}=\bm x
\end{equation*}
ise,
o zaman tan\i ma g\"ore
\begin{equation}\label{eqn:Ax}
x_1\bm a_1+\dots+x_n\bm a_n=A\bm x.
\end{equation}
Bu durumda
sistem \eqref{eqn:lin-sys} ve kar\c s\i l\i k gelen 
homojen sistem \eqref{eqn:hom},
s\i ras\i yla
\begin{align*}
A\bm x&=\bm b,&	
A\bm x&=\bm 0
\end{align*}
bi\c ciminde yaz\i labilir.

Elemanter sat\i r i\c slemleri, terslenebilir.
E\u ger bir matristen,
elemanter sat\i r i\c slemleriyle
ba\c ska bir matris elde edilirse,
bu iki matris, birbirine
\textbf{sat\i rca denktir.}
Sat\i r denkli\u gi,
bir denklik ba\u g\i nt\i s\i d\i r.
E\u ger
$\left[\begin{array}{c|c}A&\bm b\end{array}\right]$
ve
$\left[\begin{array}{c|c}C&\bm d\end{array}\right]$
birbiriyle sat\i rca denk ise,
o zaman $A\bm x=\bm b$ ve $C\bm x=\bm d$
sistemlerinin \c c\"oz\"umleri ayn\i d\i r.

Basamakl\i\ bi\c cime indirgendikten sonra,
e\u ger $A$'n\i n $j$'ninci s\"utununda
bir sat\i r\i n ba\c s eleman\i\ \emph{c\i kmazsa,}
o zaman $x_j$, $A\bm x=\bm0$ homojen sisteminin
\textbf{ba\u g\i ms\i z bir de\u gi\c skenidir.}

\begin{example}\label{ex:temel}
\.Indirgenmi\c s sistem \denkleme{eqn:mod-2} bakarak
\c c\"oz\"umleri ayn\i\ olan 
sistem \denkleme{eqn:def} kar\c s\i l\i k gelen
\begin{equation}\label{eqn:def-hom}
\left\{\begin{array}{*8{c@{\:}}c@{\;}c@{\;}c}
        & &     & &-2x_3&-& 8x_4& &    &=& 0\\
    -x_1&-& 6x_2& &     &+& 5x_4&+&2x_5&=& 0\\
   -2x_1&-&12x_2& &     &+&10x_4&+&2x_5&=& 0\\
     x_1&+& 6x_2&+&  x_3&-&  x_4&-& x_5&=& 0
  \end{array}\right.
\end{equation}
homojen sisteminin ba\u g\i ms\i z de\u g\i\c skenlerinin 
$x_2$ ve $x_4$ oldu\u gunu anlayabiliriz.
\"Ornek \numarada{ex:2}ki gibi
sistem \denklemin{eqn:def-hom}
\begin{align*}
	&(6,1,0,0,0),&(5,0,-4,1,0)	
\end{align*}
\c c\"oz\"umleri vard\i r,
ve bunlar
\begin{align*}
	&(*,1,*,0,*),&(*,0,*,1,*)
\end{align*}
bi\c cimindedir.
\end{example}

$A\bm x=\bm0$ homojen sisteminin ba\u g\i ms\i z de\u gi\c skenleri,
s\i ras\i yla
$x_{i_1}$, \dots, $x_{i_{n-\ell}}$ olsun.
Her $i$ i\c cin, $1\leq i\leq n-\ell$ ise,
homojen sistemin,
\begin{equation*}
\left.
\begin{array}{cc}
\text{$j\neq i$ durumunda }&0\\
\text{$j=i$ durumunda }&1
\end{array}\right\}
=c_{i_ji}
\end{equation*}
ko\c sulunu sa\u glayan bir
$(c_{1i},\dots,c_{ni})$ veya $\bm c_i$ \c c\"oz\"um\"u vard\i r.
O halde
$\bm c_1$, \dots, $\bm c_{n-\ell}$,
homojen sistemin \textbf{temel \c c\"oz\"umleridir.}
Sistem \denklemin{eqn:def-hom} temel \c c\"oz\"umleri,
\"Ornek \numarada{ex:temel} veriliyor.

Matris \c carpmas\i
\begin{gather}\notag
A(\bm x+\bm y)=A\bm x+A\bm y,\\\label{eqn:A(tx)}
A(t\bm x)=t(A\bm x)
\end{gather}
kurallar\i n\i\ sa\u glar.
Ayr\i ca $A\bm d=\bm b$ ise, o zaman
\begin{equation*}
A(\bm d+\bm x)=\bm b\iff A\bm x=\bm 0.
\end{equation*}
$A\bm x=\bm0$ homojen sisteminin temel \c c\"oz\"umleri
tekrar $\bm c_1$, \dots, $\bm c_{n-\ell}$ ise,
o zaman
homojen sistemin her \c c\"oz\"um\"u,
baz\i\ $t_i$ skalerleri i\c cin
\begin{equation*}
t_1\bm c_1+\dots+t_{n-\ell}\bm c_{n-\ell}
\end{equation*}
bi\c ciminde yaz\i labilir.
Bu vekt\"or,
$\bm c_i$ vekt\"orlerinin
\textbf{do\u grusal bir bile\c simidir,}
ve $\bm c_i$ vekt\"orlerinin do\u grusal bile\c simlerinden her biri,
homojen sistemin bir \c c\"oz\"um\"ud\"ur:
bunu \"Ornek \numarada{ex:2} g\"ord\"uk.
$A\bm x=\bm b$ sisteminin
\c c\"oz\"umleri,
baz\i\ $t_i$ skalerleri i\c cin
\begin{equation*}
\bm d+t_1\bm c_1+\dots+t_{n-\ell}\bm c_{n-\ell}
\end{equation*}
bi\c ciminde yaz\i labilen vekt\"orleridir:
bunu \"Ornek \numarada{ex:1} g\"ord\"uk.

Baz\i\ $A$ ve $\bm b$ i\c cin $A\bm x=\bm b$ sistemi \c c\"oz\"ulemeyebilir,
ama $A\bm x=\bm0$ homojen sisteminin her zaman \textbf{a\c sik\^ar}
$\bm x=\bm0$
\c c\"oz\"um\"u vard\i r.
E\u ger $A\bm x=\bm b$ sisteminin \c c\"oz\"um\"u varsa,
o zaman
bu sistemin \c c\"oz\"um k\"umesi
ve kar\c s\i l\i k gelen 
$A\bm x=\bm0$ homojen sisteminin \c c\"oz\"um k\"umesi,
ayn\i\ say\i dad\i r.

\begin{example}
$\Z_3$ cismi $\{-1,0,1\}$ olarak d\"u\c s\"un\"ulebilir.
Bu cisim \"uzerinde
\begin{equation}\label{eqn:Z3}
\left\{\begin{array}{*4{c@{\:}}c@{\;}c@{\;}c}
-x&+&y&+&z&=&-1\\
 x&+&y& & &=&0
\end{array}\right.
\end{equation}
sisteminin geni\c sletilmi\c s katsay\i lar matrisi
\begin{equation*}
\begin{bmatrix}-1&1&1&-1\\1&1&0&0\end{bmatrix}
\end{equation*}
olur,
ve bundan
\begin{equation*}
\xrightarrow{R_1+R_2}
\begin{bmatrix}-1&1&1&-1\\0&-1&1&-1\end{bmatrix}
\xrightarrow{R_2+R_1}
\begin{bmatrix}-1&0&-1&1\\0&-1&1&-1\end{bmatrix}
\xrightarrow[-R_2]{-R_1}
\end{equation*}
elemanter sat\i r i\c slemleriyle
\begin{equation*}
\begin{bmatrix}1&0&1&-1\\0&1&-1&1\end{bmatrix}
\end{equation*}
sat\i rca indirgenmi\c s basamakl\i\ matrisini elde ederiz,
dolay\i s\i yla sistem \denklemin{eqn:Z3} \c c\"oz\"umleri,
\begin{equation*}
\begin{bmatrix}-1\\1\\0\end{bmatrix}
+z\begin{bmatrix}-1\\1\\1\end{bmatrix}
\end{equation*}
bi\c cimli vekt\"orlerdir, ve
kar\c s\i l\i k gelen
\begin{equation}\label{eqn:Z3-hom}
\left\{\begin{array}{*4{c@{\:}}c@{\;}c@{\;}c}
-x&+&y&+&z&=&0\\
 x&+&y& & &=&0
\end{array}\right.
\end{equation}
homojen sisteminin \c c\"oz\"umleri
\begin{equation*}
z\begin{bmatrix}-1\\1\\1\end{bmatrix}
\end{equation*}
bi\c cimli vekt\"orlerdir.
Buldu\u gumuz \c c\"oz\"umleri kontrol etmek i\c cin,
iki hesaplama yeter:
\begin{compactitem}
\item
  $(-1,1,0)$ vekt\"or\"u,
  sistem \denklemin{eqn:Z3} bir \c c\"oz\"um\"ud\"ur;
\item
  $(-1,1,1)$ vekt\"or\"u,
  homojen sistem \denklemin{eqn:Z3-hom} bir \c c\"oz\"um\"ud\"ur.
\end{compactitem}
\c Simdi $z\in\{-1,0,1\}$ oldu\u gundan
sistem \denklemin{eqn:Z3} \c c\"oz\"umleri
\begin{align*}
&\begin{bmatrix}0\\0\\-1\end{bmatrix},&
&\begin{bmatrix}-1\\1\\0\end{bmatrix},&
&\begin{bmatrix}1\\-1\\1\end{bmatrix}
\end{align*}
olur, ve
kar\c s\i l\i k gelen
homojen sistem \denklemin{eqn:Z3-hom}
\c c\"oz\"umleri
\begin{align*}
&\begin{bmatrix}1\\-1\\-1\end{bmatrix},&
&\begin{bmatrix}0\\0\\0\end{bmatrix},&
&\begin{bmatrix}-1\\1\\1\end{bmatrix}.
\end{align*}
\end{example}

Bir bilgisayar, yapt\i\u g\i m\i z hesaplamalar\i\ yapabilir,
ama bilgisayar, hesaplamalar\i n\i n do\u gru olup olmad\i\u g\i n\i\ bilmiyor.
Ad\i na ra\u gmen bir bilgisayar bilgili de\u gildir.
Bilgisayar hi\c c  bir \c sey bilmiyor.
Biz, \"ornekteki gibi kontrol ederek
hesaplamalar\i m\i z\i n
do\u gru oldu\u gunu bilebiliriz.
Hesaplamalar\i n anlam\i n\i\ biliriz.
Bir matematik dersinin amac\i,
bilgisayar olman\i z de\u gil,
bilen bir insan olman\i zd\i r.

\begin{exercise}
$\Z_3$ \"uzerinde her
\begin{equation*}
\left\{\begin{array}{*4{c@{\:}}c@{\;}c@{\;}c}
-x&+&y&+&z&=&a\\
 x&+&y& & &=&b
\end{array}\right.
\end{equation*}
sistemi \c c\"oz\"ulebilir mi?
\end{exercise}

Sonsuz bir cisim \"uzerinde do\u grusal bir sistemin
\c c\"oz\"um say\i s\i\
ya s\i f\i r, ya da bir, ya da sonsuzdur.

\begin{exercise}
$\R$ \"uzerinde bir
\begin{equation*}
\left\{\begin{array}{*4{c@{\:}}c@{\;}c@{\;}c}
        x&+&2y&-&     z&=&1\\
      -2x&+&ay&+&   2 z&=&b\\
         & & y&+&(a-1)z&=&0
    \end{array}\right.
\end{equation*}
	sistemi verilirse,
  $a$'n\i n ve $b$'nin hangi de\u gerleri i\c cin
  sisteminin
  \begin{compactenum}[(a)]
  \item
    tek bir \c c\"oz\"um\"u vard\i r?
  \item
    birden fazla \c c\"oz\"um\"u vard\i r?
  \item
    hi\c c \c c\"oz\"um\"u yoktur?
  \end{compactenum}
\end{exercise}

\begin{solution}
	Sistemin geni\c sletilmi\c s katsay\i lar matrisini indirgiyoruz:
  \begin{gather*}
    \begin{bmatrix}
      1&2&-1&1\\-2&a&2&b\\0&1&a-1&0
    \end{bmatrix}\xrightarrow{2R_1+R_2}
    \begin{bmatrix}
      1&2&-1&1\\0&a+4&0&b+2\\0&1&a-1&0
    \end{bmatrix}\\\xrightarrow{R_2\leftrightarrow R_3}
    \begin{bmatrix}
      1&2&-1&1\\0&1&a-1&0\\0&a+4&0&b+2
    \end{bmatrix}.
  \end{gather*}
	Sonu\c c olarak
  \begin{compactitem}
  \item
    $a=-4$ ve $b\neq-2$ ise \c c\"oz\"um yoktur.
  \item
    $a=-4$ ve $b=-2$ ise birden fazla \c c\"oz\"um vard\i r.
  \item
    $a\neq-4$ ise
		indirgemeyi devam ettirebiliriz:
		\begin{equation*}
		\xrightarrow{-(a+4)R_2+R_3}
        \begin{bmatrix}
      1&2&-1&1\\0&1&a-1&0\\0&0&-(a-1)(a+4)&b+2
    \end{bmatrix}		
		\end{equation*}
		\begin{comment}
		
    \begin{multline*}
    \begin{bmatrix}
      1&2&-1&1\\0&1&a-1&0\\0&a+4&0&b+2
    \end{bmatrix}
		\xrightarrow{-(a+4)R_2+R_3}\\
        \begin{bmatrix}
      1&2&-1&1\\0&1&a-1&0\\0&0&-(a-1)(a+4)&b+2
    \end{bmatrix}
    \end{multline*}
		
