I seek verification of my proof for this change of bases theorem. Any help would be greatly appreciated!
The theorem is stated below:
Let $V, W$ be finite-dimensional vector spaces over a field $F$.
Let $T: V\to W$ be a linear mapping.
Let $B_{0}, B_{1}$ be distinct ordered bases for V and let $B_{2}, B_{3}$ be distinct ordered bases for $W$.
Let $M^{2}_{0}(T)$ be the matrix for $T$ with respect to $B_{0}, B_{2}$ and let $M^{3}_{1}(T)$ be the matrix for $T$ with respect to $B_{1}, B_{3}$.
Let $M^{0}_{1}(I_{V})$ be the matrix of the identity map $I_{V}:V \to V$ with respect to $B_{0}, B_{1}$ and let $M^{2}_{3}(I_{W})$ be the matrix of the identity map $I_{W}:W \to W$ with respect to $B_{2}, B_{3}$.
Then $ M^{3}_{1}(T) = M^{3}_{2}(I_{W})M^{2}_{0}(T)M^{0}_{1}(I_{V}) $.
My proof:
Take any $ v \in V$ and let $w = T(v) \in W$. I write $[v]_{0}, [v]_{1}$ for $v$ expressed in bases$B_{0}, B_{1}$, respectively. Similarly for $[w]_{2}, [w]_{3}$.
The following relations hold:
$1)$ $[v]_{0} = M^{0}_{1}(I_{V})[v]_{1} $
$2)$ $[w]_{3}= M^{3}_{2}(I_{W})[w]_{2}$.
It follows that
$ M^{3}_{1}(T)[v]_{1}=[w]_{3}= M^{3}_{2}(I_{W})[w]_{2} = M^{3}_{2}(I_{W}) M^{0}_{2}(T) [v]_0 = M^{3}_{2}(I_{W}) M^{0}_{2}(T) M^{0}_{1}(I_{V})[v]_{1}$.
Since the choice of $v$ was arbitrary, it must be that
$M^{3}_{1}(T)= M^{3}_{2}(I_{W}) M^{2}_{0}(T) M^{0}_{1}(I_{V})$. $ \blacksquare $
[Added: Essentially I have tried to rewrite the following theorem in a way that is more intuitive to me. Have I succeeded? Is my proof also correct?]
I would like to point few things about your proof:
$a)$ $v$ should be a non-zero vector for the conclusion of the statement $M^{3}_{1}(T)= M^{3}_{2}(I_{W}) M^{2}_{0}(T) M^{0}_{1}(I_{V})$.
$b)$ This statement $ M^{3}_{1}(T)[v]_{1}=[w]_{3}$ doesn't follow from relation $1)$ and $2)$. It should be made an explicit statement.
$c)$ You should take linear transformation $T$ to be a non-zero linear transformation which is inherently assumed in Theorem $45$..
Rest Your Proof works fine for me.