Linear Transformation and Basis $$
\begin{array}{l}{\text { 2.) Consider the linear transformation } T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2},\left[\begin{array}{l}{x} \\ {y}\end{array}\right] \mapsto\left[\begin{array}{c}{-2 x-5 y} \\ {2 x+4 y}\end{array}\right]} \\ {\text { Find a basis } \mathcal{B} \text { of } \mathbb{R}^{2} \text { that represents } T \text { in the form }[T]_{\mathcal{B}}=\left[\begin{array}{cc}{a} & {-b} \\ {b} & {a}\end{array}\right]}\end{array}
$$
What I tried is T$\left[\begin{array}{l}{1} \\ {0}\end{array}\right]$  = $\left[\begin{array}{l}{-2} \\ {2}\end{array}\right]$  and T$\left[\begin{array}{l}{0} \\ {1}\end{array}\right] $ = $\left[\begin{array}{l}{-5} \\ {4}\end{array}\right] $ now how to find such matrix with given form?
 A: The matrix of $T$ with respect to the canonical basis is $\left[\begin{smallmatrix}-2&-5\\2&4\end{smallmatrix}\right]$, whose trace is $2$ and whose determinant is also $2$. So, if your problem has a solution, then $a$ and $b$ will have to be such that $2a=2$ and that $a^2+b^2=2$. So, $a=1$ and $b=\pm1$. Let us search for a solution with $a=b=1$. Then we are after two vectors $v=(x_1,y_1)$ and $w=(x_2,y_2)$ such that $T(v)=v+w$ and that $T(w)=-v+w$. So, we solve the system$$\left\{\begin{array}{l}-2x_1-5x_2=x_1+y_1\\2x_1+4x_2=x_2+y_2\\-2y_1-5y_2=-x_1+y_1\\2y_1+4y_2=-x_2+y_2.\end{array}\right.$$One solution of this system is $(x_1,x_2,y_1,y_2)=(0,-1,5,-3)$. In other words, $v=(0,-1)$ and $w=(5,-3)$.
A: We have that the matrix from basis $B$ to canonical is
$$u_C=M_Bu_B$$
then
$$T_C(u)=\left[\begin{array}{cc}{-2} & {-5} \\ {2} & {4}\end{array}\right]u_C=\left[\begin{array}{cc}{-2} & {-5} \\ {2} & {4}\end{array}\right]M_Bu_B$$
and then
$$T_B(u)=M_B^{-1}T_C(u)=M_B^{-1}\left[\begin{array}{cc}{-2} & {-5} \\ {2} & {4}\end{array}\right]M_Bu_B$$
and we are looking for
$$M_B^{-1}\left[\begin{array}{cc}{-2} & {-5} \\ {2} & {4}\end{array}\right]M_B=\left[\begin{array}{cc}{a} & {-b} \\ {b} & {a}\end{array}\right] \iff \left[\begin{array}{cc}{-2} & {-5} \\ {2} & {4}\end{array}\right]M_B=M_B\left[\begin{array}{cc}{a} & {-b} \\ {b} & {a}\end{array}\right]$$
that is by $M_B=\left[\begin{array}{cc}{s} & {t} \\ {u} & {v}\end{array}\right]$ 


*

*$-2s-5u=as+bt \implies -(a+2)s-bt-5u=0$

*$-2t-5v=-bs+at \implies bs-(a+2)t-5v=0$

*$2s+4u=au+bv \implies 2s+(4-a)u-bv=0$

*$2t+4v=-bu+av \implies 2t+bu+(4-a)v=0$
$$\left[\begin{array}{cccc}{-(a+2)} & {-b}&-5&0 \\ {b} & {-(a+2)}&0&-5\\2&0&(4-a)&-b\\0&2&b&(4-a)\end{array}\right]\left[\begin{array}{c}s \\ t\\u\\v \\\end{array}\right]=\left[\begin{array}{c}0 \\ 0\\0\\0 \\\end{array}\right]$$
which can be solved with the condition $a^2+b^2=2$.
A: In the standard basis $\mathcal S$, we have $T$ represented by
$$
[T]_{\mathcal S} = \begin{bmatrix}-2&-5\\2&4\end{bmatrix}
$$
Now we need to find some invertible matrix $B$ such that $B^{-1}T_SB$ has the desired form. The columns of $B$, interpreted as vectors in the standard basis, will then be the basis $\mathcal B$, and we will have $[T]_{\mathcal B} = B^{-1}T_SB$. Inserting, expanding, and solving seems inviting because it sounds rather straight-forward, but from experience I suspect that it will not be nice. Using the fact that basis change doesn't affect trace and determinant makes things a lot easier, as José points out in his answer. But I like a more geometric approach.
The eigenvalues of $T$ are $1\pm i$. That means that the eigenvalues of $T-I$ are $\pm i$, which is to say, there is a basis where $T-I$ is represented by the matrix
$$
\begin{bmatrix}0&-1\\1&0\end{bmatrix}
$$
and then, since the identity transformation is always represented by the identity matrix, we get that $T$ is represented by
$$
\begin{bmatrix}1&-1\\1&1\end{bmatrix}
$$
in this basis. So, how do we find such a basis? Well, the eigenvectors of $T-I$ (and $T$) are, in the standard basis,
$$
\begin{bmatrix}-3\pm i\\2\end{bmatrix}\tag1
$$
We want a basis where the eigenvectors of $T-I$ are instead equal to the eigenvectors of $\left[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right]$, which are
$$
\begin{bmatrix}\pm i\\1\end{bmatrix}\tag2
$$
So what matrix transforms one basis to the other?
Well, if $v$ is an eigenvector of $T$, we want $[v]_{\mathcal S} = B[v]_{\mathcal B}$. Which is to say, $B$ describes how to take the vectors of $(2)$ and transforms them into the vectors of $(1)$.
This transformation consists of subtracting $3$ times the second component from the first component, and then double the second component. The matrix which does this is
$$
\begin{bmatrix}1&-3\\0&2\end{bmatrix}
$$
This is our matrix $B$, which again means we have our basis $\mathcal B$ from the columns of this matrix.
Swapping the order or changing the lengths of the eigenvectors in $(1)$ and $(2)$ will change which matrix $B$ we end up with. And any one of them gives a valid basis where matrix representing $T$ has the required form. You may swap the sign of $b$ as you change the eigenvectors around, but that's the only thing could happen.