		\end{comment}
dolay\i s\i yla
    \begin{compactitem}[$*$]
    \item
      $a=1$ ve $b\neq-2$ ise \c c\"oz\"um yoktur.
    \item
      $a=1$ ve $b=-2$ ise birden fazla \c c\"oz\"um vard\i r.
    \item
      $a\neq 1$ ise tek bir \c c\"oz\"um vard\i r.
    \end{compactitem}
  \end{compactitem}
	K\i saca
  \begin{compactenum}[(a)]
  \item
	$a\notin\{-4,1\}$ ise
    tek bir \c c\"oz\"um\"u vard\i r;
  \item
    $a\in\{-4,1\}$ ve $b=-2$ ise
    birden fazla \c c\"oz\"um\"u vard\i r;
  \item
    $a\in\{-4,1\}$ ve $b\neq-2$ ise
    hi\c c \c c\"oz\"um\"u yoktur.
  \end{compactenum}  
\end{solution}


\section{Matrisler}

\Sayfada{lik} dedi\u gimiz gibi
$m$-sat\i rli, $n$-s\"utunlu bir matris,
\emph{$m\times n$'liktir.}
Her $F$ cismi i\c cin
\begin{equation*}\label{Fmn}
F^{m\times n},
\end{equation*}
$F$ \"uzerinde $m\times n$'lik matrisler k\"umesi olsun.
\"Ozel bir durumda, \sayfada{F^n} verdi\u gimiz tan\i ma g\"ore
\begin{equation*}
F^{m\times 1}=F^m.
\end{equation*}
B\"oylece $m$-bile\c senlerlileri,
\textbf{s\"utun matrisi} olarak d\"u\c s\"un\"uyoruz.
\c Simdi 
$A\in F^{m\times n}$ ve $B\in F^{n\times r}$ olsun.
O zaman baz\i\ $\bm b_k$ s\"utun matrisleri i\c cin
\begin{equation}\label{eqn:B=}
  B=
  \left[\begin{array}{c|c|c}
\bm b_1&\cdots&\bm b_r
\end{array}\right]
\end{equation}
olur,
ve bu durumda
tan\i m \denklemi{eqn:Ax} kullanarak
\begin{equation}\label{eqn:AB}
AB=
\left[\begin{array}{c|c|c}
A\bm b_1&\cdots&A\bm b_r
\end{array}\right]
\end{equation}
tan\i mlar\i z.
Bu \c sekilde
\begin{equation*}
(X,Y)\mapsto XY\colon F^{m\times n}\times F^{n\times r}\to F^{m\times r}.
\end{equation*}

\begin{theorem}\label{thm:AB}
  E\u ger
  \begin{align*}
    A&=[a_{ij}]^{1\leq i\leq m}_{1\leq j\leq n},&
    B&=[b_{jk}]^{1\leq j\leq n}_{1\leq k\leq r}
  \end{align*}
  ise, o zaman
\begin{equation*}
AB
=\left[\sum_{j=1}^na_{ij}b_{jk}\right]^{1\leq i\leq m}_{1\leq k\leq r}.
\end{equation*}
\end{theorem}

\begin{proof}
  Tan\i mlar \eqref{eqn:A=}, \eqref{eqn:Ax},
  \eqref{eqn:B=}, ve \denklemde{eqn:AB}n
  \begin{gather*}
    A\bm b_k=\sum_{j=1}^nb_{jk}\bm a_j,\\
AB=
\left[\begin{array}{c|c|c}
\sum_{j=1}^nb_{j1}\bm a_j&\cdots&\sum_{j=1}^nb_{jr}\bm a_j
\end{array}\right],\\
\sum_{j=1}^nb_{jk}\bm a_j
=\sum_{j=1}^n
\begin{bmatrix}a_{1j}b_{jk}\\\vdots\\a_{mj}b_{jk}\end{bmatrix}
  \end{gather*}
ve buradan istedi\u gimiz sonu\c c \c c\i kar.
\end{proof}

\begin{theorem}\label{thm:ass}
Matrisler \c carpmas\i\ birle\c smelidir:
$A$, $m\times n$'lik; $B$, $n\times r$'lik; ve $C$, $r\times s$'lik ise
\begin{equation*}
A(BC)=(AB)C.
\end{equation*}
\end{theorem}

\begin{comment}
  \begin{proof}[Kan\i t 1.]
    \Teoremde{thm:AB}n
    \begin{multline*}
      (AB)C
      =	\left([a_{ij}]^i_j
      [b_{jk}]^j_k\right)
      [c_{k\ell}]^k_\ell
      =\left[%\textstyle
	\sum_ja_{ij}b_{jk}\right]^i_k	[c_{k\ell}]^k_\ell\\
      =\left[%\textstyle
	\sum_k\left(%\textstyle
	\sum_ja_{ij}b_{jk}\right)c_{k\ell}\right]^i_\ell.	
    \end{multline*}
    Cisimlerde \c carpma birle\c smeli ve toplam \"uzerinde da\u g\i lmal\i\ oldu\u gundan
    \begin{multline*}
      %\textstyle
      \sum_k\left(\textstyle\sum_ja_{ij}b_{jk}\right)c_{k\ell}
      =%\textstyle
      \sum_k%\textstyle
      \sum_j(a_{ij}b_{jk})c_{k\ell}\\
      =%\textstyle
      \sum_k%\textstyle
      \sum_ja_{ij}(b_{jk}c_{k\ell})
      =%\textstyle
      \sum_ja_{ij}
      %\textstyle
      \sum_kb_{jk}c_{k\ell},
    \end{multline*}
    dolay\i s\i yla
    \begin{multline*}
      (AB)C
      =\left[%\textstyle
	\sum_ja_{ij}%\textstyle
	\sum_kb_{jk}c_{k\ell}\right]^i_\ell
      =[a_{ij}]^i_j\left[%\textstyle
	\sum_kb_{jk}c_{k\ell}\right]^j_\ell\\
      %	=[a_{ij}]^i_j\left([b_{jk}]^j_k[c_{k\ell}]^k_\ell\right)
      =A(BC).\qedhere
    \end{multline*}
  \end{proof}
\end{comment}

\begin{proof}
  Tan\i mlardan
\begin{equation*}
(AB)C
=\left[\begin{array}{c|c|c}
(AB)\bm c_1&\cdots&(AB)\bm c_s
\end{array}\right],
\end{equation*}
 ve \denklemde{eqn:A(tx)}n
\begin{multline*}
(AB)\bm c_{\ell}
=\left[\begin{array}{c|c|c}
A\bm b_1&\cdots&A\bm b_r
\end{array}\right]\bm c_{\ell}\\
=c_{1\ell}(A\bm b_1)+\dots+c_{r\ell}(A\bm b_r)\\
=A(c_{1\ell}\bm b_1)+\dots+A(c_{r\ell}\bm b_r)\\
=A(c_{1\ell}\bm b_1+\dots+c_{r\ell}\bm b_r)
=A(B\bm c_{\ell}),
\end{multline*}
dolay\i syla
\begin{align*}
	(AB)C
	&=\left[\begin{array}{c|c|c}
A(B\bm c_1)&\cdots&A(B\bm c_s)
\end{array}\right]\\
	&=A\left[\begin{array}{c|c|c}
B\bm c_1&\cdots&B\bm c_s
\end{array}\right]
=A(BC).\qedhere
\end{align*}
\end{proof}

E\u ger $A=[a_{ij}]^{1\leq i\leq m}_{1\leq j\leq n}$ ise,
o zaman tan\i ma g\"ore
\begin{equation*}
A\trans
=[a_{ij}]_{1\leq i\leq m}^{1\leq j\leq n}.
%=[a_{ji}]_{1\leq j\leq m}^{1\leq i\leq n};
\end{equation*}
Bu $n\times m$'lik matris,
$A$'n\i n \textbf{devri\u gi} veya \textbf{tranzpozesidir.}
K\i saca
\begin{equation*}
\left([a_{ij}]^i_j\right)\trans=[a_{ij}]^j_i.
\end{equation*}
O zaman
\begin{equation*}
(A\trans)\trans=A.
\end{equation*}
\"Ozel bir durumda
\begin{equation*}
\begin{bmatrix}c_1\\\vdots\\c_n\end{bmatrix}\trans=
\begin{bmatrix}c_1&\cdots&c_n\end{bmatrix},
\end{equation*}
yani bir s\"utun matrisinin transpozesi,
bir \textbf{sat\i r matrisidir.}
Daha genelde
\begin{equation*}
\left[\begin{array}{c|c|c}
\bm b_1&\cdots&\bm b_r
\end{array}\right]\trans
=	\left[\begin{array}{c}
\bm b_1{}\trans\\\hline\vdots\\\hline\bm b_r{}\trans
\end{array}\right].
\end{equation*}

\begin{theorem}
$A$'n\i n s\"utunlar\i\ ve $B$'nin sat\i rlar\i\ ayn\i\ say\i da ise
\begin{equation*}
(AB)\trans=B\trans A\trans.
\end{equation*}
\end{theorem}

\begin{proof}
  \Teoremde{thm:AB}n
  \begin{align*}
    \left([a_{ij}]^i_j[b_{jk}]^j_k\right)\trans
&=\left(\left[\sum_ja_{ij}b_{jk}\right]^i_k\right)\trans\\
&=\left[\sum_ja_{ij}b_{jk}\right]^k_i\\
&=\left[\sum_jb_{jk}a_{ij}\right]^k_i
=[b_{jk}]^k_j[a_{ij}]^j_i.\qedhere
  \end{align*}
\end{proof}

Sonu\c c olarak tan\i m \denklemde{eqn:AB}n
$C=A\trans$ ise
\begin{equation*}
\left[\begin{array}{c}
\bm b_1{}\trans\\\hline\vdots\\\hline\bm b_r{}\trans
\end{array}\right]C
%=B\trans C
%=(C\trans B)\trans
%=	\left[\begin{array}{c}(C\trans\bm b_1)\trans\\\hline\vdots\\\hline(C\trans\bm b_r)\trans\end{array}\right]
=	\left[\begin{array}{c}
\bm b_1{}\trans C\\\hline\vdots\\\hline\bm b_r{}\trans C
\end{array}\right].
\end{equation*}

\c Simdi her $k$ i\c cin $k\times k$'lik olan
\begin{equation*}
  I_k=
\begin{bmatrix}
1&0&\cdots&0\\
0&1&\ddots&\vdots\\
\vdots&\ddots&\ddots&0\\
0&\cdots&0&1
\end{bmatrix}
\end{equation*}
olsun.
E\u ger $k$ say\i s\i\ anla\c s\i labilirse $I_k$ matrisi 
$I$ olarak yaz\i labilir.
$I$ matrislerine \textbf{birim matrisi} denir \c c\"unk\"u
her $m\times n$'lik $A$ matrisi i\c cin
\begin{align*}
	I_mA&=A,&AI_n&=A.
\end{align*}

\begin{theorem}\label{thm:Phi}
E\u ger $\Phi$, elemanter bir sat\i r i\c slemi ise
\begin{equation*}%\label{eqn:Phi}
\Phi(A)=\Phi(I)A.
\end{equation*}
\end{theorem}

Teoremdeki $\Phi(I)$ matrisi,
\textbf{elemanter bir matristir.}
Teorem i\c cin, genel bir kan\i t yaz\i labilir,
ama kan\i t\i n fikri,
sonraki \"ornekten anla\c s\i labilir.

\begin{example}
\begin{gather*}
	\begin{bmatrix}a&b\\ta+c&tb+d\end{bmatrix}
	=
	\begin{bmatrix}1&0\\t&1\end{bmatrix}
	\begin{bmatrix}a&b\\c&d\end{bmatrix},\\
	\begin{bmatrix}c&d\\a&b\end{bmatrix}
	=
	\begin{bmatrix}1&0\\0&1\end{bmatrix}
	\begin{bmatrix}a&b\\c&d\end{bmatrix},\\
	\begin{bmatrix}ta&tb\\c&d\end{bmatrix}
	=
	\begin{bmatrix}t&0\\0&1\end{bmatrix}
	\begin{bmatrix}a&b\\c&d\end{bmatrix}.
\end{gather*}
\end{example}

\begin{example}
\"Ornek \numaraya{ex:5} bak\i n.
E\u ger
\begin{equation*}
 A=
    \begin{bmatrix}
    0&  0&-2&-8& 0& 2\\
   -1& -6& 0& 5& 2& 1\\
   -2&-12& 0&10& 2&-2\\
    1&  6& 1&-1&-1& 0
    \end{bmatrix}
		\end{equation*}
		ve
		\begin{align*}
		E_1&=\begin{bmatrix}0&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{bmatrix},&
E_2&=
\begin{bmatrix}1&0&0&0\\1&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix},\\
E_3&=
\begin{bmatrix}1&0&0&0\\0&1&0&0\\2&0&1&0\\0&0&0&1\end{bmatrix},&
E_4&=
\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&-2&1&0\\0&0&0&1\end{bmatrix},\\
E_5&=
\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&2&0&1\end{bmatrix},&
E_6&=
\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&1&1\end{bmatrix},\\
E_7&=
\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&\nicefrac12&0\\0&0&0&1\end{bmatrix}&
\phantom{E_1}&
\phantom{{}=
\begin{bmatrix}0&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{bmatrix}}&
		\end{align*}
ise, o zaman
\begin{equation*}
E_7E_6E_5E_4E_3E_2E_1A=
  \begin{bmatrix}
    1&6&1&-1&-1& 0\\
    0&0&1& 4& 1& 1\\
    0&0&0& 0& 1& 2\\
    0&0&0& 0& 0& 0
  \end{bmatrix}.
\end{equation*}		
\end{example}

