Stuff I find slightly meaningful and close to achievable

The composition of two linear maps

A linear map and its matrix

Suppose we have a linear map \(T\colon V \to W\) where \(V,W\) are two \(\mathbb{R}\)-vector spaces with bases \(\{v_i\}_{i=1}^n\) and \(\{w_j\}_{j=1}^m\).
The matrix representation \(\mathrm{Mat}(T) = \mathrm{A}\in\mathbb{R}^{m\times n}\) of the map \(T\) is easy to find: first expand any \(v\in V\) in terms of the basis vectors:

T( v ) = T \left(\sum_{i=1}^n \alpha_i\,v_i\right)

Then use the linearity of the map \(T\) to obtain:

T( v ) = \sum_{i=1}^n \alpha_i\,T(v_i)

finally, expand each vector \(T(v_i)\) in terms of the basis vectors of \(W\):

T(v_i) = \sum_{j=1}^m a_{ji} w_j

This gives us the equality for any \(v\in V\):
T(v) = \sum_{j=1}^m \sum_{i=1}^n \alpha_i a_{ji} w_j

The coefficients \(\vec{\alpha} = [\alpha_1\,,\, \dots \,,\, \alpha_n]\) uniquely characterize the vector \(v\), and similarly the coefficients \(a_{ij}\) are uniquely characterize the map \(T\) given the bases of \(V,W\).
Because the expansion of \(T(v)\) is unique in the basis \(\{w_j\}_j\), we obtain that \(T(v)\) has the coefficients \(\beta_j = \sum_{k=1}^n a_{jk}\alpha_k\).
We can then only think of linear maps as multiplication and addition of the coefficients \(\alpha_i\) with the entries \(a_{ij}\).

We can then define the array \(\mathrm{A} = \left[ a_{ij} \right]_{ij}\) and matrix multiplication \(\mathrm{A} \vec{\alpha}\) as expected.

Composition as matrix multiplication

Let \(T,S,R\) be linear maps with matrices \(\mathrm{A},\mathrm{B},\mathrm{C}\). The third map \(R\) is the composition \(S\circ T\).
Given bases for the spaces \(V\xrightarrow{T} W \xrightarrow{S} Z\), let us denote

From the first section, we know that \(T\) has matrix representation \(\mathrm{A} = \left[a_{ij}\right]_{ij}\) so that the ouput coefficients are

$$ \beta_j = \sum_{k=1}^n a_{jk} \alpha_k $$

and that \(S\) has representation \(\mathrm{B} = \left[b_{ij}\right]_{ij}\) so that

$$ \gamma_q = \sum_{j=1}^m b_{qj} \beta_j $$

therefore, we have the equalities

\gamma_q &= \sum_{j=1}^m b_{qj} \sum_{k=1}^n a_{jk} \alpha_k \\
&= \sum_{k=1}^n \left(\sum_{j=1}^m b_{qj}a_{jk}\right) \alpha_k \\
&= \sum_{k=1}^n c_{qk}\alpha_k

thus we obtain that the matrix \(\mathrm{C}\) has coefficients \(c_{qk} = \sum_{j=1}^m b_{qj}a_{jk} = \left[ \mathrm{BA} \right]_{qk}\).
We thusly proved that \(\mathrm{C} = \mathrm{Mat}(S\circ T) = \mathrm{BA} = \mathrm{Mat}(S)\mathrm{Mat}(T)\).