The linear transformation $S : \mathbb{R}^2 \to \mathbb{R}^2$ is given by the standard matrix $A_\epsilon = \begin{bmatrix} 1 & 2 \\ 1 & 2\end{bmatrix} $. Find the matrix for S in the basis $ F = \{ \begin{bmatrix}1 \\ 1 \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \end{bmatrix} \}$
How do you get the answer $\begin{bmatrix} 3 & 0 \\ 0 & 0\end{bmatrix}$?
On
Let ${\rm std}$ be the standard basis and $\mathcal{F}$ be the new basis you have. One has $$[S]_{\mathcal{F}} = [{\rm Id}]_{{\rm std},\mathcal{F}}[S]_{\rm std}[{\rm Id}]_{\mathcal{F}, {\rm std}}.$$What is $[{\rm Id}]_{\mathcal{F}, {\rm std}}$? Evaluate ${\rm Id}$ in the elements of $\mathcal{F}$, write the results as combinations of elements in ${\rm std}$ and place the coefficients in columns. In this case it just amounts to putting the elements of $\mathcal{F}$ themselves in columns. So $$[S]_{\mathcal{F}} = \begin{pmatrix} 1& -2 \\ 1 & 1\end{pmatrix}^{-1}\begin{pmatrix}1 & 2 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 1& -2 \\ 1 & 1\end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 0\end{pmatrix}.$$
Recall that, relative to an arbitrary ordered basis, the coordinates of the basis vectors themselves are our old friends $(1,0,0,0,\dots)^T$, $(0,1,0,0,\dots)^T$, $(0,0,1,0,\dots)^T$ and so on. Recall also that the columns of a transformation matrix are just the images of the basis vectors expressed relative to that basis. So, you just need to work out what the coordinates of those images are. I’ll do the first one for you: $S(1,1)^T = (3,3)^T = 3(1,1)^T+0(-2,1)^T$, so the first column is $(3,0)^T$. The second column is even simpler.