\begin{theorem}
$\left[\begin{array}{c|c}A&I_n\end{array}\right]$
ve
$\left[\begin{array}{c|c}C&P\end{array}\right]$
sat\i rca denk ise, o zaman
\begin{equation*}
PA=C.
\end{equation*}
\end{theorem}

\begin{proof}
Varsay\i ma g\"ore baz\i\ $E_i$ elemanter matrisleri i\c cin
\begin{align*}
	\left[\begin{array}{c|c}C&P\end{array}\right]
&=E_{\ell}\cdots E_1	\left[\begin{array}{c|c}A&I_n\end{array}\right]\\	
&=\left[\begin{array}{c|c}E_{\ell}\cdots E_1A&E_{\ell}\cdots E_1I_n\end{array}\right]\\	
&=\left[\begin{array}{c|c}E_{\ell}\cdots E_1A&E_{\ell}\cdots E_1\end{array}\right],
\end{align*}
dolay\i s\i yla
\begin{align*}
	P&=E_{\ell}\cdots E_1,&C&=E_{\ell}\cdots E_1A=PA.\qedhere
\end{align*}
\end{proof}

\begin{example}\label{ex:inv}
  $A=\begin{bmatrix}
    1& 3&-3\\
    4&-1&-1\\
   -2& 1&-2
  \end{bmatrix}$
  ve
  $\bm b=
  \begin{bmatrix}
    6\\-2\\2
  \end{bmatrix}$
  ise
%  Tablo \numarada{tab:inv}ki hesaplamalara g\"ore,
%  \begin{table}
%    \caption{\.Indirgeme}
%    \label{tab:inv}   
    \begin{multline*}
      \left[\begin{array}{c|c|c}A&\bm b&I\end{array}\right]
      =
      \begin{bmatrix}
        1& 3&-3& 6&1&0&0\\
        4&-1&-1&-2&0&1&0\\
        -2& 1&-2& 2&0&0&1
      \end{bmatrix}\\
      \xrightarrow[2R_1+R_3]{-4R_1+R_2}
      \begin{bmatrix}
        1&  3&-3&  6& 1&0&0\\
        0&-13&11&-26&-4&1&0\\
        0&  7&-8& 14& 2&0&1
      \end{bmatrix}\\
      \xrightarrow{-\frac1{13}R_2}
      \begin{bmatrix}
        1&3&          -3     & 6&        1    &          0     &0\\
        0&1&\matfrac{-11}{13}& 2&\matfrac4{13}&\matfrac{-1}{13}&0\\
        0&7&          -8     &14&        2    &          0     &1
      \end{bmatrix}\\
      \xrightarrow{-7R_2+R_3}
      \begin{bmatrix}
        1&3&         -3      &6&          1    &           0     &0\\
        0&1&\matfrac{-11}{13}&2&\matfrac  4 {13}&\matfrac{-1}{13}&0\\
        0&0&\matfrac{-27}{13}&0&\matfrac{-2}{13}&\matfrac  7 {13}&1
      \end{bmatrix}\\
      \xrightarrow{-\frac{13}{27}R_3}
      \begin{bmatrix}
        1&3&          -3     &6&        1    &          0     &           0     \\
        0&1&\matfrac{-11}{13}&2&\matfrac4{13}&\matfrac{-1}{13}&           0     \\
        0&0&           1     &0&\matfrac2{27}&\matfrac{-7}{27}&\matfrac{-13}{27}
      \end{bmatrix}\\
      \xrightarrow[3R_3+R_1]{\frac{11}{13}R_3+R_2}
      \begin{bmatrix}
        1&3&0&6&\matfrac{33}{27}&\matfrac{-21}{27}&\matfrac{-39}{27}\\
        0&1&0&2&\matfrac{10}{27}&\matfrac {-8}{27}&\matfrac{-11}{27}\\
        0&0&1&0&\matfrac  2 {27}&\matfrac {-7}{27}&\matfrac{-13}{27}
      \end{bmatrix}\\
      \xrightarrow{-3R_2+R_1}
      \begin{bmatrix}
        1&0&0&0&\matfrac  3 {27}&\matfrac   3 {27}&\matfrac{ -6}{27}\\
        0&1&0&2&\matfrac{10}{27}&\matfrac {-8}{27}&\matfrac{-11}{27}\\
        0&0&1&0&\matfrac  2 {27}&\matfrac {-7}{27}&\matfrac{-13}{27}
      \end{bmatrix}
    \end{multline*}
    %  \end{table}
    dolay\i s\i yla
 $\bm d=(0,2,0)$ olmak \"uzere
\begin{equation*}
\frac1{27}\begin{bmatrix}3&3&-6\\10&-8&-11\\ 2&-7&-13\end{bmatrix}
  \cdot
  \left[
    \begin{array}{c|c}
      A&\bm b
    \end{array}
    \right]
  =\left[
    \begin{array}{c|c}
      I_3&\bm d
    \end{array}
    \right].
\end{equation*}
Kontrol ederiz:
\begin{multline}\label{eqn:kontrol}
\begin{bmatrix}3&3&-6\\10&-8&-11\\ 2&-7&-13\end{bmatrix}
\begin{bmatrix}1&3&-3&6\\ 4&-1& -1&-2\\-2& 1& -2&2\end{bmatrix}=\\
  \begin{bmatrix}
   3+12+12& 9-3- 6&- 9-3+12&18 -6-12\\
  10-32+22&30+8-11&-30+8+22&60+16-22\\
   2-28+26& 6+7-13&- 6+7+26&12+14-26
  \end{bmatrix}\\
  =\begin{bmatrix}
  27&0&0&0\\
  0&27&0&54\\
  0&0&27&0
  \end{bmatrix}.
\end{multline}
\end{example}

\section{Kare matrisler}

Matris \c carpmas\i n\i n \"ozel bir durumunda
%\Teoremde{thm:AB}n
\begin{equation*}
(X,Y)\mapsto XY\colon F^{n\times n}\times F^{n\times n}\to F^{n\times n}.
\end{equation*}
\Teoreme{thm:ass} g\"ore bu i\c slem birle\c smelidir.

\begin{example}
  Matrisler \c carpmas\i\
  %her $F^{n\times n}$ k\"umesinde
  de\u gi\c smeli de\u gildir:
\begin{align*}
	\begin{bmatrix}0&1\\1&0\end{bmatrix}
	\begin{bmatrix}1&1\\0&0\end{bmatrix}
	&=	\begin{bmatrix}0&0\\1&1\end{bmatrix},&
	\begin{bmatrix}1&1\\0&0\end{bmatrix}
	\begin{bmatrix}0&1\\1&0\end{bmatrix}
&=	\begin{bmatrix}1&1\\0&0\end{bmatrix}.
\end{align*}
\end{example}


E\u ger
\begin{align*}
	AB&=I_n,&BA&=I_n
\end{align*}
ise,
o zaman
$A$ ve $B$'den her biri, di\u gerinin \textbf{tersidir,}
ve
\begin{align*}
	B&=A\inv,&A&=B\inv
\end{align*}
yaz\i l\i r.

\begin{example}\label{ex:2x2-inv}
E\u ger $ad\neq bc$ ise, o zaman
\begin{equation*}
\begin{bmatrix}a&b\\c&d\end{bmatrix}\inv
=\frac1{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.
\end{equation*}
\end{example}

\begin{example}\label{ex:invv}
$I_n=I_n{}\inv$.
Ayr\i ca her elemanter matrisin tersi vard\i r.
E\u ger $A$ ve $B$'nin tersleri varsa, o zaman
\begin{equation*}
B\inv A\inv=(AB)\inv.
\end{equation*}
E\u ger $E_i$ matrisleri elemanter ise, o zaman
\begin{equation*}
E_1{}\inv\cdots E_{\ell}{}\inv=(E_{\ell}\cdots E_1)\inv.
\end{equation*}
Ama $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ matrisinin tersi yoktur.
\end{example}

Bir
\begin{equation*}
\begin{bmatrix}
d_1&0&\cdots&0\\
0&d_2&\ddots&\vdots\\
\vdots&\ddots&\ddots&0\\
0&\cdots&0&d_n
\end{bmatrix}
\end{equation*}
matrisi,
bir \textbf{k\"o\c segen matrisidir,}
ve
\begin{equation*}
\diag(d_1,\dots,d_n)
\end{equation*}
olarak yaz\i labilir.
O halde
\begin{equation*}
I_n=\diag(1,\dots,1).
\end{equation*}
Ayr\i ca
\begin{equation*}
\diag(a_1,\dots,a_n)\cdot\diag(b_1,\dots,b_n)
=\diag(a_1b_1,\dots,a_nb_n).
\end{equation*}
Bundan dolay\i\
$\diag(d_1,\dots,d_n)$ matrisinin tersi var olmas\i\ i\c cin
gerek ve yeter bir ko\c sul,
\begin{equation*}
d_1\cdots d_n\neq0
\end{equation*}
olur.  Bu durumda
\begin{equation*}
\diag(d_1,\dots,d_n)\inv=\diag(d_1{}\inv,\dots,d_n{}\inv).
\end{equation*}

\begin{theorem}%\label{thm:BA}
$F^{n\times n}$ k\"umesinde $AB=I_n$ ise,
o zaman
\begin{equation}\label{eqn:BA=I}
BA=I_n.
\end{equation}
\end{theorem}

\begin{proof}
$AB=I_n$ olsun.
Terslenebilir bir $P$ matrisi i\c cin $PA$,
sat\i rca indirgenmi\c s basamakl\i\ bir $C$ matrisidir.
O zaman
\begin{gather*}
	CB=(PA)B=P(AB)=PI_n=P,\\
	C(BP\inv)=(CB)P\inv=PP\inv=I_n.
\end{gather*}
Bundan dolay\i\ $C$'nin her sat\i r\i n\i n ba\c s eleman\i\ vard\i r.
Asl\i nda $C=I_n$ olmal\i,
dolay\i s\i yla $B=P$, ve \eqref{eqn:BA=I} \c c\i kar.
\end{proof}

\begin{example}
\"Ornek \numarada{ex:inv}ki hesaplamalara g\"ore
\begin{equation*}
\begin{bmatrix}1&3&-3\\ 4&-1& -1\\-2& 1& -2\end{bmatrix}\inv
=\frac1{27}
\begin{bmatrix}3&3&-6\\10&-8&-11\\ 2&-7&-13\end{bmatrix}.
\end{equation*}
\end{example}

\begin{example}
\.Iki-elemanl\i\ $\Z_2$ cismi \"uzerinde,
  \begin{gather*}
	\begin{bmatrix}
    1&0&1&0&1&0&0&0\\
		0&1&0&0&0&1&0&0\\
		1&1&0&1&0&0&1&0\\
		1&1&1&1&0&0&0&1
  \end{bmatrix} 
	\xrightarrow[R_2+R_4]{R_2+R_3}
	\begin{bmatrix}
    1&0&1&0&1& 0&0&0\\
		0&1&0&0&0& 1&0&0\\
		1&0&0&1&0&1&1&0\\
		1&0&1&1&0&1&0&1
  \end{bmatrix}\\
		\xrightarrow{R_1+R_4}
	\begin{bmatrix}
    1&0&1&0& 1& 0&0&0\\
		0&1&0&0& 0& 1&0&0\\
		1&0&0&1& 0&1&1&0\\
		0&0&0&1&1&1&0&1
  \end{bmatrix}
	\xrightarrow{R_4+R_3}\\
	\begin{bmatrix}
    1&0&1&0& 1& 0&0& 0\\
		0&1&0&0& 0& 1&0& 0\\
		1&0&0&0& 1& 0&1&1\\
		0&0&0&1&1&1&0& 1
  \end{bmatrix} 
	\xrightarrow{R_3+R_1}
	\begin{bmatrix}
    0&0&1&0& 0& 0&1& 1\\
		0&1&0&0& 0& 1& 0& 0\\
		1&0&0&0& 1& 0& 1&1\\
		0&0&0&1&1&1& 0& 1
  \end{bmatrix}\\ 	
	\xrightarrow{R_1\leftrightarrow R_3}
	\begin{bmatrix}
		1&0&0&0& 1& 0& 1&1\\
		0&1&0&0& 0& 1& 0& 0\\
    0&0&1&0& 0& 0&1& 1\\
		0&0&0&1&1&1& 0& 1
  \end{bmatrix}
  \end{gather*}
	dolay\i s\i yla
	\begin{equation*}
	\begin{bmatrix}
    1&0&1&0\\
		0&1&0&0\\
		1&1&0&1\\
		1&1&1&1
  \end{bmatrix}\inv=
	\begin{bmatrix}
		 1& 0& 1& 1\\
		 0& 1& 0& 0\\
     0& 0& 1& 1\\
		 1& 1& 0& 1
  \end{bmatrix}.	
	\end{equation*}
	Kontrol edelim:
	\begin{equation*}
\begin{bmatrix}
		 1& 0& 1&1\\
		 0& 1& 0& 0\\
     0& 0&1& 1\\
		1&1& 0& 1
  \end{bmatrix} 			
	\begin{bmatrix}
    1&0&1&0\\
		0&1&0&0\\
		1&1&0&1\\
		1&1&1&1
  \end{bmatrix}
		=
			\begin{bmatrix}
			1&0&0&0\\
			0&1&0&0\\
			0&0&1&0\\
			0&0&0&1
  \end{bmatrix}
	\end{equation*}
\end{example}

\begin{exercise}
Baz\i\ elemanter sat\i r i\c slemleriyle,
birim matrisinden terslenebilir bir matris elde edin.
Bu matrisin tersini bulun,
ve sonu\c cunuzu kontrol edin.
\end{exercise}

\begin{theorem}\label{thm:inv}
Bir $A$ kare matrisi i\c cin a\c sa\u g\i daki ko\c sullar birbirine denktir.
\begin{compactenum}
\item\label{item:E}
$A$, baz\i\ elemanter matrislerin \c carp\i m\i d\i r.
\item\label{item:inv}
$A$'n\i n tersi vard\i r.
\item\label{item:triv}
$A\bm x=\bm0$ sisteminin a\c sik\^ar olmayan \c c\"oz\"um\"u yoktur.
\item\label{item:zero}
$A$, sat\i r\i\ s\i f\i r olan bir matrise sat\i rca denk de\u gildir.
\end{compactenum}
\end{theorem}

\begin{proof}
$\eqref{item:E}\Rightarrow\eqref{item:inv}$.
Bunu \"Ornek \numarada{ex:invv} g\"ord\"uk.

