Matrix Multiplication and Function Composition Given the vector space $F^n$ and two linear function $T,S:F^n \rightarrow F^n$ is it true that multiplying the representative matrices according to the standard basis of $T$ and $S$ is equivalent to the composition of their explicit formula's?  I.E. Given a vector v in $F^n$ is $[T]_E[S]_Ev = [TS]_Ev = T(S(v)) $ ?
 A: I'm reading this question as: "Does matrix multiplication really give rise to the matrix for the composition of maps that the matrices represent?" I'm always using matrices operating on the left of column vectors, in my notation below.
Define $\pi_i:F^n\rightarrow F$ by projecting onto the $i$th coordinate. Define $\mu_j:F\rightarrow F^n$ by inserting the input value into the $j$th coordinate. 
How do these maps relate to the matrix representation of a function? Let $f:F^n\rightarrow F^n$ be represented by $A=[A]_{ij}$ in the standard basis. If you compute $\mu_j(1)$, you get the unit vector with a 1 in the $j$th coordinate. To affect $f$ on this, you multiply on the left with its matrix $A$. So, $A\mu_j(1)$ is the $j$th column of $A$. Now by applying $\pi_i$ after this, you will pick out the $i$th entry of the $j$th row. So, $\pi_i(A\mu_j(1))=A_{ij}$. This shows that $\pi_i  f\mu_j=A_{ij}$. This last $A_{ij}$ is the $1\times 1$ matrix representation of a linear transformation of $F$ into $F$.
Can you see that $\sum_{k=1}^n\mu_k(\pi_k(v))=v$ for all $v\in F^n$? This means that $\sum_{k=1}^n\mu_k\pi_k=Id_V$, the identity on $V$. 
Let a transformation $g:F^n\rightarrow F^n$ be represented by the matrix $B=[B]_{ij}$ in standard basis. We will now show that $[BA]_{ij}$ is the matrix representing the composition $gf$. This will be true if $\pi_i gf \mu_j=[BA]_{ij}$, by our discussion before.
We compute that $\pi_i gf \mu_j=\pi_i g \mathrm{Id}_V f \mu_j=\pi_i g(\sum_{k=1}^n \mu_k\pi_k)f \mu_j=\sum_{k=1}^n\pi_i g \mu_k\pi_kf \mu_j=\sum_{k=1}^n B_{ik}A_{kj}=[BA]_{ij}$. So there you have it: the components of $gf$ are given by matrix multiplication!
A: More pedantically:
Let $[T]_{ij}$ be the representation in the standard basis $\{e_i\}$ of the operator $T$. Suppose $x\in F^n$, then we can write $x = \sum_j [x]_j e_j$.
The aim is to show that $[TS]_{ij} = \sum_k [T]_{ik} [S]_{kj}$. Note that since 
$\{e_i\}$ is a basis, the matrix representation is unique, ie, if the matrix $\alpha_1$ and $\alpha_2$ represent the operator $A$, then $\alpha_1 = \alpha_2$.
Then we have
\begin{align} T(Sx) & = T(\sum_j [x]_j S e_j) \\
 & = T(\sum_j [x]_j \sum_k [S]_{kj} e_k) \\
& = \sum_j [x]_j \sum_k [S]_{kj} T e_k \\
& = \sum_j [x]_j \sum_k [S]_{kj} \sum_i [T]_{ik} e_i \\
& = \sum_j [x]_j  \sum_i \sum_k [T]_{ik} [S]_{kj} e_i \\
& = (TS)x \\
& = \sum_j [x]_j  \sum_i [TS]_{ij} e_i.
\end{align}
The desired result follows from uniqueness.
