Question

Please explain with example: follow the comment:

If A is an n×n matrix with the property that Ax = 0
for all x ∈ Rn, show that A = O. Hint: Let x = ej
for j = 1, . . . , n.

Ax = 0 for all XEO" Let A-(a,a,. Let e.-| | | ← jth element Comments (1) Anonymous why there is 1 in side ej Submit Step 2 of 2 Hence Ae,-a Since AX-0 for all Xe0 Ae, = 0 (Since e, e。") a0 for 1sjS

Why ej=(1,....n) then it comes out it is a column vector and all zero except 1 inside, i don't get it

Answer 1

$For~n=2~then~{\bf A}=\begin{pmatrix} a_{11}& a_{12}\\ a_{21} & a_{22} \end{pmatrix}\\ {\bf x}\in \mathbb{R}^2~means~{\bf x}=\begin{pmatrix} x_1\\ x_2 \end{pmatrix}~where~x_1,x_2\in \mathbb{R}.\\ where~ \mathbb{R}^2=\left\{ \begin{pmatrix} x_1\\ x_2 \end{pmatrix}:~x_1,x_2\in \mathbb{R}\right \}.\\ So,~\begin{pmatrix} 1\\ 0 \end{pmatrix}~and~\begin{pmatrix} 0\\ 1 \end{pmatrix}\in \mathbb{R}^2\\ Now,~{\bf A}{\bf x}={\bf 0}~\forall~{\bf x}\in \mathbb{R}^2$

$Choose~{\bf x}=\begin{pmatrix} 1\\ 0 \end{pmatrix}~and~\begin{pmatrix} 0\\ 1 \end{pmatrix} ~then\\ {\bf A}\begin{pmatrix} 1\\ 0 \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix}....(1)~\\ and\\ ~{\bf A}\begin{pmatrix} 0\\ 1 \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix}.....(2)\\ From~(1)\Rightarrow ~\begin{pmatrix} a_{11}& a_{12}\\ a_{21} & a_{22} \end{pmatrix}\begin{pmatrix} 1\\ 0 \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix}~or,\begin{pmatrix} a_{11}\\ a_{21} \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix}\\ \Leftrightarrow a_{11}=0,~a_{21}=0.....(3)\\$

$From~(2)\Rightarrow ~\begin{pmatrix} a_{11}& a_{12}\\ a_{21} & a_{22} \end{pmatrix}\begin{pmatrix} 0\\ 1 \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix}~or,\begin{pmatrix} a_{12}\\ a_{22} \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix}\\ \Leftrightarrow a_{12}=0,~a_{22}=0.......(4)\\$

$Hence~from~(3)~and~(4),~{\bf A}={\bf 0}.\\ We~generally~denote~\begin{pmatrix} 1\\ 0 \end{pmatrix}~by~{\bf e}_1~and~\begin{pmatrix} 0\\ 1 \end{pmatrix}~by~{\bf e}_2.~These~are~unit~vectors.\\ We~denote~the~first~colunm~of~{\bf A}~i.e.~\begin{pmatrix} a_{11}\\ a_{21} \end{pmatrix}~by~{\bf a}_1~and~second~column~i.e.~\begin{pmatrix} a_{12}\\ a_{22} \end{pmatrix}~by~{\bf a}_2\\ So~from~(3)~and~(4),~{\bf a}_1={\bf 0},~{\bf a}_2={\bf 0}.$

$For~n=3,~then\\ \mathbb{R}^3=\left\{ \begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}:x_1,x_2,x_3\in \mathbb{R}\right\}.\\ {\bf A}^{3\times 3}=\begin{pmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\ a_{31}& a_{32} &a_{33} \end{pmatrix}\\ {\bf A}{\bf x}={\bf 0}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}~\forall~{\bf x}\in \mathbb{R}^3~(given)\\ Then~take~{\bf x}={\bf e}_1, ~{\bf e}_2~and~{\bf e}_3~respectively~as~n=2~where$

$Since~{\bf A}{\bf x}={\bf 0}~\forall~{\bf x}\in \mathbb{R}^3~so\\ {\bf A}{\bf e}_1={\bf 0},~{\bf A}{\bf e}_2={\bf 0},~{\bf A}{\bf e}_3={\bf 0}~since~{\bf e}_1,{\bf e}_2,{\bf e}_3\in \mathbb{R}^3.$

Again

$Now,~{\bf A}^{n\times n}=\left({\bf a}_1,{\bf a}_2,...,{\bf a}_j,....,{\bf a}_n \right )\\ where,~{\bf a}_{j}=\begin{pmatrix} a_{1j}\\ a_{2j}\\ \vdots\\ a_{ij}\\ \vdots\\ a_{nj} \end{pmatrix},~j=1(1)n\\ Now,~{\bf A}{\bf x}={\bf 0}~\forall~{\bf x}\in \mathbb{R}^n\\ then~{\bf A}{\bf e}_j={\bf 0},~j=1(1)n\\ where,$

