# How to find coordinate vector of a linear transformation

I'm trying to find the coordinate vector of the following linear transformation

$$T \bigr(a , b \bigl) \quad=\quad \begin{pmatrix} 3a-b \\ a+3b \end{pmatrix}$$

in the following basis

$$\beta \quad=\quad \bigr\{ \, (1,1) \, , \, (1,-1) \, \bigl\}$$

I thought that i should be the coefficients in the transformation matrix, i.e

$$[T]_{\beta} \quad=\quad \begin{pmatrix} 3 & -1 \\ 1 & 3 \end{pmatrix}$$

$$\begin{pmatrix} 3 & 1 \\ -1 & 3 \end{pmatrix}$$

What did I do wrong?

You have that $$T(1,1)=(2,4)=3(1,1)-(1,-1)\quad \text{and}\quad T(1,-1)=(4,-2)=(1,1)+3(1,-1).$$ Therefore $$[T]_{\beta \beta }=\begin{pmatrix}3&1\\-1&3\end{pmatrix},$$ is correct.
Notice that the matrix of $$T$$ that you wrote is $$[T]_{\alpha \alpha }$$ where $$\alpha =\{(1,0),(0,1)\}$$ is the canonical basis.
Your solution is correct if you were working with a basis of the form $$\beta'=\{(1,0),(0,1)\}$$. However, it is not the case because we are working with $$\beta=\{(1,1),(1,-1)\}$$, but we always can use the definition of matrix representation for $$T$$ respect to the ordered base $$\beta$$,$$[T]_{\beta}=\begin{pmatrix}\uparrow&\uparrow\\ [T(1,1)]_{\beta}&[T(1,-1)]_{\beta}\\\downarrow&\downarrow \end{pmatrix},$$ where $$[T(1,1)]_{\beta}$$ and $$[T(1,-1)]_{\beta}$$ they are the coordinates respect to the basis $$\beta$$. The coordinates we can find directly by definition.
Using the rule given by $$T$$ we have that $$T(1,1)=\begin{pmatrix}2\\4\end{pmatrix}$$ then since $$\begin{pmatrix}2\\4\end{pmatrix}=3\begin{pmatrix}1\\1\end{pmatrix}+(-1)\begin{pmatrix}1\\-1\end{pmatrix}$$ we conclude that $$[T(1,1)]_{\beta}=\begin{bmatrix}3\\-1\end{bmatrix}$$. Similarly, we find the coordinate $$[T(1,-1)]_{\beta}=\begin{bmatrix}1\\3\end{bmatrix}$$.
Therefore $$[T]_{\beta}=\begin{pmatrix}3&1\\-1&3\end{pmatrix}.$$
Using $$\beta'$$ we have $$[T]_{\beta'}=\begin{pmatrix}3&-1\\1&3\end{pmatrix}.$$ That is one of the advantages of working with canonical bases (as $$\beta'$$).