$\eqref{item:inv}\Rightarrow\eqref{item:triv}$.
$A$'n\i n tersi varsa
\begin{align*}
A\bm x=\bm0
&\implies A\inv A\bm x=A\inv\bm0\\
&\implies\bm x=\bm0.
\end{align*}

$\eqref{item:triv}\Rightarrow\eqref{item:zero}$.
E\u ger $A$, sat\i r\i\ s\i f\i r olan bir matrise sat\i rca denk ise,
o zaman $A\bm x=\bm0$ sisteminin serbest de\u gi\c skeni vard\i r,
dolay\i s\i yla sistemin a\c sik\^ar olmayan \c c\"oz\"um\"u vard\i r.

$\eqref{item:zero}\Rightarrow\eqref{item:E}$.
\Teoreme{thm:Phi} g\"ore 
baz\i\ elemanter matrislerin $P$ \c carp\i m\i\ i\c cin $PA$,
sat\i rca indirgenmi\c s basamakl\i\ bir matristir.
Bu matrisin s\i f\i r olan sat\i r\i\ yoksa,
o zaman $PA=I$, dolay\i s\i yla $A=P\inv$.
\end{proof}




\section{Determinantlar}

\Teoreme{thm:inv} bir ko\c sul ekleyece\u giz.
\Sayfada{eqn:det-def},
her $A$ kare matrisi i\c cin,
\begin{equation}\label{eqn:det=0}
\det A\neq0\iff\text{$A$'n\i n tersi vard\i r}
\end{equation}
kural\i n\i\ sa\u glayan $\det A$ skalerini tan\i mlayaca\u g\i z.
Bu $\det A$ skaleri,
$A$'n\i n \emph{belirleyicisi} veya \textbf{determinant\i d\i r.}
B\"oylece her $F$ cismi i\c cin,
her $n$ sayma say\i s\i\ i\c cin,
\begin{equation*}
X\mapsto\det X\colon F^{n\times n}\to F.
\end{equation*}
\.Iki-elemanl\i\ $\Z_2$ cismi \"uzerinde
\begin{equation*}
\left.
\begin{array}{cc}
\text{$A$'n\i n tersi varsa}&1\\
\text{$A$'n\i n tersi yoksa}&0\\
\end{array}\right\}=\det A.
\end{equation*}
Bu durumda
\begin{equation}\label{eqn:det-mul}
\det(AB)=\det A\cdot\det B.
\end{equation}
Herhangi cisim \"uzerinde,
determinant\i n verece\u gimiz tan\i m\i,
bu kural\i\ sa\u glayacak.

\begin{theorem}
E\u ger
\begin{equation}\label{eqn:2-det}
\det\begin{bmatrix}a&b\\c&d\end{bmatrix}=ad-bc
\end{equation}
tan\i mlan\i rsa,
o zaman $2\times 2$'lik matrisler i\c cin 
kurallar \eqref{eqn:det=0} ve \eqref{eqn:det-mul}
sa\u glan\i r.
\end{theorem}

\begin{proof}
\"Ornek \numarada{ex:2x2-inv} g\"ord\"u\u g\"um\"uz gibi
$ad-bc\neq0$ ise 
$\left[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right]$
matrisinin tersi vard\i r.
\c Simdi $ad-bc=0$ olsun.
\begin{compactitem}
\item
E\u ger $ad=0$ ise, o zaman $bc=0$,
ve bu durumda
$\left[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right]$
matrisinin bir sat\i r\i\ veya bir s\"utunu s\i f\i rd\i r,
dolay\i s\i yla
\begin{equation}\label{eqn:abcd}
\left\{
\begin{array}{*2{c@{\:}}*2{c@{\;}}c}
ax&+&by&=&0\\
cx&+&dy&=&0
\end{array}\right.
\end{equation}
sisteminin a\c sik\^ar olmayan \c c\"oz\"um\"u vard\i r.
\item
E\u ger $ad\neq0$ ise
\begin{equation*}
\begin{bmatrix}a&b\\c&d\end{bmatrix}\xrightarrow{-\frac caR_1+R_2}
\begin{bmatrix}a&b\\0&0\end{bmatrix}
\end{equation*}
dolay\i s\i yla tekrar sistem \denklemin{eqn:abcd}
a\c sik\^ar olmayan \c c\"oz\"um\"u vard\i r.
\end{compactitem}
Son olarak tan\i m \denkleme{eqn:2-det} g\"ore
\begin{align*}
	\det\left(
	\begin{bmatrix}a&b\\c&d\end{bmatrix}
	\begin{bmatrix}e&f\\g&h\end{bmatrix}
	\right)
	&=\det\begin{bmatrix}ae+bg&af+bh\\ce+dg&cf+dh\end{bmatrix}\\
	&=aedh+bgcf-afdg-bhce\\
	&=(ad-bc)(eh-fg)\\
	&=
	\det
	\begin{bmatrix}a&b\\c&d\end{bmatrix}\cdot
	\det
	\begin{bmatrix}e&f\\g&h\end{bmatrix}.
          \qedhere
\end{align*}
\end{proof}

E\u ger kurallar \eqref{eqn:det=0} ve \eqref{eqn:det-mul}
do\u gru olacaksa
\begin{equation}\label{eqn:det-I}
\det I=1.
\end{equation}
\c Simdilik baz\i\ $i$ ve $j$ i\c cin
\begin{align*}
\Phi_t&=tR_i+R_j,&
\Psi&=R_i\leftrightarrow R_j,&
\Theta_t&=tR_i
\end{align*} 
ise, o zaman
\begin{align*}
	\Phi_t\circ\Phi_s&=\Phi_{t+s},&
	\Psi\circ\Psi&=\Phi_0,&
	\Theta_t\circ\Theta_s&=\Theta_{ts}.
\end{align*}
E\u ger
\begin{align}\label{eqn:E-det}
	\det(\Phi_t(I))&=1,&
	\det(\Psi(I))&=-1,&
	\det(\Theta_t(I))&=t
\end{align}
tan\i mlan\i rsa,
o zaman baz\i\ durumlarda \eqref{eqn:det-mul} sa\u glan\i r.
Her $E_i$ elemanter olmak \"uzere
e\u ger
\begin{equation}\label{eqn:det-E}
\det(E_r\cdots E_1)=\det E_r\cdots\det E_1
\end{equation}
tan\i mlanabilirse, o zaman
\Teorem{thm:inv} sayesinde,
terslenebilir matrisler i\c cin
\eqref{eqn:det-mul} sa\u glan\i r.

\begin{example}\label{ex:det}
\"Ornek \numarada{ex:inv}ki hesaplamara g\"ore
\begin{equation*}
\det\begin{bmatrix} 1& 3&-3\\
                  4&-1&-1\\
								 -2& 1&-2\end{bmatrix}
								=1\cdot(-13)\cdot1\cdot\left(-\frac{27}{13}\right)\cdot1\cdot1=27.
\end{equation*}
\end{example}

Determinantlar i\c cin
h\^al\^a kesin bir tan\i m\i m\i z yoktur,
\c c\"unk\"u ayn\i\ matris,
birden fazla \c sekilde
elemanter matrislerin \c carp\i m\i\ olarak yaz\i labilir.

\begin{example}
\begin{multline*}
	\begin{bmatrix}1&0\\0&1\end{bmatrix}
\xrightarrow{2R_1+R_2}\begin{bmatrix}1&0\\2&1\end{bmatrix}
\xrightarrow{\frac12R_2}\begin{bmatrix}1&0\\1&\nicefrac12\end{bmatrix}
\xrightarrow{-R_2+R_1}\begin{bmatrix}0&\nicefrac{-1}2\\1&\nicefrac12\end{bmatrix}\\
\xrightarrow{R_1+R_2}\begin{bmatrix}0&\nicefrac{-1}2\\1&0\end{bmatrix}
\xrightarrow{-2R_1}\begin{bmatrix}0&1\\1&0\end{bmatrix}
\xrightarrow{R_1\leftrightarrow R_2}\begin{bmatrix}1&0\\0&1\end{bmatrix},
\end{multline*}
dolay\i s\i yla
\begin{equation*}
	\begin{bmatrix}1&0\\0&1\end{bmatrix}
=
	\begin{bmatrix}0&1\\1&0\end{bmatrix}
	\begin{bmatrix}-2&0\\0&1\end{bmatrix}
		\begin{bmatrix}1&0\\1&1\end{bmatrix}
		\begin{bmatrix}1&-1\\0&1\end{bmatrix}
				\begin{bmatrix}1&0\\0&\nicefrac12\end{bmatrix}
						\begin{bmatrix}1&0\\2&1\end{bmatrix}.
\end{equation*}
Bu e\c sitlik,
istedi\u gimiz
e\c sitlikler \eqref{eqn:det-I}, \eqref{eqn:E-det},
ve \eqref{eqn:det-E} ile \c celi\c smez
\c c\"unk\"u
\begin{equation*}
1=-1\cdot(-2)\cdot1\cdot1\cdot\frac12\cdot1.
\end{equation*}
\end{example}

\.Istedi\u gimiz kurallara g\"ore
\begin{equation*}
\det\diag(d_1,\dots,d_n)=d_1\cdots d_n.
\end{equation*}
Ayr\i ca
\begin{multline*}
\det\begin{bmatrix}
d_1& * &\cdots&*\\
 0 &d_2&\ddots&\vdots\\
\vdots&\ddots&\ddots&*\\
0&\cdots&0&d_n
\end{bmatrix}\\
=d_1d_2\cdots d_n
=\det\begin{bmatrix}
d_1& 0 &\cdots&0\\
 * &d_2&\ddots&\vdots\\
\vdots&\ddots&\ddots&0\\
*&\cdots&*&d_n
\end{bmatrix}.
\end{multline*}
Buradaki matrisler, 
\textbf{\"u\c cgen matrisidir.}
\begin{comment}

\textbf{\"Ust \"u\c cgen matrisi,}
\begin{equation*}
i>j\implies a_{ij}=0
\end{equation*}
ko\c sulunu sa\u glayan bir $[a_{ij}]^{1\leq i\leq n}_{1\leq j\leq n}$ matrisidir;
bu durumda matrisin determinant\i,
\begin{equation*}
a_{11}\cdots a_{nn}.
\end{equation*}

\end{comment}

\begin{theorem}\label{thm:3x3}
E\u ger
\begin{align*}
\det\begin{bmatrix}a&0&0\\0&b&0\\0&0&c\end{bmatrix}
=\det\begin{bmatrix}0&0&c\\a&0&0\\0&b&0\end{bmatrix}
=\det\begin{bmatrix}0&b&0\\0&0&c\\a&0&0\end{bmatrix}
&=abc,\\
\det\begin{bmatrix}a&0&0\\0&0&c\\0&b&0\end{bmatrix}
=\det\begin{bmatrix}0&b&0\\a&0&0\\0&0&c\end{bmatrix}
=\det\begin{bmatrix}0&0&c\\0&b&0\\a&0&0\end{bmatrix}
&=-abc,
\end{align*}
ve
\begin{multline}\label{eqn:3-det}
	\det\begin{bmatrix}a&b&c\\d&e&f\\g&h&k\end{bmatrix}\\
	=
    \det\begin{bmatrix}a&0&0\\0&e&0\\0&0&k\end{bmatrix}
  +	\det\begin{bmatrix}0&0&c\\d&0&0\\0&h&0\end{bmatrix}
	+	\det\begin{bmatrix}0&b&0\\0&0&f\\g&0&0\end{bmatrix}\\
	+	\det\begin{bmatrix}a&0&0\\0&0&f\\0&h&0\end{bmatrix}	
	+	\det\begin{bmatrix}0&b&0\\d&0&0\\0&0&k\end{bmatrix}
	+	\det\begin{bmatrix}0&0&c\\0&e&0\\g&0&0\end{bmatrix}
\end{multline}
tan\i mlan\i rsa,
o zaman $3\times 3$'lik matrisler i\c cin 
kurallar \eqref{eqn:det=0} ve \eqref{eqn:det-mul}
sa\u glan\i r.
\end{theorem}

\begin{proof}
$3\times 3$'lik matrisler i\c cin,
verilen tan\i m ile
e\c sitlikler \eqref{eqn:E-det} do\u grudur.
\"Orne\u gin
\begin{gather*}
	\det\begin{bmatrix}1&0&0\\t&1&0\\0&0&1\end{bmatrix}
	=\det\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
	+\det\begin{bmatrix}0&0&0\\t&0&0\\0&0&1\end{bmatrix}
	=1,\\
	\det\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}=-1,\\
	\det\begin{bmatrix}1&0&0\\0&1&0\\0&0&t\end{bmatrix}=t.
\end{gather*}
Benzer \c sekilde
$E$ elemanter bir matris olmak \"uzere
\begin{equation*}
\det(EA)=\det E\cdot\det A.
\end{equation*}
Sonu\c c olarak,
\Teoreme{thm:inv} g\"ore
\begin{compactitem}
\item
$A$ ve $B$'nin tersleri varsa,
kural \eqref{eqn:det-mul} do\u grudur;
\item
$A$ ve $B$ sat\i rca denk ise
\begin{equation*}
\det A=0\iff\det B=0.
\end{equation*}
\end{compactitem}
Ayr\i ca $A$'n\i n tersi yoksa $A$, 
sat\i r\i\ s\i f\i r olan bir matrise sat\i rca denktir;
ve bu matrisin determinant\i\ $0$ olur,
dolay\i s\i yla $\det A=0$.
B\"oylece kural \eqref{eqn:det=0} do\u grudur,
ve
kural \eqref{eqn:det-mul} her durumuda do\u grudur.
\end{proof}

