characteristic polynomial with rows as coordinates

Question

Suppose $V$ is a finite dimensional vector space with basis $E$, $f$ an endomorphism of $V$ and $B$ another basis of $V$. Denote by $B'$ the matrix whose rows are the coordinates of $B$ and by $C'$ the matrix whose rows are the coordinates of the image $f(B)$ (in both cases with respect to the basis $E$). Is it true that one may compute the characteristic polynomial of $f$ as the characteristic polynomial of a matrix A satisfying $C' = A\cdot B'$?

Answer 1

Yes. Let's denote by $[f]^B_E$ the matrix representing the endomorphism $f$ with repsect to the basis $B$ in the domain and $E$ in the range. More explicitly, if $B = (v_1, \dots, v_n)$, the columns of $[f]^B_E$ are the coordinate vectors of $f(v_1), \dots, f(v_n)$ ("the image of $f$"). Then, your assumptions give us $$ B' = \left( [\operatorname{id}]^{B}_{E} \right)^{T}, C' = \left( [f]^{B}_{E}\right)^T $$ and so $$ A = B'^{-1} C' = \left( \left( [\operatorname{id}]^{B}_{E} \right)^{T} \right)^{-1} \left( [f]^{B}_{E}\right)^T = \left( \left( [\operatorname{id}]^{B}_{E} \right)^{-1} \right)^{T} \left( [f]^{B}_{E}\right)^T =\\ \left( [\operatorname{id}]^{E}_{B} \right)^{T} \left( [f]^{B}_{E}\right)^T = \left( [f]^{B}_{E} \cdot [\operatorname{id}]^{E}_{B} \right)^T = \left( [f]^E_E \right)^T. $$

That is, your assumptions are just a somewhat obscure way of saying that $A$ is the transpose of the matrix representing $f$ with respect to the basis $E$ (both in the domain and in the range). Since the characteristic polynomial of $f$ is the characteristic polynomial of any matrix representing $f$ with respect to some basis, and the characteristic polynomial of a matrix is identical to the characteristic polynomial of its transpose, the characteristic polynomial of $f$ is the characteristic polynomial of $A$.