Find matrix of linear transformation

Question

A linear transformation $T: \Bbb R^2\to \Bbb R^2$ is given by

$T(i) = i+j; T(j) = 2i -j$.

With respect to the basis $\{i,j\}$, The matrix of $T = \left(\begin{array}{cc}1 & 2 \\1 & -1\end{array}\right).$

What is the matrix of $T$ if the basis $\{i,j\}$ is replaced by $e_1 = (i-j)$
and $e_2=3i+j$?

The solution is: $\left(\begin{array}{cc}-7/4 & -1/4 \\1/4 & 7/4\end{array}\right)$.

I am not able to come up with this solution. Here's my attempt:

$T(e_1) = T(i) - T(j) = i+j -(2i -j) = -i + 2j$

$T(e_2) = 3T(i) + T(j) = 3(i+j) +2i-j = 5i+2j$.

Thus, $T$ would be $\left(\begin{array}{cc}-1 & 5 \\2 & 2\end{array}\right)$.

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Second Question:

If you let $T = \left(\begin{array}{cc}1 & 2 \\1 & -1\end{array}\right).$ Then what is $T$ relative to $e_1,e_2$?

Answer 1

If $T(i)=(1,1)$ and $T(j)=(2,-1)$ and $e_1=i-j=(1,-1)$ and $e_2=3i+j=(3,1)$, then $$T(e_1)=T(i-j)=T(i)-T(j)=(-1,2)=a_1e_1+b_1e_2 \,\text{(say)}$$ and $$T(e_2)=T(3i+j)=3(1,1)+(2,-1)=(5,2)=a_2e_1+b_2e_2\,\text{(say)}.$$ Thus the matrix of $T$ w.r.t. the new basis $\{e_1,e_2\}$ is $\begin{pmatrix} a_1 & a_2\\ b_1 & b_2 \end{pmatrix}$ and you need to find the values of $a_1,a_2,b_1$ and $b_2$. The above systems of equations reduces to $$a_1+3b_1=-1\\ -a_1+b_1=2$$ and

$$a_2+3b_2=5\\ -a_2+b_2=2.$$ Solve these equations to obtain $a_1=-\frac{7}{4},b_1=\frac{1}{4},a_2=-\frac{1}{4}$ and $b_2=\frac{7}{4}$.

Answer 2

$e_1+e_2=4i$;$i=\frac{e_1+e_2}{4} = P(i |j)^T$;$e_2-3e_1=4j$;$j=\frac{e_2-3e_1}{4} = P(i| j)^T$.

Now compute the Matrix $P^{-1}$ and your $T$ in the new Basis is given by $P^{-1}TP$.

Answer 3

The matrix of the transformation of basis is: $$ P= \begin{bmatrix} 1&3\\ -1&1 \end{bmatrix} $$ and $$ P^{-1}= \dfrac{1}{4} \begin{bmatrix} 1&-3\\ 1&1 \end{bmatrix} $$ Your transformed matrix is: $$ P^{-1}TP= \dfrac{1}{4} \begin{bmatrix} 1&-3\\ 1&1 \end{bmatrix} \begin{bmatrix} 1&2\\ 1&-1 \end{bmatrix} \begin{bmatrix} 1&3\\ -1&1 \end{bmatrix}= \dfrac{1}{4} \begin{bmatrix} -7&-1\\ 1&7 \end{bmatrix} $$

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Added after comments.

The figure illustrate how operate the given transformation in the two basis.

Indices $\{i,j\}$ refer to the canonical basis, indices $\{1,2\}$ to the new basis $\mathbf{e_1},\mathbf{e_2}$.

In the figure we have the vector $\mathbf{v}=\mathbf{v_{i,j}}=[1,2]^T$ (in canonical basis), that is transformed in $\mathbf{v'}=\mathbf{v'_{i,j}}=[5,-1]^T$.

In the new basis we have: $$ \mathbf{v_{1,2}}= P^{-1}\mathbf{v_{i,j}}= \dfrac{1}{4}\left[ \begin{array}{cccc} -5\\ 3 \end {array} \right] $$ and, for the tranformed vector: $$ \mathbf{v'_{1,2}}= P^{-1}\mathbf{v'_{ij}}= \left[ \begin{array}{cccc} 2\\ 1 \end {array} \right]= P^{-1}T_{i,j}\mathbf{v_{i,j}}=P^{-1}T_{i,j}P\,\mathbf{v_{1,2}}=T_{1,2}\mathbf{v_{1,2}} $$ so we have: $$ T_{1,2}=P^{-1}T_{i,j}P= \dfrac{1}{4} \left[ \begin{array}{cccc} -7&-1\\ 1&7 \end {array} \right] $$

I hope this can be helpful.