Bir k\"umeden ayn\i\ k\"umeye giden birebir ve \"orten bir fonksiyon,
k\"umenin perm\"utasyonu veya \textbf{simetrisidir.}
$\{1,\dots,n\}$ k\"umesinin simetrileri
\begin{equation*}
\Sym(n)
\end{equation*}
k\"umesini olu\c stursun.
E\u ger $\sigma\in\Sym(n)$ ise
\begin{equation*}
\sgn(\sigma)=\prod_{1\leq i<j\leq n}\frac{\sigma(j)-\sigma(i)}{j-i}
\end{equation*}
olsun: bu say\i\ ya $1$ ya da $-1$ olur,
ve $\sigma$'n\i n \textbf{i\c saretidir.}
(\.Isaretin Latincesi \emph{signum,} \.Ingilizcesi \emph{sign} olur.)
\begin{comment}

\begin{theorem}
E\u ger $\sigma\in\Sym(n)$
ve $\tau\in\Sym(n)$ ise
\begin{equation*}
\sgn(\sigma\circ\tau)=\sgn(\sigma)\sgn(\tau).
\end{equation*}
\end{theorem}

\begin{proof}
\begin{align*}
	\sgn(\sigma\circ\tau)
	&=\prod_{1\leq i<j\leq n}\frac{\sigma(\tau(j))-\sigma(\tau(i))}{j-i}\\
	&=\prod_{1\leq i<j\leq n}\left(\frac{\sigma(\tau(j))-\sigma(\tau(i))}{\tau(j)-\tau(i)}
	\cdot\frac{\tau(j)-\tau(i)}{j-i}\right)\\
	&=\prod_{1\leq i<j\leq n}\frac{\sigma(\tau((j))-\sigma(\tau(i))}{\tau(j)-\tau(i)}
	\prod_{1\leq i<j\leq n}\frac{\tau(j)-\tau(i)}{j-i}\\
	&=\prod_{1\leq i<j\leq n}\frac{\sigma(j)-\sigma(i)}{j-i}
	\prod_{1\leq i<j\leq n}\frac{\tau(j)-\tau(i)}{j-i}\\
	&=\sgn(\sigma)\sgn(\tau).\qedhere
\end{align*}
\end{proof}

\end{comment}
A\c sa\u g\i da verece\u gimiz iki tan\i m i\c cin
\begin{equation}\label{eqn:A-square}
A=[a_{ij}]^{1\leq i\leq n}_{1\leq j\leq n}
\end{equation}
olsun.
\begin{compactenum}
\item
Resmi bir tan\i m olarak 
\begin{equation}\label{eqn:det-def}
\det A=\sum_{\sigma\in\Sym(n)}\sgn(\sigma)\prod_{j=1}^na_{\sigma(j)j}
\end{equation}
olsun.
O zaman \eqref{eqn:2-det} ve \eqref{eqn:3-det},
bu tan\i m\i n iki durumudur.
\item
\c Sapkal\i\ girdilerin silindi\u gi
\begin{equation}\label{eqn:M}
\begin{bmatrix}     
         a_{11} &\cdots&\widehat{a_{1j}}&\cdots&     a_{1n}     \\
          \vdots&      &     \vdots     &      &     \vdots     \\
\widehat{a_{i1}}&\cdots&\widehat{a_{ij}}&\cdots&\widehat{a_{in}}\\
          \vdots&      &     \vdots     &      &     \vdots     \\
  			 a_{n1} &\cdots&\widehat{a_{nj}}&\cdots&     a_{nn}
										\end{bmatrix}
										=M_{ij}
\end{equation}
olmak \"uzere
\begin{equation*}
\Adj A=\left[(-1)^{i+j}\det M_{ij}\right]^{1\leq j\leq n}_{1\leq i\leq n}
\end{equation*}
olsun: bu matris, $A$'n\i n \textbf{ekmatrisidir.}
$\Adj A$ matrisinin $(i,j)$ girdisi,
$(-1)^{i+j}\det M_{ji}$ olur.
\end{compactenum}

\begin{example}
$\Adj\begin{bmatrix}a&b\\c&d\end{bmatrix}
=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$.
\end{example}

\begin{theorem}
Tan\i m \eqref{eqn:det-def} ile
t\"um matrisler i\c cin
kurallar
\eqref{eqn:det=0} ve \eqref{eqn:det-mul}
sa\u glan\i r.
\end{theorem}

\begin{proof}
\Teoremin{thm:3x3} kan\i t\i n\i\ izleyin.
\end{proof}

\begin{theorem}
  Tan\i mlar
  \eqref{eqn:A-square} ve \denklemi{eqn:M} kullanarak
\begin{compactenum}
\item
$1\leq k\leq n$ ise
\begin{equation*}
\det A=\sum_{j=1}^n(-1)^{k+j}a_{kj}\det M_{kj},
\end{equation*}
\item
$1\leq\ell\leq n$ ise
\begin{equation}\label{eqn:det-cof}
\det A=\sum_{i=1}^n(-1)^{i+\ell}a_{i\ell}\det M_{i\ell}.
\end{equation}
\end{compactenum}
Sonu\c c olarak
\begin{equation*}
A\Adj A
=\Adj A\cdot A
=\diag(\det A,\dots,\det A)=\det A\cdot I_n,
\end{equation*}
dolay\i s\i yla $\det A\neq0$ ise
\begin{equation}\label{eqn:A-inv-Adj}
A\inv=\frac1{\det A}\Adj A.
\end{equation}
\end{theorem}

\begin{proof}
\"Orne\u gin
\begin{equation*}
\det[\cdots]=\lvert\cdots\rvert
\end{equation*}
olmak \"uzere
\begin{gather*}
	\det\begin{bmatrix}
	a_{11}&a_{12}&a_{13}\\
	a_{21}&a_{22}&a_{23}\\
	a_{31}&a_{32}&a_{33}
	\end{bmatrix}\\
	=\begin{vmatrix}
	a_{11}&0&0\\
	0&a_{22}&a_{23}\\
	0&a_{32}&a_{33}
	\end{vmatrix}
	+\begin{vmatrix}
	0&a_{12}&a_{13}\\
	a_{21}&0&0\\
	0&a_{32}&a_{33}
	\end{vmatrix}
	+\begin{vmatrix}
	0&a_{12}&a_{13}\\
	0&a_{22}&a_{23}\\
	a_{31}&0&0
	\end{vmatrix}\\
	=a_{11}\begin{vmatrix}
	a_{22}&a_{23}\\
	a_{32}&a_{33}
	\end{vmatrix}
	-a_{21}\begin{vmatrix}
	a_{12}&a_{13}\\
	a_{32}&a_{33}
	\end{vmatrix}
	+a_{31}\begin{vmatrix}
	a_{12}&a_{13}\\
	a_{22}&a_{23}\\
	\end{vmatrix}.\qedhere
\end{gather*}
\end{proof}

\begin{comment}

\begin{example}
$ad\neq bc$ ise
$\begin{bmatrix}a&b\\c&d\end{bmatrix}\inv
=\begin{bmatrix}\nicefrac d{(ad-bc)}&\nicefrac{-b}{(ad-bc)}\\\nicefrac{-c}{(ad-bc)}&\nicefrac a{(ad-bc)}\end{bmatrix}$.
\end{example}

\end{comment}

\begin{example}
\"Ornek \ref{ex:inv} ve \numaraya{ex:det} g\"ore
\begin{equation*}
	\Adj\begin{bmatrix}1&3&-3\\4&-1&-1\\-2&1&-2\end{bmatrix}
	=\begin{bmatrix}3&3&-6\\10&-8&-11\\2&-7&-13\end{bmatrix}.
\end{equation*}
Ayr\i ca, direkt hesaplama ile
\begin{multline*}
	\Adj\begin{bmatrix}1&3&-3\\4&-1&-1\\-2&1&-2\end{bmatrix}\\
	=\begin{bmatrix}
	 \begin{vmatrix}-1&-1\\ 1&-2\end{vmatrix}&-\begin{vmatrix}3&-3\\ 1&-2\end{vmatrix}&\begin{vmatrix}3&-3\\-1&-1\end{vmatrix}\\
	-\begin{vmatrix} 4&-1\\-2&-2\end{vmatrix}& \begin{vmatrix}1&-3\\-2&-2\end{vmatrix}&-\begin{vmatrix}1&-3\\ 4&-1\end{vmatrix}\\
	 \begin{vmatrix} 4&-1\\-2& 1\end{vmatrix}&-\begin{vmatrix}1& 3\\-2& 1\end{vmatrix}&\begin{vmatrix}1& 3\\ 4&-1\end{vmatrix}
	\end{bmatrix}\\
	=\begin{bmatrix}3&3&-6\\10&-8&-11\\2&-7&-13\end{bmatrix}.
\end{multline*}
Zaten yapt\i\u g\i m\i z kontrol \denkleme{eqn:kontrol} g\"ore
\begin{equation*}
	\det\begin{bmatrix}1&3&-3\\4&-1&-1\\-2&1&-2\end{bmatrix}=27,
\end{equation*}
dolay\i s\i yla
\begin{equation*}
A\inv=\frac1{27}\Adj A
=\begin{bmatrix}\matfrac  3 {27}&\matfrac  3 {27}&\matfrac{- 6}{27}\\
                \matfrac{10}{27}&\matfrac{-8}{27}&\matfrac{-11}{27}\\
                \matfrac  2 {27}&\matfrac{-7}{27}&\matfrac{-13}{27}\end{bmatrix}.
\end{equation*}
\end{example}

\begin{theorem}
$\det(A\trans)=\det A$.
\end{theorem}

\begin{theorem}
$A\in F^{n\times k}$, $B\in F^{n\times(n-k-1)}$ olsun.
\begin{equation*}
F(\bm x)=
\det\left[\begin{array}{c|c|c}A&\bm x&B\end{array}\right]
\end{equation*}
olmak \"uzere
\begin{gather*}
	F(t\bm x)=t\cdot F(\bm x),\\
	F(\bm x+\bm y)=F(\bm x)+F(\bm y).
\end{gather*}
\end{theorem}

Teoremde $F$ \textbf{do\u grusald\i r,}
ve sonu\c c olarak
\begin{equation*}
(\bm x_1,\dots,\bm x_n)\mapsto
\det\left[\begin{array}{c|c|c}\bm x_1&\cdots&\bm x_n\end{array}\right]
\end{equation*}
g\"ondermesi \textbf{$n$-do\u grusald\i r.}
Sonraki teoreme g\"ore ayn\i\ g\"onderme \textbf{alma\c s\i kt\i r.}

\begin{theorem}
$\det\left[
\begin{array}{c|c|c|c|c}\cdots&\bm x&\cdots&\bm x&\cdots\end{array}
\right]=0$.
\end{theorem}

\begin{theorem}[Cramer Kural\i]
E\u ger
\begin{equation*}
A=\left[\begin{array}{c|c|c}\bm a_1&\cdots&\bm a_n\end{array}\right]
\end{equation*}
ve terslenebilir ise,
o zaman
$A\bm x=\bm b$ sisteminin tek \c c\"oz\"um\"unde
her $j$ i\c cin
\begin{equation*}
x_j=\frac1{\det A}
\det
\left[\begin{array}{c|c|c|c|c|c|c}\bm a_1&\cdots&\bm a_{j-1}&\bm b&\bm a_{j+1}&\cdots&\bm a_n\end{array}\right].
\end{equation*}
\end{theorem}

\begin{proof}
$A$'n\i n tersi var oldu\u gundan
\begin{equation*}
A\bm x=\bm b\iff\bm x= A\inv\bm b.
\end{equation*}
Ayr\i ca sonu\c c \denkleme{eqn:A-inv-Adj} g\"ore
$A\inv\bm b$ vekt\"or\"un\"un $j$'ninci bile\c seni,
\begin{equation*}
\frac1{\det A}
\begin{bmatrix}(-1)^{1+j}M_{1j}&\cdots&(-1)^{n+j}M_{nj}\end{bmatrix}
\bm b
\end{equation*}
olur.  Bir de sonu\c c \denkleme{eqn:det-cof} g\"ore
\begin{multline*}
\begin{bmatrix}(-1)^{1+j}M_{1j}&\cdots&(-1)^{n+j}M_{nj}\end{bmatrix}
\bm b
=\sum_{i=1}^n(-1)^{i+j}b_iM_{ij}\\
=\det
\left[\begin{array}{c|c|c|c|c|c|c}\bm a_1&\cdots&\bm a_{j-1}&\bm b&\bm a_{j+1}&\cdots&\bm a_n\end{array}\right].
\qedhere
\end{multline*}
\end{proof}

