Determine second base of matrix $A=\:\begin{pmatrix}2&-3\\ 1&1\end{pmatrix}$ if second base exists.

Question

Let it $A=\begin{pmatrix}2&-3\\ 1&1\end{pmatrix}$ matrix of a linear operator L by standard base of space $R^2$.Does second base of same space exists such that linear operator has same matrix according to that second base. I know that is $L(x,y)=(2x-3y,x+y)$ and that second base exists,but I don't know how to prove that.

Answer 1

Hint If the matrix is the same in the base $(u, v)$, it means that $L(u) = 2 u + v$ and $L(v) = -3 u + v$