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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!

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    $\begingroup$ You are missing the key thing to be shown. Why is $AB$ the matrix of the transformation $T \circ S$? (Also I think you want your first $T$ to be $S$, and you mean $BA$ (because $A$ is $m \times n$ and $B$ is $l \times m$). $\endgroup$
    – arkeet
    Sep 30, 2016 at 20:41

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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.

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