Answer to Question #152541 in Linear Algebra for Sourav Mondal

Question

a) Let T = R2–>R3 be a linear
transformation given by :
T(x1,x2) =(x1 + X2,x1, x1-x2 )
Show that {(1,1), (1, 0)} and {(1, 1, 0), (1, 0, 1), (0, 1, 1)) are bases of R2 and R3
respectively.Find the matrix of T with respect to these ordered basis.

Answer 1

$Solution: (a)~Let ~S={(1,1),(1,0)}=v_{1},v_{2}~~~where ~v_{1}=(1,1),~v_{2}=(1,0)$

$we~have,~S ~ is ~basis ~for~R^2 (i)S ~is~linearly~independent.(ii)S ~is~spanning~set~of~R^2$ $(i)To ~check:S~is~linearly~independent.$

$Consider,~ linear~ combination ~k_{1}v_{1}+k_{2}v_{2}=0~where~k_{1},k_{2}\in R^2$

$\ \therefore k_{1}(1,1)+k_{2}(1,0)=0$

$\Rightarrow (k_{1},k_{1})+(k_{2},0)=0$

$\Rightarrow ~(k_{1}+k_{2},k_{1})=0$

$\therefore ~we~ get~ two~ homogeneous ~equations ~as$

$k_{1}+k_{2}=0 ........................(1)\\ and~ k_{1}=0.......................... (2)\\ put~ k_{1}=0~in~equation~(1),we~get~ k_{2}=0\\Hence~we~get~ k_{1}= k_{2}=0 \\ \therefore ~S ~is~ linearly~ independent Set.$

$(ii) To~ check~:~S~ is~spanning~set~of~R^2.~\\ ~~~~~Let~V=(a,b,c)\in R^2~be~any~element~and~v_{1}=(1,1)~and ~v_{1}=(1,0).\\~~~~~Let~if~possible~V=k_{1}v_{1}+k_{2}v_{2}~where~k_{1},k_{2}\in R^2 ~\\ ~\ \therefore (a,b)=k_{1}(1,1)+k_{2}(1,0)~\\(a,b)=(k_{1},k_{1})+(k_{2},0)~\\(a,b)=(k_{1}+k_{2},k_{1})~\\k_{1}+k_{2}=a~and~k_{1}=b$ $(ii)~ To~check~:~S~spanning~set~of~R^2.\\~~~~~~~Let~V=(a,b)\in R^2~be~any~elenent~and~v_{1}=(1,1),v_{2}=(1,0).\\~~~~~~~Let~if~possible~V=k_{1}v_{1}+k_{2}v_{2}~where~k_{1},k_{2}\in R^2\\ \therefore (a,b)=k_{1}v_{1}+k_{2}v_{2}=k_{1}(1,1)+k_{2}(1,0)=(k_{1}+k_{2},k_{1})\\k_{1}+k_{2}=a~and~k_{1}=b$

$Here, ~~~Number ~of~ equations~=~Number~ of~ variables\\ \therefore A=\begin{bmatrix} 1 & 1\\1 & 0 \end{bmatrix}\\ ~|A|=-1\neq0\\ ~\therefore A ~is~ non-singular.\\ ~A ~is~ invertible.\\ \therefore System~ is~ consistent.\\ \therefore Linear~ system~ AX=B~has~solution.\\ \therefore Every~~element~of~R^2~can~be~expressed~as~a~linear~combination~of~v_{1}~and~v_{2}.\\ S~spans~R^2.\\ \therefore~ All~ two~ conditions~ hold.\\ \therefore S~is~basis~for~R^2$

$(b)~Let ~P={(1,1,0),(1,0,1),(0,1,1)}=u_{1},u_{2},u_{3}~~~where ~u_{1}=(1,1,1),~u_{2}=(1,0,1),u_{3}=(0,1,1)$

$we~have,P ~is ~basis ~for~R^3~if (i)P ~is~linearly~independent.(ii)P ~is~spanning~set~of~R^3$

$(i)To ~check:P~is~linearly~independent.$

$Consider,~ linear~ combination ~k_{1}v_{1}+k_{2}v_{2}+k_{3}v_{3}=0~where~k_{1},k_{2},k_{3}\in R^3$

$\ \therefore k_{1}(1,1,0)+k_{2}(1,0,1)+k_{3}(0,1,1)=0$

$\Rightarrow (k_{1},k_{1},0,)+(k_{2},0,k_{2})+(0,k_{3},k_{3})=0 \\ \Rightarrow k_{1}+k_{2}=0 \\~~~~~k_{1}+k_{3}=0 \\~~~~~k_{2}+k_{3}=0 \\Here, ~~~Number ~of~ equations~=~Number~ of~ variables\\ \therefore A=\begin{bmatrix}1 & 1&0\\1 & 0 &1\\0&1&1\end{bmatrix}\\ ~|A|=-2\neq0\\ ~\therefore A ~is~ non-singular.\\ ~A ~is~ invertible.\\ \therefore System~ is~ consistent.\\ \therefore Linear~ system~ AX=0~has~unique~trivial~solution.\\ ~~k_{1}=k_{2}=k_{3}=0\\ \therefore ~P~ is ~linearly~ independent ~set$

