Change of basis of linear transformations with matrix representation let $B$  be a basis  for a vector space $V$ and let $F:V\rightarrow V$ be a linear map. Prove that if $C$ is a matrix similar to $M_{B}^{B}(F)$ then there exists a basis $B'$ of $V$ such that $C=M_{B'}^{B'}(F)$.
If $C$ is a similar to $M_{B}^{B}(F)$ then there exists a non singular matrix $P$ such that $C=P^{-1}M_{B}^{B}(F)P$. But how can we prove that there exists a basis $B'$ as the above?
 A: Think about the reverse problem: if you have bases $B$ and $B'$, how do you show that $M_B^B(F)$ and $M_{B'}^{B'}$ are similar? What is the non-singular matrix $P$? Essentially $P$ is a change-of-basis matrix, meaning a matrix of the identity transformation from one basis to another. We have
$$M_{B'}^{B'}(F) = M_B^{B'}(I) M_B^B(F) M^B_{B'}(I) = \left(M^B_{B'}(I)\right)^{-1} M_B^B(F) M^B_{B'}(I).$$
That is, we may take $P = M^B_{B'}(I)$.
Now, if we start with a matrix $P$, but don't necessarily have a matrix $B'$, it stands to reason that we require a basis $B'$ such that $P = M^B_{B'}(I)$.
Think about how $M^B_{B'}(I)$ is defined: you take each of the basis vectors in $B'$, transform them under $I$ (i.e. do nothing to them), then express each basis vector of $B'$ as a linear combination of the vectors in $B$. The coefficients of these linear combinations are the columns of $P$. So, we should be able to recover the vectors in $B'$ by simply expanding the columns of $P$ as coordinate column vectors in terms of $B$.
That is, if $B = (v_1, \ldots, v_n)$, then we should define, for each $1 \le k \le n$,
$$w_k = \sum_{i=1}^n (P)_{ik} v_i, \tag{1}$$
then $B' = (w_1, \ldots, w_n)$ should be the basis you want.
Note that we haven't exactly proven that $B'$ is the basis you want, or even a basis in the first place! From our construction, we can easily see that the unique linear map $T : V \to V$, taking $v_k$ to $w_k$ for each $k$, satisfies
$$M_B^B(T) = P.$$
Since $P$ is invertible, it follows that $T$ is invertible, and hence maps bases to bases. In particular, this implies $B'$, the image of $B$ under $T$, is a basis.
Is $B'$ the basis we want? Well, now that $B'$ is a confirmed basis, we can use $(1)$ to conclude that
$$P = M^B_{B'}(I).$$
From this, it follows that $P^{-1} = M_B^{B'}(I)$, and hence
$$M_{B'}^{B'}(F) = M_B^{B'}(I) M_B^B(F) M^B_{B'}(I) = P^{-1} M_B^B(F) P = C,$$
as required.
A: Let $B'$ be the set of vectors whose coordinates in the basis $B$ are the columns of $P$. More precisely, if $B$ is $\{b_1,\dots,b_n\}$, define $b_i'$ by specifying that $$
[b_i']_B=\text{col}_i(P)
$$
As $P$ is invertible, its columns form a basis (of $\mathbb{F}^n)$, so that $\{b_1',\dots,b_n'\}$ is a basis of $V$.
Note that $P=M_{B'}^B(I)$ (I'm taking a gamble regarding how you order these), where $I:V\to V$ is the identity. Can you see where to go from here?