\chapter{Soyut}

\section{Vekt\"or Uzaylar\i}

\begin{definition}\label{def:field}
  A\c sa\u g\i da,
  bir \textbf{cismin} aksiyomlar\i\ soldad\i r;
bu cisim \"uzerinde
bir \textbf{vekt\"or uzay\i n\i n} aksiyomlar\i,
sa\u gdad\i r:
\begin{align*}
       x+(y+z)&=(x+y)+z,&\bm u+(\bm v+\bm w)&=(\bm u+\bm v)+\bm w,\\
           x+0&=x      ,&         \bm x+\bm0&=\bm x              ,\\
        x+(-x)&=0      ,&     \bm x+(-\bm x)&=\bm0               ,\\
           x+y&=y+x    ,&        \bm u+\bm v&=\bm v+\bm u        ,\\
 x\cdot(y+z)&=x\cdot y+x\cdot z,&x\cdot(\bm u+\bm v)&=x\cdot\bm u+x\cdot\bm v,\\
  (x+y)\cdot z&=x\cdot z+y\cdot z  ,&(x+y)\cdot\bm u&=x\cdot\bm u+y\cdot\bm u,\\
(x\cdot y)\cdot z&=x\cdot(y\cdot z),&(x\cdot y)\cdot\bm u&=x\cdot(y\cdot\bm u),\\
      1\cdot x&=x      ,&             1\cdot\bm u&=\bm u              .\\
            x\cdot y&=y\cdot x                       ,&&\\
        \exists y\;(x\cdot y&=1\lor x=0),&&
\end{align*}
\Sayfada{ex:fields}ki \"Ornek \numaraya{ex:fields} bak\i n.
\end{definition}

\begin{example}
$F$'nin bir cisim oldu\u gu $F^n$ ve $F^{m\times n}$ k\"umeleri,
vekt\"or uzay\i d\i r.
(Sayfa \pageref{F^n} ve \sayfanumaraya{Fmn} bak\i n.)
\end{example}

\begin{example}\label{ex:pos}
  Toplama $(u,v)\mapsto u\cdot v$ ve \c carpma $(x,u)\mapsto u^x$ olmak \"uzere
  $(0,\infty)$ aral\i\u g\i, $\R$ veya $\Q$ \"uzerinde bir vekt\"or uzay\i d\i r.
\end{example}

\begin{example}
  Bir $A$ k\"umesinden $\R$'ye giden fonksiyonlar
  $\R$ \"uzerinde
  \begin{equation*}
    \mathscr F(A,\R)
  \end{equation*}
  vekt\"or uzay\i n\i\ olu\c sturur.
\end{example}

\begin{example}
  Her $F$ cismi i\c cin
  katsay\i lar\i\ $F$'den gelen,
  de\u gi\c skeni $X$ olan polinomlar
  $F$ \"uzerinde
  \begin{equation*}
    F[X]
  \end{equation*}
  vekt\"or uzay\i n\i\ olu\c sturur.
\end{example}

\begin{theorem}
  Her vekt\"or uzay\i nda
  \begin{align*}
    x\cdot\bm0&=\bm0,&
    0\cdot\bm u&=\bm0,&
    -1\cdot\bm u&=-\bm u.
  \end{align*}
\end{theorem}

\section{Uzaylar\i n Altuzaylar\i}

\begin{definition}
  Bir vekt\"or uzay\i n\i n bir altk\"umesi
  bo\c s de\u gilse ve
  toplamaya ve \c carpmaya g\"ore kapal\i\ ise
  altk\"ume, vekt\"or uzay\i n bir \textbf{altuzay\i d\i r.}
\end{definition}

\begin{example}
  Her vekt\"or uzay\i n\i n,
  tek eleman\i\ $\bm0$ olan a\c sik\^ar altuzay\i d\i r.
\end{example}

\begin{example}%\sloppy
  E\u ger $X$, \"ozde\c slik fonksiyonu olarak anla\c s\i l\i rsa,
  o zaman $\R[X]$,
  $\mathscr F[\R,\R]$ uzay\i n\i n bir altuzay\i\ olur.
\end{example}

\begin{example}
  Her $n$ do\u gal say\i s\i\ i\c cin
  \begin{equation*}
    \{f\in\R[X]\colon\deg(f)\leq n\}
  \end{equation*}
  k\"umesi $\R[X]$ uzay\i n\i n bir altuzay\i d\i r.
\end{example}

\begin{theorem}
  $A\in F^{m\times n}$ ise
  \begin{equation*}
    \{\bm u\in F^n\colon A\bm u=\bm0\}
  \end{equation*}
\c c\"oz\"um k\"umesi, $F^n$'nin bir altuzay\i d\i r.
\end{theorem}

\begin{example}
  $\R^2$ uzay\i n\i n
  \begin{gather*}
    \{(x,y)\in\R^2\colon x\geq0\And y\geq0\},\\
    \{(x,y)\in\R^2\colon xy\geq0\}
  \end{gather*}
  altk\"umeleri,
$\R^2$ uzay\i n\i n altuzay\i\ de\u gildir.
\end{example}

\begin{example}
$\Q$ \"uzerinde
\"Ornek \numarada{ex:pos} tan\i mlanm\i\c s $(0,\infty)$ uzay\i n\i n
\begin{equation*}
\left\{\left(\frac xy\right)^{1/n}\colon\{x,y,n\}\included\N\right\}
\end{equation*}
altuzay\i\ vard\i r.
\end{example}

\begin{definition}
  $F$ bir cisim ve $U$, $F$ \"uzerinde bir vekt\"or uzay\i\ olsun.
  E\u ger $B$, $U$'nun bir $\{\bm b_1,\dots,\bm b_n\}$ altk\"umesi ise,
  o zaman $F^n$'nin her $(a_1,\dots,a_n)$ eleman\i\ i\c cin
  \begin{equation*}
    a_1\cdot\bm b_1+\dots+a_n\cdot\bm b_n
  \end{equation*}
  toplam\i, $B$'nin bir \textbf{lineer bile\c simidir.}
 (E\u ger $n=0$ ise verilen lineer bile\c sim, $\bm0$ olur.)
  $B$'nin b\"ut\"un lineer bile\c simleri, $B$'nin \textbf{(lineer) gergisini}
  olu\c sturur.  Bu gergi
  \begin{align*}
    \text{ya}
    \qquad
    \lspan{\bm  b_1,\dots,\bm b_n}
    \qquad
    \text{ya da}
    \qquad
    \lspan B
  \end{align*}
  olarak yaz\i labilir.
  Elemanlar\i\ vekt\"or olan bir k\"ume,
  gergisini \textbf{\"uretir}
  ve gergisinin \textbf{\"urete\c c k\"umesidir.}
\end{definition}

\begin{comment}
  \begin{theorem}
    $F^m$ uzay\i n\i n her $\{\bm a_1,\dots,\bm a_n\}$ altk\"umesi i\c cin,
    $F^{m\times n}$ uzay\i n\i n bir $A$ eleman\i\ i\c cin
    \begin{equation*}
      \lspan{\bm a_1,\dots,\bm a_n}=\{\bm u\in F^m\colon A\bm u=\bm0\}.
    \end{equation*}
  \end{theorem}

  \begin{theorem}
    $F^{m\times n}$ uzay\i n\i n her $A$ eleman\i\ i\c cin,
    $F^m$ uzay\i n\i n bir $\{\bm a_1,\dots,\bm a_n\}$ altk\"umesi i\c cin,
    \begin{equation*}
      \{\bm u\in F^m\colon A\bm u=\bm0\}
      =\lspan{\bm a_1,\dots,\bm a_n}.
    \end{equation*}
  \end{theorem}
\end{comment}

\begin{theorem}
  Bir vekt\"or uzay\i n\i n her altk\"umesinin gergisi,
  verilen uzay\i n altuzay\i d\i r.
\end{theorem}

\begin{definition}
  Bir $F$ cismi \"uzerinde $A$,
  $m\times n$'lik bir matris olsun.
  $A$'n\i n
  \begin{enumerate}[1)]
  \item
    \textbf{sat\i r uzay\i,} $A$'n\i n sat\i rlar\i n\i n gergisidir;
  \item
    \textbf{s\"utun uzay\i,} $A$'n\i n s\"utunlar\i n\i n gergisidir;
  \item
    \textbf{s\i f\i r uzay\i,}
    $\{\bm u\in F^n\colon A\bm u=\bm0\}$ uzay\i d\i r.
  \end{enumerate}
  Bu uzaylar s\i ras\i yla
  \begin{align*}
    &\sat(A),&&\sut(A),&&\sif(A)
  \end{align*}
  olarak yaz\i ls\i n.
  Burada $\sat(A)$ ve $\sif(A)$, $F^n$ uzay\i n\i n altuzay\i d\i r;
  ve $\sut(A)$,  $F^m$ uzay\i n\i n altuzay\i d\i r.
\end{definition}

\section{Tabanlar}

\begin{definition}
  $U$ bir vekt\"or uzay\i\ ve $B$, $U$'nun bir $\{\bm b_1,\dots,\bm b_n\}$
  altk\"umesi olsun.
  E\u ger
  \begin{equation*}
    x_1\bm b_1+\dots+ x_n\bm b_n=\bm0
  \end{equation*}
  denkleminin sadece a\c sik\^ar \c c\"oz\"um\"u varsa,
  o zaman
  $B$ \textbf{lineer ba\u g\i ms\i zd\i r.}
  E\u ger $B$ lineer ba\u g\i ms\i z ve $U$'yu \"uretirse,
  o zaman $B$, $U$'nun bir \textbf{taban\i d\i r.}
\end{definition}

\begin{theorem}
  E\c sitlik \denklemde{eqn:A=}ki gibi
  $A=\left[
    \begin{array}{c|c|c}
      \bm a_1&\cdots&\bm a_n
    \end{array}
    \right]$
  olsun ve $B$,
  $A$'ya sat\i rca denk olan, basamakl\i\ bir $[b_{i\,j}]^i_j$ matrisi olsun.
  E\u ger bir $\ell$ i\c cin ve
  \begin{equation*}
    k_1<\cdots<k_{\ell}
  \end{equation*}
  ko\c sulunu sa\u glayan baz\i\ $k_i$ i\c cin
  $B$'nin sat\i rlar\i n\i n ba\c s elemanlar\i\
  $b_{i\,k_i}$ girdileri ise,
%  $b_{1\,k_1}$, \dots, $b_{\ell\,k_{\ell}}$ ise,
  o zaman
  \begin{enumerate}[1)]
  \item
    $\sat(A)$ i\c cin
    $B$'nin ilk $\ell$-tane sat\i r\i,
  \item
    $\sut(A)$ i\c cin
 $\bm a_{k_1}$, \dots, $\bm a_{k_{\ell}}$,
  \item
    $\sif(A)$ i\c cin
    $B\bm x=\bm0$ denkleminin temel \c c\"oz\"umleri,
  \end{enumerate}
  bir taban
  olu\c sturur.
\end{theorem}

\begin{corollary}
  $A$,
    $F$ \"uzerinde
$m\times n$'lik
$\left[
    \begin{array}{c|c|c}
      \bm a_1&\cdots&\bm a_n
    \end{array}
    \right]$
  matrisi
  olsun.
  \begin{compactenum}
  \item
    $\{\bm a_1,\dots,\bm a_n\}$ k\"umesinin
    \begin{compactitem}
    \item
      lineer ba\u g\i ms\i z olan
    \item
      gergisi ayn\i\ olan
    \end{compactitem}
    bir altk\"umesi elde etmek i\c cin,
$\sut(A)$ uzay\i n\i n bir taban\i\ al\i nabilir.
  \item
    E\u ger
    $\{\bm a_1,\dots,\bm a_n\}$ k\"umesi zaten lineer ba\u g\i ms\i z ise,
    o zaman 
    \begin{equation*}
      \left[
        \begin{array}{c|c}
          A&I
        \end{array}
        \right]
    \end{equation*}
    matrisinin s\"utun uzay\i n\i n
    teoremdeki gibi elde edilen taban\i,
    \begin{compactitem}
    \item
      $\{\bm a_1,\dots,\bm a_n\}$ k\"umesini kapsar ve
    \item
      $F^m$ uzay\i n\i n bir taban\i d\i r.
    \end{compactitem}
  \end{compactenum}
\end{corollary}

\begin{corollary}
  $F^m$ uzay\i n\i n bir altuzay\i n\i n her taban\i,
  ayn\i\ say\i da elemana sahiptir.
\end{corollary}

\begin{definition}
  $F^m$ uzay\i n\i n bir $U$ altuzay\i n\i n bir taban\i n\i n say\i s\i,
  $U$'nun \textbf{boyutudur,} ve bu boyut,
  \begin{equation*}
    \boy(U)
  \end{equation*}
  olarak yaz\i labilir.
\end{definition}

\begin{corollary}
  Herhangi $A$ matrisi i\c cin
  \begin{equation*}
    \boy\bigl(\sat(A)\bigr)=\boy\bigl(\sut(A)\bigr).
  \end{equation*}
  E\u ger $A$'n\i n $n$ tane s\"utunu varsa
  \begin{equation*}
    \boy\bigl((\sat(A)\bigr)+\boy\bigl(\sif(A)\bigr)=n.
  \end{equation*}
\end{corollary}

