A question regarding a proof based on similarity between two matrices I've got another question I'd like to try to prove.
"Let $\Bbb{R^n} → \Bbb{R^n}$ be a linear transformation. If $A$ is the standard matrix representation of $L$, then an n x n matrix $B$ will also be a matrix representation of $L$ iff $B$ is similar to $A$."
This is true, and I'm trying to understand why. Here's what I've got so far:

If $B$ is similar to $A$, then $B = S^{-1}AS$, where $S$ = some transition matrix from one ordered basis to another. 
Assume $A$ is the standard representation of $L$. Then there exists a $B$ similar to $A$ such that $B = S^{-1}AS = $ another matrix representation of L.

I don't know exactly if that's correct. It kind of feels correct, but also not enough for me. I just don't know where else I can go with it, because other methods feel like dead-ends to me.
Please respond if you have the time. Thanks!
-Jon 
 A: The proof below contains an argument for sufficiency, i.e., if $B$ is similar to $A$, then $B$ is a matrix representation for the linear transformation $L$.
Suppose that $B= S^{-1} A S$. Let $s_i$ denote the $i^{th}$ column of $S$ and let $\mathcal{B} := \{s_1,\dots,s_n\}$. Since $S$ is invertible, it follows that  $\mathcal{B}$ is a basis of $\mathbb{R}^n$. 
Suppose that $L(x) = y \in \mathbb{R}^n$. Then $Ax = y$. Since $S$ is invertible, there is a unique vector $[x]_\mathcal{B} \in \mathbb{R}^n$ such that $x = S[x]_\mathcal{B}$ (this is the coordinate vector of $x$ with respect to the basis $\mathcal{B}$). Similarly, there is a unique vector $[y]_\mathcal{B} \in \mathbb{R}^n$ such that $y=S[y]_\mathcal{B}$. Notice that 
\begin{align*}
y= Ax 
&\Longleftrightarrow S[y]_\mathcal{B} = A(S[x]_\mathcal{B})         \\
&\Longleftrightarrow [y]_\mathcal{B} = (S^{-1} A S) [x]_\mathcal{B}  \\
&\Longleftrightarrow [y]_\mathcal{B} = B [x]_\mathcal{B}.
\end{align*}
The last expression is the representation of $L$ with respect to the basis $\mathcal{B}$.
