Show that multiplication of matrices corresponds to composition of linear transformations.
My approach is that there are two matrices A and B. Let S be a linear transformation that maps a vector in $\mathbb{R}^n$ to a vector in $\mathbb{R}^m$ and is represented by the matrix A. Let T be a linear transformation that maps a vector in $\mathbb{R}^m$ to a vector in $\mathbb{R}^l$ and is represented by the matrix B. Therefore, BA maps a vector in $\mathbb{R}^n$ to a vector in $\mathbb{R}^l$ and correspond a composition of linear transformation $T \circ S$.
Is this a legitimate proof? If not, can you please tell me how else to approach this?
Thanks!
Best Answer
I'd say that, more or less, that is the idea. At present, though, it is not clear why $AB$ should correspond precisely to $T\circ S$: yes, the domain and codomain are the same, but there are many maps with domain and codomain coinciding with those of $T\circ S$.
Here is what I'd do. $S(x)=Bx$, $T(y)=Ay$, so $T\circ S(x)=A\cdot(Bx)$. So it remains to be shown that $A\cdot(Bx)=(AB)\cdot x$. But this is a special case of the associativity of the matrix product. More precisely, I mean the following theorem concludes the argument.
Theorem (Associativity of matrix product)
Let $A$ be an $m\times n$ matrix, $B$ an $n\times\ell$ one, $C$ an $\ell\times k$ one. Then:
$$(AB)C=A(BC).$$
Can you prove that? If not, I can always add a proof on your request.
Note
I automatically adjusted the matching between map names and matrix names, which caused $A$ and $B$ to have opposite meanings as to those of your post. In other words, my $A$ is your $B$ and viceversa.