\section{Lineer D\"on\"u\c s\"umler}

\begin{definition}
  Bir $U$ vekt\"or uzay\i ndan
  bir $V$ vekt\"or uzay\i na giden,
\begin{align*}
  L(\bm u_1+\bm u_2)&=L(\bm u_1)+L(\bm u_2),\\
  L(t\cdot\bm u)&=t\cdot L(\bm u)
\end{align*}
kurallar\i n\i\ sa\u glayan bir $L$ g\"ondermesine
\textbf{lineer d\"on\"u\c s\"um} denir.
Bu durumda
\begin{compactenum}[1)]
\item
  $L$'nin \textbf{\c cekirde\u gi,} $U$'nun
  \begin{equation*}
    \{\bm u\in U\colon L(\bm u)=\bm0\}
  \end{equation*}
  altk\"umesidir,
\item
  $L$'nin \textbf{imgesi,} $V$'nin
  \begin{equation*}
    \{L(\bm u)\colon\bm u\in U\}
  \end{equation*}
  altk\"umesidir.
\end{compactenum}
Bu k\"umeler s\i ras\i yla
\begin{align*}
  &\cek(L),&&L[U]
\end{align*}
olarak yaz\i labilir.
E\u ger $L$ birebir ve $L[U]=V$ ise,
o zaman $L$ bir \textbf{izomorf\/izimdir.}
\end{definition}

\begin{example}
  $A\in F^{m\times n}$ ise
$F^n$ uzay\i ndan $F^m$ uzay\i na giden
  $\bm u\mapsto A\bm u$ fonksiyonu,
  lineer d\"on\"u\c s\"umd\"ur.
\end{example}

\begin{theorem}
  $L\colon U\to V$ ve lineer ise
  \begin{compactenum}
    \item
      $\cek(L)$, $U$'nun bir altuzay\i d\i r,
      \item
$L[U]$, $V$'nun bir altuzay\i d\i r.
  \end{compactenum}
\end{theorem}

\begin{theorem}
  Bir $F$ cismi \"uzerinde
   $\{\bm a_1,\dots,\bm a_n\}$,
  bir $U$ uzay\i n\i n altk\"umesi olsun.
  O zaman
  \begin{equation*}
    (x_1,\dots,x_n)\mapsto x_1\cdot\bm a_1+\cdots+ x_n\cdot\bm a_n,
  \end{equation*}
  $F^n$ uzay\i ndan $U$'ya giden lineer bir d\"on\"u\c s\"umd\"ur.
  E\u ger $\bm a_i$ vekt\"orleri
  lineer ba\u g\i ms\i z ise, o zaman verilen d\"on\"u\c s\"um
  bir izomorf\/izimdir.
\end{theorem}

\begin{corollary}
  Bir vekt\"or uzay\i n\i n sonlu bir taban\i\ varsa,
  o zaman uzay\i n her taban\i\
  ayn\i\ say\i da elemana sahiptir.
\end{corollary}

\begin{definition}
  Teoremde $\{\bm a_1,\dots,\bm a_n\}$, $U$'nun bir $B$ taban\i\ ise
  \begin{equation*}
    [x_1\cdot\bm a_1+\cdots+ x_n\cdot\bm a_n]_B
    =(x_1,\dots,x_n)
  \end{equation*}
  olsun.
  Bu vekt\"or, $x_1\cdot\bm a_1+\cdots+ x_n\cdot\bm a_n$ vekt\"or\"un\"un
  $B$'ye g\"ore
  \textbf{koordinat vekt\"or\"ud\"ur.}
\end{definition}

\begin{theorem}
  $B$, bir $U$ uzay\i n\i n bir taban\i;
  $C$, bir $V$ uzay\i n\i n bir taban\i;
  $L\colon U\to V$ ve lineer; olsun.
  O zaman bir $A$ matrisi i\c cin
  $U$'nun her $\bm u$ eleman\i\ i\c cin
  \begin{equation*}
    [L(\bm u)]_C=A[\bm u]_B.
  \end{equation*}
  Asl\i nda $B=\{\bm b_1,\dots, \bm b_n\}$ ise
  \begin{equation*}
    A=\left[
      \begin{array}{c|c|c}
        [L(\bm b_1)]_C&\cdots&[L(\bm b_n)]_C
      \end{array}
      \right].
  \end{equation*}
\end{theorem}

\begin{definition}
  Teoremde e\u ger $U=V$ ve $L$, \"ozde\c slik fonksiyonu ise $A$,
  $B$'den $C$'ye \textbf{ge\c ci\c s matrisidir.}
\end{definition}

\begin{definition}
  $\bm u,\bm v\in F^n$ ise
  \begin{equation*}
    \bm u\cdot\bm v=\bm u^{\mathrm t}\bm v=u_1v_1+\cdots+u_nv_n
  \end{equation*}
  olsun.
\end{definition}

\begin{theorem}
  $\bm a\in F^n$ ise $\bm u\mapsto\bm a\cdot\bm u$,
  $F^n$ uzay\i ndan $F$'ye giden lineer bir d\"on\"u\c s\"umd\"ur.
  Ayr\i ca
  \begin{equation*}
    \bm u\cdot\bm v=\bm v\cdot\bm u.
  \end{equation*}
  $A\in F^{m\times n}$, $\bm u\in\sif(A)$, ve $\bm v\in\sat(A)$ ise
  \begin{equation*}
    \bm u\cdot\bm v=0.
  \end{equation*}
\end{theorem}

\begin{theorem}
  $A\in\R^{m\times n}$;
  $B$, $\sif(A)$ uzay\i n\i n bir taban\i;
  ve $C$, $\sat(A)$ uzay\i n\i n bir taban\i\ ise,
  o zaman $B\cup C$ birle\c simi,
  $\R^n$ uzay\i n\i n bir taban\i d\i r.
\end{theorem}

\begin{proof}
  $B=\{\bm b_1,\dots,\bm b_k\}$, $C=\{\bm c_1,\dots,\bm c_{\ell}\}$ olsun.
  O zaman $k+\ell=n$, dolay\i s\i yla
  $\{\bm b_1,\dots,\bm b_k,\bm c_1,\dots,\bm c_{\ell}\}$
  k\"umesinin lineer ba\u g\i ms\i z oldu\u gunu kan\i tlamak yeter.
  Baz\i\ $x_i$ ve $y_j$ i\c cin
  \begin{equation*}
    x_1\bm b_1+\cdots+ x_k\bm b_k+y_1\bm c_1+\cdots+\bm_{\ell}\bm c_{\ell}=\bm0
  \end{equation*}
  olsun.
  O zaman
  \begin{equation*}
    x_1\bm b_1+\cdots+ x_k\bm b_k=-y_1\bm c_1-\cdots-\bm_{\ell}\bm c_{\ell}.
  \end{equation*}
  \"Ozel olarak $\bm d=x_1\bm b_1+\cdots+ x_k\bm b_k$ ise
  \begin{equation*}
    \bm d\in\sif(A)\cap\sat(A).
  \end{equation*}
  Bu durumda $\bm d\cdot\bm d=0$, yani
  \begin{equation*}
    d_1{}^2+\cdots+d_n{}^2=0.
  \end{equation*}
  Her $d_i$ ger\c cel say\i\ oldu\u gundan $d_i=0$,
  dolay\i s\i yla $\bm d=\bm0$.
  $B$ lineer ba\u g\i ms\i z oldu\u gundan $x_i=0$;
  ayn\i\ \c sekilde $y_j=0$.
\end{proof}

\begin{example}
  $\R$ \"uzerinde $A=
  \begin{bmatrix}
    \mathrm 1&1
  \end{bmatrix}$ ise
  $\sif(A)=\lspan{(-1,1)}$ ve $\sat(A)=\lspan{(1,1)}$,
  ve $\{(-1,1),(1,1)\}$ lineer ba\u g\i ms\i zd\i r \c c\"unk\"u
  \begin{equation*}
    \det
    \begin{bmatrix}
      -1&1\\1&1
    \end{bmatrix}\neq0.
  \end{equation*}
\end{example}

\begin{example}
  $\Z_2$ \"uzerinde $A=
  \begin{bmatrix}
    \mathrm 1&1
  \end{bmatrix}$ ise $\sif(A)=\sat(A)$.
\end{example}

\begin{example}
  $\C$ \"uzerinde $A=
  \begin{bmatrix}
    \mathrm i&1
  \end{bmatrix}$ ise $\sif(A)=\sat(A)$.
\end{example}

\section{Toplamlar}

$U$, $V$, ve $W$,
$F$ cismi \"uzerinde vekt\"or uzay\i\ olsun.

\begin{definition}
E\u ger $V$ ve $W$,
$U$'nun altuzay\i\ ise,
onlar\i n \textbf{toplam\i,}
\begin{equation*}
  \{\bm v+\bm w\colon\bm v\in V\And \bm w\in W\}
\end{equation*}
k\"umesidir.  Bu k\"ume,
\begin{equation*}
V+W
\end{equation*}
olarak yaz\i l\i r.
\end{definition}

\begin{example}\label{ex:VW}
  $F^4$ uzay\i nda
  \begin{align*}
    V&=\lspan{(1,1,1,0),(1,-1,1,0)},\\
    W&=\lspan{(1,0,1,1),(1,0,1,-1)}
  \end{align*}
  ise
  \begin{equation*}
    V+W=\lspan{(1,1,1,0),(1,-1,1,0),(1,0,1,1),(1,0,1,-1)}.
  \end{equation*}
Bu uzay,
  \begin{equation*}
    \begin{bmatrix}
      1&1&1&1\\1&-1&0&0\\1&1&1&1\\0&0&1&-1
    \end{bmatrix}
  \end{equation*}
  matrisinin s\"utun uzay\i d\i r.
  Ayr\i ca
  \begin{multline*}
    \begin{bmatrix}
      1&1&1&1\\1&-1&0&0\\1&1&1&1\\0&0&1&-1
    \end{bmatrix}
    \xrightarrow[-R_1+R_3]{-R_1+R_2}
    \begin{bmatrix}
      1&1&1&1\\0&-2&-1&-1\\0&0&0&0\\0&0&1&-1
    \end{bmatrix}\\
    \xrightarrow{R_3\leftrightarrow R_4}
    \begin{bmatrix}
      1&1&1&1\\0&-2&-1&-1\\0&0&1&-1\\0&0&0&0
    \end{bmatrix}    
  \end{multline*}
  oldu\u gundan
	\begin{enumerate}[(a)]
	\item
	$F$'de $2\neq0$ ise
    $\{(1,1,1,0),(1,-1,1,0),(1,0,1,1)\}$,
	\item
	$F$'de $2=0$ ise
    $\{(1,1,1,0),(1,0,1,1)\}$,
        \end{enumerate}
 $V+W$'nin bir taban\i d\i r.
\end{example}

\begin{theorem}
  E\u ger $V\included U$ ve $W\included U$ ise,
  o zaman $V+W$ k\"umesi,
	\begin{enumerate}[1)]
	\item
	$U$'nun bir altuzay\i d\i r,
	\item
	$U$'nun $V$'yi ve $W$'yi kapsayan en k\"u\c c\"uk altuzay\i d\i r.
	\end{enumerate}
\end{theorem}

\begin{proof}
\begin{asparaenum}
\item
Kan\i tlanacak \"u\c c ko\c sul vard\i r.
  \begin{compactenum}[(a)]
  \item
    $V+W$ bo\c s de\u gildir, \c c\"unk\"u
    \begin{align*}
      \bm0&\in V\cap W,&
      \bm0+\bm0&=\bm0
    \end{align*}
    oldu\u gundan
    $\bm0\in V+W$.
  \item
    $\bm u_1\in V+W$ ve $\bm u_2\in V+W$ ise
    $V$'nin baz\i\ $\bm v_i$ elemanlar\i\
    ve $W$'nin baz\i\ $\bm w_i$ elemanlar\i\ i\c cin
    \begin{equation*}
      \bm u_i=\bm v_i+\bm w_i,
    \end{equation*}
    dolay\i s\i yla
    (vekt\"orler toplamas\i\ birle\c smeli ve de\u gi\c smeli oldu\u gundan)
    \begin{equation*}
      \bm u_1+\bm u_2
      =(\bm v_1+\bm w_1)+(\bm v_2+\bm w_2)
      =(\bm v_1+\bm v_2)+(\bm w_1+\bm w_2).
    \end{equation*}
    $U$ ve $V$'nin her biri toplama alt\i nda kapal\i\ oldu\u gundan
    \begin{align*}
      \bm v_1+\bm v_2&\in V,&
      \bm w_1+\bm w_2&\in W,
    \end{align*}
    dolay\i s\i yla $\bm u_1+\bm u_2\in V+W$.
  \item
    Benzer \c sekilde $\bm u\in V+W$ ve $t\in F$ ise
    $V$'nin bir $\bm v$ eleman\i\
    ve $W$'nin bir $\bm w$ eleman\i\ i\c cin
    \begin{equation*}
      \bm u=\bm v+\bm w,
    \end{equation*}
    dolay\i s\i yla
    (skalerle \c carpma da\u g\i lmal\i\ oldu\u gundan)
    \begin{equation*}
t\bm u=t(\bm v+\bm w)=t\bm v+t\bm w.
    \end{equation*}
    $U$ ve $V$'nin her biri \c carpma alt\i nda kapal\i\ oldu\u gundan
    \begin{align*}
      t\bm v&\in V,&
      t\bm w&\in W,
    \end{align*}
    dolay\i s\i yla $t\bm u\in V+W$.
  \end{compactenum}
	\item
$U'$, $U$'nun $V$'yi ve $W$'yi kapsayan bir altuzay\i\ olsun.
E\u ger $\bm u\in V+W$ ise, o zaman
    $V$'nin bir $\bm v$ eleman\i\
    ve $W$'nin bir $\bm w$ eleman\i\ i\c cin
		\begin{align*}
      \bm u&=\bm v+\bm w,&\bm v&\in U',&\bm w&\in U',&\bm v+\bm w&\in U',
		\end{align*}
    dolay\i s\i yla $\bm u\in U'$.\qedhere
	\end{asparaenum}
\end{proof}