${\bf e}_1=\begin{pmatrix} 1\\ 0\\ \vdots\\ 0\\ \vdots\\ 0 \end{pmatrix},~{\bf e}_2=\begin{pmatrix} 0\\ 1\\ 0\\ \vdots\\ 0\\ \vdots\\ 0 \end{pmatrix},\ldots,{\bf e}_j=\begin{pmatrix} 0\\ 0\\ \vdots\\ 0\\ 1\\ 0\\ \vdots\\ 0 \end{pmatrix},\ldots,{\bf e}_n=\begin{pmatrix} 0\\ 0\\ \vdots\\ 0\\ \vdots\\ 0\\ 1 \end{pmatrix}\\ Note~that~"1"~appears~in~{\bf e}_j~at~jth~position.\\ then~{\bf a}_j={\bf 0}~\forall~j=1(1)n.\\ Hence~{\bf A}={\bf 0}$

Alternative proof:

$Let~{\bf A}~is~a~matrix~of~order~n\times n~(i.e.~{\bf A}~has~n~rows~and~n~columns)~then~we~write~{\bf A}~as\\ {\bf A}^{n\times n}=\begin{pmatrix} a_{11} & a_{12} & \ldots & a_{1j} &\ldots&a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2j} &\ldots&a_{2n} \\ \vdots& \vdots & \vdots & \vdots &\vdots&\vdots \\ a_{i1} & a_{i2} & \ldots & a_{ij} &\ldots&a_{in} \\ \vdots& \vdots & \vdots & \vdots &\vdots&\vdots \\ a_{n1} & a_{n2} & \ldots & a_{nj} &\ldots&a_{nn} \end{pmatrix}\\ Given,\\ {\bf A}^{n\times n}{\bf x}^{n\times 1}={\bf 0}^{n\times 1}~\forall~{\bf x}^{n\times 1}\in \mathbb{R}^n\\ where,~{\bf x}^{n\times 1}=\begin{pmatrix} x_1\\ x_2\\ \vdots\\ x_i\\ \vdots\\ x_n \end{pmatrix}\\ =column~vector~which~has~n~components~(since~{\bf x}\in \mathbb{R}^n),\\ ~{\bf 0 }^{n\times 1}=\begin{pmatrix} 0\\ 0\\ \vdots\\ 0 \end{pmatrix}\\$

$Let~{\bf A}~is~a~matrix~of~order~n\times n~(i.e.~{\bf A}~has~n~rows~and~n~columns)~then~we~write~{\bf A}~as\\ {\bf A}{\bf x}=\begin{pmatrix} a_{11}x_1+ a_{12}x_2 + \ldots + a_{1j} x_j+\ldots+a_{1n}x_n \\ a_{21}x_1+ a_{22}x_2+ \ldots + a_{2j}x_j +\ldots+a_{2n}x_n \\ \vdots \\ a_{i1}x_1+ a_{i2}x_2+ \ldots +a_{ij}x_j +\ldots+a_{in}x_n \\ \vdots \\ a_{n1}x_1+ a_{n2} x_2+ \ldots + a_{nj} x_j+\ldots+a_{nn} x_n \end{pmatrix}={\bf 0}=\begin{pmatrix} 0\\ 0\\ \vdots\\ 0\\ \vdots\\ 0 \end{pmatrix}\\\\ \Rightarrow a_{i1}x_1+ a_{i2}x_2+ \ldots +a_{ij}x_j +\ldots+a_{in}x_n=0~\forall~i=1,2,...,n........(1)\\ Since~{\bf A}{\bf x}={\bf 0}~\forall~{\bf x}\in\mathbb{R}^n\\ Now~we~choose~x_1=1,~x_2=x_3=\ldots=x_n=0~then~\begin{pmatrix} 1\\ 0\\ \vdots\\ 0\\ \vdots\\ 0 \end{pmatrix}\in \mathbb{R}^n\\ then~from~(1)\Rightarrow a_{i1}\times 1+ a_{i2}\times 0+ \ldots +a_{ij}\times 0 +\ldots+a_{in}\times 0=0\\ or,~a_{i1}=0~\forall~i=1(1)n~i.e.~a_{11}=a_{21}=\ldots=a_{n1}=0....(2)\\$

$Now~we~choose~x_1=0,~x_2=1,~=x_3=\ldots=x_n=0~then~\begin{pmatrix} 0\\ 1\\ 0\\ \vdots\\ 0\\ \vdots\\ 0 \end{pmatrix}\in \mathbb{R}^n\\ then~from~(1)\Rightarrow a_{i1}\times 0+ a_{i2}\times 1+a_{i3}\times 0+ \ldots +a_{ij}\times 0 +\ldots+a_{in}\times 0=0\\ or,~a_{i2}=0~\forall~i=1(1)n~i.e.~a_{12}=a_{22}=\ldots=a_{n2}=0....(3)\\ Proceeding~in~this~way,~we~can~show~that~for~the~choice\\ x_1=\ldots=x_{j-1}=0,x_j=1,x_{j+1}=\ldots=x_n=0\\ a_{ij}=0~\forall~i=1(1)n~i.e.~i.e.~a_{1j}=a_{2j}=\ldots=a_{nj}=0.....(4)\\ and~lastly~for~the~choice\\ x_1=\ldots=x_{n-1}=0,~x_n=1\\ a_{in}=0~\forall~i=1(1)v~i.e.~a_{1n}=a_{2n}=\ldots=a_{nn}=0......(5)\\ Combining~(2)-(5),~we~get\\ {\bf A}={\bf 0}=matrix~of~order~n\times n~whose~all~elements~are~zero.$