\documentclass{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amssymb} \usepackage{amsmath} \setcounter{MaxMatrixCols}{10} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2570} %TCIDATA{} %TCIDATA{Created=Thursday, February 21, 2002 14:53:39} %TCIDATA{LastRevised=Monday, October 03, 2005 11:13:38} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=On line bluem.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{AllPages= %F=36,\PARA{038
Scientific Notebook: On Line Mathematics\dotfill \thepage} %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \section{Matice} \subsection{D\^{o}kaz vety} Nech matica $\mathbf{A}$ m\'{a} riadky $\mathbf{a_{1}},\mathbf{a_{2}},\ldots ,\mathbf{a_{m}}$ a nech matica $\mathbf{B}$ vznikla z matice $\mathbf{A}$ pomocou ERO $\mathbf{a_{2}}:=\mathbf{a_{2}}+\alpha \mathbf{a_{1}}$. Potom riadkami matice $\mathbf{B}$ s\'{u} $\mathbf{b_{1}}=\mathbf{a_{1}},\ \mathbf{% b_{2}}=\mathbf{a_{2}}+\alpha \mathbf{a_{1}},\ \mathbf{b_{3}}=\mathbf{a_{3}}% ,\ldots ,$ $\mathbf{b_{m}}=\mathbf{a_{m}}$.\newline \begin{enumerate} \item[(1)] Nech riadky matice $\mathbf{A}$ s\'{u} line\'{a}rne nez\'{a}visl% \'{e}.\newline Zistime, pre ak\'{e} \v{c}\'{\i}sla $\alpha _{1},\ \ldots ,\ \alpha _{m}$, plat\'{\i} \begin{equation*} \alpha _{1}\mathbf{a_{1}}+\alpha _{2}(\mathbf{a_{2}}+\alpha \mathbf{a_{1}}% )+\alpha _{3}\mathbf{a_{3}}+\cdots +\alpha _{m}\mathbf{a_{m}}=\mathbf{0} \end{equation*}% T\'{u}to rovnos\v{t} m\^{o}\v{z}eme upravi\v{t} na \begin{equation*} (\alpha _{1}+\alpha _{2}\alpha )\mathbf{a_{1}}+\alpha _{2}\mathbf{a_{2}}% +\alpha _{3}\mathbf{a_{3}}+\cdots +\alpha _{m}\mathbf{a_{m}}=\mathbf{0} \end{equation*}% Z line\'{a}rnej nez\'{a}vislosti vektorov $\mathbf{a_{1}},\mathbf{a_{2}}% ,\ldots ,\mathbf{a_{m}}$ vypl\'{y}va \begin{equation*} \alpha _{1}+\alpha _{2}\alpha =0,\ \alpha _{2}=0,\ \alpha _{3}=0,\ \ldots ,\ \alpha _{m}=0 \end{equation*}% odkia\v{l} \begin{equation*} \alpha _{1}=0,\ \alpha _{2}=0,\ \alpha _{3}=0,\ \ldots ,\ \alpha _{m}=0 \end{equation*}% To v\v{s}ak znamen\'{a}, \v{z}e riadky matice $\mathbf{B}$ s\'{u} line\'{a}% rne nez\'{a}visl\'{e}. \item[(2)] Nech teraz riadky matice $\mathbf{A}$ s\'{u} line\'{a}rne z\'{a}% visl\'{e}. Existuj\'{u} teda \v{c}\'{\i}sla $\alpha _{1},\ \ldots ,\ \alpha _{m}$, z ktor\'{y}ch aspo\v{n} jedno je r\^{o}zne od nuly tak, \v{z}e \begin{equation*} \alpha _{1}\mathbf{a_{1}}+\alpha _{2}\mathbf{a_{2}}+\alpha _{3}\mathbf{a_{3}}% +\cdots +\alpha _{m}\mathbf{a_{m}}=\mathbf{0} \end{equation*}% T\'{a}to rovnos\v{t} sa d\'{a} upravi\v{t} na \begin{equation*} (\alpha _{1}-\alpha _{2}\alpha )\mathbf{a_{1}}+\alpha _{2}(\mathbf{a_{2}}% +\alpha \mathbf{a_{1}})+\alpha _{3}\mathbf{a_{3}}+\cdots +\alpha _{m}\mathbf{% a_{m}}=\mathbf{0} \end{equation*}% a pritom aspo\v{n} jedno z \v{c}\'{\i}sel $\alpha _{1}-\alpha _{2}\alpha ,\ \alpha _{2},\ \ldots ,\ \alpha _{m}$ je r\^{o}zne od nuly. To v\v{s}ak znamen% \'{a}, \v{z}e riadky matice $\mathbf{B}$ s\'{u} line\'{a}rne z\'{a}visl\'{e}. \item[(3)] Nech maxim\'{a}lny po\v{c}et line\'{a}rne nez\'{a}visl\'{y}ch riadkov v matici $\mathbf{A}$ je $r$ a v matici $\mathbf{B}$ je $s$. Nech v skupine $r$ line\'{a}rne nez\'{a}visl\'{y}ch riadkov matice $\mathbf{A}$ nie je druh\'{y} riadok. Potom tieto riadky s\'{u} aj v matici $\mathbf{B}$, a preto $r\leq s$. Ak sa v tejto skupine $r$ line\'{a}rne nez\'{a}visl\'{y}ch riadkov matice $\mathbf{A}$ druh\'{y} riadok nach\'{a}dza, sta\v{c}\'{\i} ho nahradi\v{t} riadkom $\mathbf{a_{2}}+\alpha \mathbf{a_{1}}$ a dostaneme $r$ line\'{a}rne nez\'{a}visl\'{y}ch riadkov matice $\mathbf{B}$. Tak\v{z}e op% \"{a}\v{t} $r\leq s$. Na druhej strane matica $\mathbf{A}$ vznikne z matice $% \mathbf{B}$ pomocou ERO $\mathbf{b_{2}}:=\mathbf{b_{2}}-\alpha \mathbf{b_{1}} $, a tak na z\'{a}klade predo\v{s}l\'{y}ch \'{u}vah $s\leq r$. Potom v\v{s}% ak mus\'{\i} plati\v{t} $r=s$. \end{enumerate} Nech teraz matica $\mathbf{B}$ vznikla z matice $\mathbf{A}$ pomocou ESO $% \mathbf{s_{2}}:=\mathbf{s_{2}}+\alpha \mathbf{s_{1}}$, kde $\mathbf{s_{1}},% \mathbf{s_{2}}$ s\'{u} prv\'{y} a druh\'{y} st\'{l}pec matice $\mathbf{A}$. Pre line\'{a}rnu z\'{a}vislos\v{t} \v{c}i nez\'{a}vislos\v{t} riadkov matice $\mathbf{A}$ resp. $\mathbf{B}$ je rozhoduj\'{u}ce \v{c}i rovnos\v{t} \begin{equation*} \alpha _{1}\mathbf{a_{1}}+\alpha _{2}\mathbf{a_{2}}+\alpha _{3}\mathbf{a_{3}}% +\cdots +\alpha _{m}\mathbf{a_{m}}=\mathbf{0} \end{equation*}% resp. \begin{equation*} \alpha _{1}\mathbf{b_{1}}+\alpha _{2}\mathbf{b_{2}}+\alpha _{3}\mathbf{b_{3}}% +\cdots +\alpha _{m}\mathbf{b_{m}}=\mathbf{0} \end{equation*}% je splnen\'{a} aj pre nenulov\'{e} \v{c}i len pre nulov\'{e} \v{c}\'{\i}sla $% \alpha _{1},\ \ldots ,\ \alpha _{m}$. Rozp\'{\i}\v{s}me tieto dve rovnosti po zlo\v{z}k\'{a}ch. Pritom si treba uvedomi\v{t}, \v{z}e ak $\mathbf{a}% _{j}=(a_{j1},a_{j2},a_{j3},\ldots ,a_{jn})$, tak $\mathbf{b}% _{j}=(a_{j1},a_{j2}+\alpha a_{j1},a_{j3},\ldots ,a_{jn})$. Z\'{\i}skame tak dve s\'{u}stavy line\'{a}rnych rovn\'{\i}c o nezn\'{a}mych $\alpha _{1},\ \ldots ,\ \alpha _{m}$. \begin{equation*} \begin{tabular}{l} $a_{11}\alpha _{1}+a_{21}\alpha _{2}+a_{31}\alpha _{3}+\cdots +a_{m1}\alpha _{m}=0$ \\ $a_{12}\alpha _{1}+a_{22}\alpha _{2}+a_{32}\alpha _{3}+\cdots +a_{m2}\alpha _{m}=0$ \\ $a_{13}\alpha _{1}+a_{23}\alpha _{2}+a_{33}\alpha _{3}+\cdots +a_{m3}\alpha _{m}=0$ \\ ${\Large ....................................