\begin{corollary}
$C\included U$ ve $D\included U$ ise
\begin{equation*}
\lspan C+\lspan D=\lspan{C\cup D}.
\end{equation*}
\end{corollary}

\begin{proof}
  Al\i\c st\i rma.
\end{proof}

\begin{theorem}
  $V\times W$ kartezyan \c carp\i m\i,
  \begin{align*}
    (\bm v_1,\bm w_1)+(\bm v_2,\bm w_2)
    &=(\bm v_1+\bm v_2,\bm w_1+\bm w_2),\\
t(\bm v,\bm w)&=(t\bm v,t\bm w)
  \end{align*}
  kurallar\i yla vekt\"or uzay\i\ oluyor.
\end{theorem}

\begin{proof}
  Al\i\c st\i rma.
\end{proof}

\begin{definition}
Vekt\"or uzay\i\ olarak anla\c s\i l\i nca $V\times W$ \c carp\i m\i
\begin{equation*}
  V\oplus W
\end{equation*}
olarak yaz\i l\i r
ve ona
$V$ ve $W$'nin \textbf{direkt toplam\i} denir.
\end{definition}

\begin{example}
$F^2=F\oplus F$.
\end{example}

\begin{example}
$F^{k+m}=F^k\oplus F^m$.
\end{example}

\begin{example}
  $F^n=\underbrace{F\oplus\dots\oplus F}_n$.
\end{example}

\begin{example}
  $\Z$ cisim de\u gildir,
  \c c\"unk\"u \c carpmaya g\"ore sadece $1$ ve $-1$'in tersleri vard\i r;
  ama $\underbrace{\Z\oplus\dots\oplus\Z}_n$ toplamlar\i\ olu\c sturulabilir.
\end{example}

\begin{comment}
  \begin{definition}
    $U$'dan $V$'ye giden,
    \begin{align*}
      h(\bm u_1+\bm u_2)&=h(\bm u_1)+h(\bm u_2),\\
      h(t\bm u)&=t\cdot h(\bm u)
    \end{align*}
    kurallar\i n\i\ sa\u glayan bir $h$ g\"ondermesine
    \textbf{do\u grusal d\"on\"u\c s\"um} denir.
    E\u ger $h$ birebir ve $V$'yi \"orten ise,
    o zaman $h$ alt\i nda $U$ ve $V$ \textbf{izomorftur} ve
    \begin{equation*}
      U\cong V
    \end{equation*}
    yaz\i l\i r.
    Bazen izomorfluk, e\c sitlik olarak anla\c s\i l\i r.
  \end{definition}
\end{comment}

\begin{theorem}
E\u ger $V\included U$ ve $W\included U$ ise,
o zaman
$V\times W$'den $V+W$'ye giden
\begin{equation*}
  (\bm v,\bm w)\mapsto\bm v+\bm w
\end{equation*}
g\"ondermesi,
$V+W$'yi \"orten bir $h$ lineer d\"on\"u\c s\"um\"ud\"ur,
ve bu d\"on\"u\c s\"um i\c cin
a\c sa\u g\i daki ko\c sullar birbirine denktir:
\begin{enumerate}
\item
  $h$ birebirdir.
\item
  $h(\bm v,\bm w)=\bm 0$ denkleminin 
	sadece a\c sik\^ar $(\bm0,\bm0)$ \c c\"oz\"um\"u vard\i r.
\item
  E\u ger $B$, $V$'nin bir taban\i\ ve $C$, $W$'nin bir taban\i\ ise,
  o zaman $B\cup C$, $V+W$'nin bir taban\i d\i r.
\item
  E\u ger $B\included V$, $C\included W$, ve her biri lineer ba\u g\i ms\i z ise,
  o zaman $B\cup C$ de lineer ba\u g\i ms\i zd\i r.
\end{enumerate}
\end{theorem}

\begin{proof}
  Al\i\c st\i rma.
\end{proof}

\begin{definition}
Teoremde $h$ birebir ise $V+W$ toplam\i, 
$V$ ve $W$'nin \textbf{i\c c direkt toplam\i d\i r}
ve $V\oplus W$ olarak yaz\i labilir.
\end{definition}

\begin{example}
  \"Ornek \numarada{ex:VW}
\begin{equation*}
V+W=V\oplus\lspan{(1,0,1,1)};
\end{equation*}
$F$'de $2=0$ ise $V+W=V\oplus W$.
\end{example}

\section{B\"ol\"umler}

\begin{definition}
  $n\in\N$ ve $a\in\Z$ ise
  \begin{align*}
    n\Z&=\{nx\colon x\in\Z\},&
    a+n\Z&=\{a+x\colon x\in n\Z\},
  \end{align*}
  ve
  \begin{equation*}
    \Zmod n=\{x+n\Z\colon x\in\Z\}
  \end{equation*}
  olsun.
  Ayr\i ca
  \begin{equation*}
    [a]=a+n\Z
  \end{equation*}
  olsun.
\end{definition}

\begin{theorem}
  $\Zmod n$ k\"umesinde toplama ve \c carpma
  \begin{align*}
    [x]+[y]&=[x+y],&
    [x]\cdot[y]&=[xy]
  \end{align*}
  ile tan\i mlanabilir.
\end{theorem}

\begin{proof}
  Tan\i mdan
  \begin{equation*}
    [a]=[b]\iff a-b\in n\Z.
  \end{equation*}
  \c Simdi
  \begin{align*}
    [a_1]&=[a_2],&[b_1]&=[b_2]
  \end{align*}
  olsun.  O zaman
  \begin{align*}
    a_1-a_2&\in n\Z,&b_1-b_2&\in n\Z.
  \end{align*}
  Sonu\c c olarak
  \begin{gather*}
    (a_1-a_2)+(b_1-b_2)\in n\Z,\\
    (a_1+b_1)-(a_2+b_2)\in n\Z,\\
    [a_1+b_1]=[a_2+b_2],
  \end{gather*}
ve  ayr\i ca
  \begin{gather*}
    (a_1-a_2)\cdot b_1+ a_2\cdot(b_1-b_2)\in n\Z,\\
    a_1b_1-a_2b_2\in n\Z,\\
      [a_1b_1]=[a_2b_2].\qedhere
  \end{gather*}
\end{proof}

\begin{definition}
  $\Zmod n$'de
	\begin{align*}
[x]^1&=[x],&[x]^{k+1}&=[x]^k\cdot[x]		
	\end{align*}
        olsun.
        \"Oyleyse
\begin{equation*}
  ([x],y)\mapsto[x]^y\colon(\Zmod n)\times\N\to\Zmod n.
\end{equation*}
\end{definition}

\begin{theorem}
  $\Zmod n$'de her $k$ sayma say\i s\i\ i\c cin
  \begin{equation*}
    [x]^k=[x^k].
  \end{equation*}
\end{theorem}

\begin{proof}
  T\"umevar\i m:
  \begin{compactenum}
  \item
    $[x]^1=[x]=[x^1]$.
  \item
    E\u ger $[x]^j=[x^j]$ ise, o zaman
    \begin{equation*}
      [x]^{j+1}
      =[x]^j\cdot[x]
      =[x^j]\cdot[x]
      =[x^j\cdot x]
      =[x^{j+1}].\qedhere
    \end{equation*}
  \end{compactenum}
\end{proof}

\begin{example}
  $\Zmod3$'de
  \begin{align*}
  [1]&=[4],&[2^1]&\neq[2^4].
  \end{align*}
  B\"oylece
  \begin{equation*}
    [x]^{[y]}=[x^y]
  \end{equation*}
  kural\i\ ile
  $([x],[y])\mapsto[x]^{[y]}$ i\c slemi \textbf{iyitan\i ml\i}
de\u gildir:
b\"oyle bir i\c slem yoktur.
\end{example}

Tekrar $U$, $V$, ve $W$,
$F$ cismi \"uzerinde vekt\"or uzay\i\ olsun.

\begin{definition}
  E\u ger $V\included U$ ve $\bm a\in U$ ise
  \begin{gather*}
    \bm a+V=\{\bm a+\bm v\colon v\in V\},\\
  U/V=\{\bm u+V\colon\bm u\in U\}
  \end{gather*}
  olsun.
\end{definition}

\begin{theorem}
  $V\included U$ ise
  \begin{align*}
    (\bm u_1+V)+(\bm u_2+V)&=(\bm u_1+\bm u_2)+V,\\
    t(\bm u_1+V)&=t\bm u_1+V
  \end{align*}
  kurallar\i yla $U/V$ b\"ol\"um\"u,
  iyitan\i ml\i\ bir vekt\"or uzay\i d\i r.
\end{theorem}

\begin{proof}
  Al\i\c st\i rma.
\end{proof}

\begin{example}
  \"Ornek \numarada{ex:VW}
\begin{equation*}
(V+W)/V\cong\lspan{(1,0,1,1)};
\end{equation*}
$F$'de $2=0$ ise $(V+W)/V\cong W$.
\end{example}

\begin{example}
  E\u ger
  \begin{equation*}
    U=\lspan{(1,1,1,0),(1,-1,1,0),(1,0,1,1),(1,0,1,-1)}
  \end{equation*}
  ise
  \begin{multline*}
    \begin{bmatrix}
      1& 1&1& 1&1&0&0&0\\
			1&-1&0& 0&0&1&0&0\\
			1& 1&1& 1&0&0&1&0\\
			0& 0&1&-1&0&0&0&1
    \end{bmatrix}
    \xrightarrow[-R_1+R_3]{-R_1+R_2}\\
    \begin{bmatrix}
      1& 1& 1& 1& 1&0&0&0\\
			0&-2&-1&-1&-1&1&0&0\\ 
			0& 0& 0& 0&-1&0&1&0\\
			0& 0& 1&-1& 0&0&0&1
    \end{bmatrix}
    \xrightarrow{R_3\leftrightarrow R_4}\\
    \begin{bmatrix}
      1& 1& 1& 1& 1&0&0&0\\
			0&-2&-1&-1&-1&1&0&0\\ 
			0& 0& 1&-1& 0&0&0&1\\
			0& 0& 0& 0&-1&0&1&0
    \end{bmatrix}    
  \end{multline*}
  oldu\u gundan
  \begin{equation*}
    F^4/U\cong\lspan{(1,0,0,0)}.
  \end{equation*}
\end{example}

\begin{theorem}
  E\u ger $ B\cap C=\emptyset$
  ve $B\cup C$ lineer ba\u g\i ms\i z ise
  \begin{equation*}
    \lspan{B\cup C}/\lspan B\cong\lspan C.
  \end{equation*}
\end{theorem}

\begin{proof}
$B=\{\bm b_1,\dots,\bm b_{\ell}\}$ ve
$C=\{\bm c_1,\dots,\bm c_m\}$
  olsun.
O zaman $\lspan{B\cup C}$ gergisinin her $\bm a$ eleman\i,
  tek bir \c sekilde,
  $u_i\in F$ ve $v_j\in F$ olmak \"uzere
  bir
  \begin{equation*}
    u_1\bm b_1+\dots+u_{\ell}\bm b_{\ell}+v_1\bm c_1+\dots+v_m\bm c_m
  \end{equation*}
  lineer bile\c simi olarak yaz\i labilir.
  Bu durumda
  \begin{equation*}
    h(\bm a)=v_1\bm c_1+\dots+v_m\bm c_m
  \end{equation*}
  e\c sitli\u gi ile $\lspan{B\cup C}$ gergisinden
  $\lspan C$ gergisine giden bir $h$ lineer d\"on\"u\c s\"um\"u
  tan\i mlanabilir.
  \c Simdi
  \begin{equation*}
    \bm a'=
    u_1'\bm b_1+\dots+u_{\ell}'\bm b_{\ell}+v_1'\bm c_1+\dots+v_m'\bm c_m
  \end{equation*}
  olsun.
  E\u ger
  \begin{equation}\label{eqn:a+B}
   \bm a'+\lspan B=\bm a+\lspan B 
  \end{equation}
  ise, o zaman $\bm a'-\bm a\in\lspan B$.
  Bundan dolay\i
  \begin{multline*}
    (u_1'-u_1)\bm b_1+\dots+(u_{\ell}'-u_{\ell})\bm b_{\ell}\\
    +(v_1'-v_1)\bm c_1+\dots+(v_m'-v_m)\bm c_m\in\lspan B.
  \end{multline*}
  Bu durumda $B\cup C$ lineer ba\u g\i ms\i z oldu\u gundan
  \begin{equation*}
    v_1'-v_1=\dots=v_m'-v_m=0,
  \end{equation*}
  dolay\i s\i yla $v_1'=v_1$, \dots, $v_m'=v_m$, ve sonu\c c olarak
  \begin{equation}\label{eqn:h(a)}
    h(\bm a')=h(\bm a).
  \end{equation}
  Bu \c sekilde $\lspan{B\cup C}/\lspan B$ b\"ol\"um\"unden
  $\lspan C$ gergisine giden
  \begin{equation*}
    \tilde h(\bm u+\lspan B)=h(\bm u)
  \end{equation*}
  e\c sitli\u gi ile iyitan\i mlanm\i\c s lineer d\"on\"u\c s\"um\"u
  vard\i r.
  Benzer \c sekilde e\c sitlik \eqref{eqn:h(a)} do\u gru ise
   e\c sitlik \eqref{eqn:a+B} de do\u grudur,
  dolay\i s\i yla $\tilde h$, birebirdir.
  Son olarak $\tilde h$ d\"on\"u\c s\"um\"un\"un
  $\lspan C$ gergisini \"orten oldu\u gu apa\c c\i kt\i r.
\end{proof}


\end{document}