I'm self-studying some linear algebra from the Schaum's Outline of Linear Algebra by Lipschutz and I came across a problem where I don't fully follow the key step. Below I make a note with respect to where I don't follow:
Problem 6.26: (Note that the notation $[T]_S$ and $[v]_S$ refer to the matrix representation of the operator $T$ and the coordinate vector of the vector $v$, respectively). The proof is clear to me until it comes time to make the key step to conclude that the equality after "Hence," implies that the two matrices acting on the coordinate vector are the same. I understand that the coordinate mapping is surjective (and, indeed, an isomorphism) as stated, and (I think) that this implies that there is a $v$ in $V$ such that this equality holds - and since it holds for all $v$ in $V$ the proof is complete? I feel like I'm missing something here.
I think we can complete the argument as follows:
Once we have that $$P^{-1}[T]_SP[v]_{S'} = [T]_{S'}[v]_{S'} \quad \textrm{for any $v \in V$;} \tag{$*$}$$ call $A = P^{-1}[T]_SP$ and $B = [T]_{S'}$, and suppose that $S'$ consist of the $n$ ordered vectors $v_1,\dots,v_n$. Now observe that, for all $j$ between $1$ and $n$, the coordinate vector $[v_j]_{S'}$ is the column vector in $K^n$ whose only nonzero entry is a $1$ in the $j$-th place, let's denote this vector as $[v_j]_{S'} = e_j$.
Applying $(*)$ with $v = v_j$ we have $Ae_j = Be_j$. Since the left hand side is just the $j$-th column of $A$ and the right hand side the $j$-th column of $B$, we conclude that all the columns of $A$ and $B$ coincide, hence, they are equal.