$\begin{bmatrix} T \\ \end{bmatrix}_\mathscr{F}$ = $P^{-1}$ $\begin{bmatrix} T \\ \end{bmatrix}_\mathscr{E}$ $P$
Where $\mathscr{F}$ is the new basis and $\mathscr{E}$ is the old basis. $P$ represents the change-of-basis matrix, from $\mathscr{E}$ to $\mathscr{F}$. $\begin{bmatrix} T \\ \end{bmatrix}_\mathscr{F}$ is the matrix representation of a linear transformation $T$, with respect to the basis $\mathscr{F}$. Same thing goes for $\begin{bmatrix} T \\ \end{bmatrix}_\mathscr{E}$
Can someone prove why that equation is true? I have no idea how to show that it's true.
Warning the $\cong$ symbol is symmetric, yet when writing $\varphi: K^n \cong V$ I explicitly mean that it is an isomorphism going from $K^n$ to V.
A choice of basis of an $n$-dimensional $K$-vector space is nothing else than a choice of isomorphism $\varphi:K^n \cong V$ (the standard basis of $K^n$ determines your basis of $V$). A linear transformation $T:V\rightarrow V$ becomes a linear transformation $T_\varphi = \varphi^{-1}T\varphi:K^n \rightarrow K^n$, which in turn bijectively corresponds to a matrix $[T]_\varphi\in K^{n\times n}$.
Now, given another basis/isomorphism $\psi:K^n \cong V$ we get another linear transformation $T_\psi=\psi^{-1}T\psi:K^n \rightarrow K^n$ with corresponding matrix representation $[T]_\psi$. Now note that we get a commuting diagram $$\begin{array}{ccccc} K^n &\overset{\varphi}{\cong} &V& \overset{\psi}{\cong} & K^n\\ \downarrow T_\varphi &&\downarrow T && \downarrow T_\psi\\ K^n &\overset{\varphi}{\cong} &V& \overset{\psi}{\cong} & K^n \end{array}$$
Finally note that the composite $\psi^{-1}\varphi:K^n \cong K^n$ is corresponds precisely to the matrix $P$ changing from the basis $\varphi$ to the basis $\psi$. The commutativity of the diagram thus corresponds to the matrix identity $P^{-1}[T]_\psi P = [T]_\varphi$...