}$ \\ $a_{1n}\alpha _{1}+a_{2n}\alpha _{2}+a_{3n}\alpha _{3}+\cdots +a_{mn}\alpha _{m}=0$% \end{tabular}% \end{equation*}% resp. \begin{equation*} \begin{array}{cc@{\ }cc@{\ }cc@{\ + \cdots\ }cc@{\ =\ }c} a_{11}\alpha _{1} & + & a_{21}\alpha _{2} & + & a_{31}\alpha _{3} & + & a_{m1}\alpha _{m} & = & 0 \\ (a_{12}+\alpha a_{11})\alpha _{1} & + & (a_{22}+\alpha a_{21})\alpha _{2} & + & (a_{32}+\alpha a_{31})\alpha _{3} & + & (a_{m2}+\alpha a_{11})\alpha _{m} & = & 0 \\ a_{13}\alpha _{1} & + & a_{23}\alpha _{2} & + & a_{33}\alpha _{3} & + & a_{m3}\alpha _{m} & = & 0 \\ \multicolumn{9}{c}{\Large % .......................................................................} \\ a_{1n}\alpha _{1} & + & a_{2n}\alpha _{2} & + & +\ a_{3n}\alpha _{3} & + & a_{mn}\alpha _{m} & = & 0% \end{array}% \end{equation*}% Jasne vidie\v{t}, \v{z}e tieto s\'{u}stavy s\'{u} ekvivaletn\'{e}, ke\v{d}% \v{z}e druh\'{a} s\'{u}stava vznikne z prvej pripo\v{c}\'{\i}tan\'{\i}m $% \alpha $-n\'{a}sobku prvej rovnice k druhej. To ale znamen\'{a}, \v{z}e obidve s\'{u}stavy maj\'{u} rovnak\'{e} rie\v{s}enia. Riadky matice $\mathbf{% A}$ s\'{u} teda line\'{a}rne z\'{a}visl\'{e} pr\'{a}ve vtedy, ke\v{d} s\'{u} line\'{a}rne z\'{a}visl\'{e} riadky matica $\mathbf{B}$. Keby sme v tejto \v{c}asti d\^{o}kazu skupinu v\v{s}etk\'{y}ch riadkov matice $\mathbf{A}$ nahradili skupinou, ozn\v{c}me ju $M_{A}$, jej niektor\'{y}ch $r$ riadkov, $% r\leq m$, a skupinu v\v{s}etk\'{y}ch riadkov matice $\mathbf{B}$ skupinou $% M_{B}$ jej riadkov, pri\v{c}om $M_{B}=M_{A}$, ak $M_{A}$ neobsahuje riadok $% A_{2}$, v opa\v{c}nom pr\'{\i}pade $M_{B}$ vnikne z $M_{A}$ nahraden\'{\i}m riadku $A_{2}$ riadkom $A_{2}+\alpha A_{1}$, tak pr\'{\i}deme k z\'{a}veru, \v{z}e $M_{A}$, $M_{B}$ s\'{u} s\'{u}\v{c}asne line\'{a}rne z\'{a}visl\'{e} alebo s\'{u}\v{c}asne line\'{a}rne nez\'{a}visl\'{e}. Z toho potom vypl\'{y}% va, \v{z}e maxim\'{a}lny po\v{c}et line\'{a}rne nez\'{a}visl\'{y}ch riadkov v oboch maticiach je rovnak\'{y}. D\^{o}kaz pre in\'{e} ERO \v{c}i ESO je podobn\'{y}. $\blacksquare $ \begin{center} \begin{tabular}{|c|} \hline $\,$\textbf{% %TCIMACRO{\hyperref{Sp\"{a}\v{t}}{}{}{A31.tex#1}}% %BeginExpansion \msihyperref{Sp\"{a}\v{t}}{}{}{A31.tex#1}% %EndExpansion } \\ \hline \end{tabular} \end{center} \end{document}