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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?

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The second column of your matrix of $T$ with respect to $B_2$ is wrong.

Indeed, $T(f_2) = T(1,2) = (3,-1) \neq 3f_1 - f_2 = (2,1).$

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  • $\begingroup$ Oh, silly me, I wrote the coordinates with respect to the canonical basis...thank you very much! $\endgroup$
    – TheZone
    Jan 26 at 19:47
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    $\begingroup$ Yes, a common and forgivable mistake. Don't worry about it :) $\endgroup$ Jan 26 at 19:47

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