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This video here seems to suggest that if a vector $v = (c_1, \dots, c_n)$ is given with coordinates in some basis $b_1, \dots, b_n$ and $B$ is the matrix with columns $b_1, \dots, b_n$ then $Bv$ is the vector $v$ given with coordinates in the standard basis. (watch 0 to 3:37)

In Tapp's book on matrix groups on page 19 he denotes $B$ by $g$ and some given linear transformation by $f$. As I understand from reading the page, $A$ is the matrix of $f$ in the standard basis and the goal of the page is to find the matrix of $f$ in the basis $B$.

What I don't get is, he states that this matrix is $gAg^{-1}$. Shouldn't it be $g^{-1}Ag$?

Please could someone help me and tell me what it is that I misunderstand?

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Note $g\in GL_n(\mathbb{K})$ is defined with rows -- not columns -- $(v_i)$ (here $(b_i)$); this is in fact identical to $B^T=B^{-1}$ in your example (i.e. the map which changes from the standard basis to our basis $(b_i)$). Hence $gAg^{-1}$ corresponds to your $B^{-1}AB$.

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  • $\begingroup$ Great, thanks for helping me, your answer makes it very clear! $\endgroup$ – learner Dec 13 '14 at 6:36

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