Let $T:V \to V$ be a linear transformation, and let $B$ and $C$ be two bases for $V$. Let $M$ be the matrix of $T$ with respect to the basis $B$, and let $N$ be the matrix of $T$ with respect to the basis $C$. Prove that $\det(M) = \det(N)$.
So far, I have "By the Diagonal Matrix Representation, $T = MNM^{-1}$, where $N$ is a diagonal $n \times n$ matrix. If $B$ is the basis for $\mathbb{R}^n$ formed from the columns of $N$, then $N$ is the B-matrix for the transformation $x \to Tx$." I'm not sure what to do next.
Additional info: I know that $\det(AB) = \det(A)\det(B)$ applies here, I'm not sure what $A$ and $B$ would be though. Also what I have so far, is the notation rather $T=PCP^{-1}$, where $B$ is the basis formed from columns of $P$?
Hint: Do you know an equation for the determinant of the product of two matrices?
Further hint: ... or maybe I should have said, three matrices.