$(ii)~ To~check~:~P~is~spanning~set~of~R^3.\\~~~~~~~Let~V=(a,b,c)\in R^3~be~any~elenent~and~u_{1}=(1,1,0),u_{2}=(1,0,1),u_{3}=(0,1,1).\\~~~~~~~Let~if~possible~V=k_{1}v_{1}+k_{2}v_{2}+k_{3}v_{3}~where~k_{1},k_{2},k_{3}\in R^3\\ \therefore (a,b,c)=k_{1}v_{1}+k_{2}v_{2}+k_{3}v_{3}~=k_{1}(1,1,0)+k_{2}(1,0,1)+k_{3}(0,1,1)\\ \therefore k_{1}+k_{2}=a~\\~~~~ k_{1}+k_{3}=b~~and\\~~~~k_{2}+k_{3}=c~\\which ~is~non-homogenous~system.$

$\\Here, ~~~Number ~of~ equations~=~Number~ of~ variables\\ \therefore~Let~ A=\begin{bmatrix}1 & 1&0\\1 & 0 &1\\0&1&1\end{bmatrix}\\ ~|A|=-2\neq0\\ ~\therefore A ~is~ non-singular.\\ ~A ~is~ invertible.\\ \therefore System~ is~ consistent.\\ \therefore Linear~ system~ AX=B~has~solution.\\ \therefore Every~element~in~R^3~can~be~expressed~as~a~linear~combination~of~u_{1},u_{2},u_{3}\\ \therefore ~P~spans~R^3\\ \therefore All~two~conditions~satisfied.\\ \\ \therefore P~is~basis~for~R^3$

Now , To find the matrix of T w.r.t. these ordered basis.

$Let ~B={u_1,u_2}=\begin{bmatrix}1 \\1 \end{bmatrix},\begin{bmatrix}1 \\0 \end{bmatrix} ~and ~B'=v_1,v_2,v_3=\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix},\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix},\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}$

$T:R^2 \rightarrow R^3~ defined~ by ~T(\begin{bmatrix}x_1 \\x_2 \end{bmatrix}) =(\begin{bmatrix}x_1+x_2 \\x_1\\x_1-x_2 \end{bmatrix})\\ T\begin{bmatrix}1 \\1 \end{bmatrix} =\begin{bmatrix}1+1 \\1\\1-1 \end{bmatrix}=\begin{bmatrix}2 \\1\\0 \end{bmatrix}\\T\begin{bmatrix}1 \\0 \end{bmatrix} =\begin{bmatrix}1+0 \\1\\1-0 \end{bmatrix}=\begin{bmatrix}1 \\1\\1 \end{bmatrix}$

To find the matrix of T w.r.t. bases B and B' construct matrix $[v_1,v_2,v_3|T(u_1),T(u_2)]$ and transformed to reduced row echelon form as

$\begin{bmatrix}1&1&0 ~| ~2&1\\1 & 0 & 1|~1&1\\0&1&1|~0&~1\end{bmatrix}\\Applying~R_2-R_1\rightarrow~R_2\\\begin{bmatrix}1&1&0 |~~~~~2&1\\0 & -1 &1|-1&0\\0&1&1|0&1\end{bmatrix} \\Applying~-R_2\rightarrow~R_2\\\begin{bmatrix}1&1&~~~0| 2&1\\0&1& -1|1&0\\0&1&~~~~1|~0&~1\end{bmatrix}$

$\\Applying~R_3-R_2\rightarrow~R_3\\\begin{bmatrix}1&1&0| 2&1\\0&1& -1|1&0\\~0&~0&~2|-1&1\end{bmatrix}$

$\\Applying~\frac{1}{2}R_3\rightarrow~R_3\\\begin{bmatrix}1&1&~~~0| 2&1\\0&1& -1|1&0\\0&0&~~~~1|~-\frac{1}{2}&~\frac{1}{2}\end{bmatrix}$

$\\Applying~R_2+R_3\rightarrow~R_2\\\begin{bmatrix}1&1&~~~0| 2&1\\0&1& 0|\frac{1}{2}&\frac{1}{2}\\0&0&~~~~1|~-\frac{1}{2}&~\frac{1}{2}\end{bmatrix}$

$\\Applying~R_1-R_2\rightarrow~R_1\\\begin{bmatrix}1&0&~~~0| \frac{3}{2}&\frac{1}{2}\\0&1& 0|\frac{1}{2}&\frac{1}{2}\\0&0&~~~~1|~-\frac{1}{2}&~\frac{1}{2}\end{bmatrix}=[I_3|A]$

$\therefore$ The matrix of T w.r.t. Bases B and B' is

$A=[T]_{B,B'}=\begin{bmatrix}\frac{3}{2}&\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}\\-\frac{1}{2}&\frac{1}{2}\end{bmatrix}$