# The matrix of $T(x_1,x_2)=(x_1+x_2,x_1-x_2)$ with respect to a basis

Consider the linear map $$T:\mathbb{R}^2\to \mathbb{R}^2, T(x_1, x_2)=(x_1+x_2,x_1-x_2)$$. Let $$B_1$$ be the canonical base of $$\mathbb{R}^2$$ and consider another basis $$B_2=\{f_1,f_2\}$$, where $$f_1=(1,1)$$ and $$f_2=(1,2)$$.
So, according to my computations, the matrix of $$T$$ with respect to $$B_1$$ is $$\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}$$ and the matrix of $$T$$ with respect to $$B_2$$ is $$\begin{pmatrix} 2 & 3 \\ 0 & -1 \end{pmatrix}$$. However, I tried computing the transition matrices. I got that the transition matrix from $$B_1$$ to $$B_2$$ is $$\begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix}$$ and the transition matrix from $$B_2$$ to $$B_1$$ is $$\begin{pmatrix} 2 & -1\\ -1 & 1 \end{pmatrix}$$. I should have that $$\begin{pmatrix} 2 & 3 \\ 0 & -1 \end{pmatrix}= \begin{pmatrix} 2 & -1\\ -1 & 1 \end{pmatrix}\cdot \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}\cdot \begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix}$$, but this is not true. Where did I go wrong?

The second column of your matrix of $$T$$ with respect to $$B_2$$ is wrong.
Indeed, $$T(f_2) = T(1,2) = (3,-1) \neq 3f_1 - f_2 = (2,1).$$