<s>[INST] 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?
 [/INST] 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 $a^2+b^2=2$.
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