Let matrix $A:= \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}$ And consider the linear map $L$ (restricted to A) : $\mathbb{R}^2 \rightarrow \mathbb{R}^2$, $x \rightarrow Ax$ with respect to the standard basis \begin{array}{c c c} \begin{pmatrix} 1 \\ 0 \end{pmatrix} & \text{and} &\begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{array} for $\mathbb{R}^2$ .
Find the transformed matrix $T$ such that the same map can be expressed as $LA : \mathbb{R}^2 \rightarrow \mathbb{R}^2$, $ \rightarrow Tx$ with respect to the non-standard basis \begin{array}{c c c} \begin{pmatrix} 1 \\ 1 \end{pmatrix} & \text{and} &\begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{array} for $\mathbb{R}^2$ (use this non-standard basis on both sides of the arrow, domain and codomain)
Here I have no idea how to proceed. I tried drawing a matrix transformation and computing the transformed map T with respect to vectors and bases and the known matrix A however I am not sure this is the correct way. Thanks a lot in advance.
Let $P=\begin{bmatrix} 1&0 \\ 1&1 \end{bmatrix} $... Then $B=P^{-1}AP$ will be the matrix of the linear transformation expressed in the new basis...
I get $\begin{bmatrix} 2 &1\\3&2\end{bmatrix}$...