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?


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 $B^{-1}AB$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.