TITLE: Need help proving blockwise property of matrix multiplication.
QUESTION [6 upvotes]: I want to prove the following:

If $X$ and $Y$ are $n \times n$ matrices, and $$ X = \left[\begin{matrix} A
& B\\ C & D \end{matrix}\right],  Y = \left[\begin{matrix} E & F\\ G & H \end{matrix}\right] $$
where A, B, C, D, E, F, G, and H are $n/2 \times n/2$ submatrices, then
  the product $XY$ can be expressed in terms of these blocks:
$$ XY = Z = \left[\begin{matrix} AE + BG & AF + BH\\ CE + DG & CF + DH \end{matrix}\right] $$

My initial thought was to use the matrix multiplication definition: 
$$Z_{ij} = \sum_{k=1}^n X_{ik} Y_{kj}$$
and show that each $Z_{ij}$ equals the element in $Z$ by going case by case.
Case1 would be something like: $1 \leq i \leq(n/2), 1 \leq j \leq (n/2)$.
So in this case, $X_{ij} = A_{ij}$ and $Y_{ij} = E_{ij}$
I am stuck here and not sure if I am on the right track. Please advise me on how to proceed from here (or suggest an alternative method).

REPLY [6 votes]: I assume $n$ is even.  :-) Your method will definitely work.  For example:
$$
Z_{1,1} = \sum_{k=1}^n X_{1,k}Y_{k,1} = \sum_{k=1}^{n/2}X_{1,k}Y_{k,1} + \sum_{k=n/2+1}^{n} X_{1,k}Y_{k,1} =\\
\sum_{k=1}^{n/2} A_{1,k}E_{k,1} + \sum_{k=1}^{n/2} B_{1,k}G_{k,1} = (AE)_{1,1} + (BG)_{1,1}
$$
Another way to look at it is to realize that matrices represent linear transformations and multiplication of matrices corresponds to composition.  Now identify $\mathbb{R}^n$ with $\mathbb{R}^{n/2} \times \mathbb{R}^{n/2}$.  Then $X: \mathbb{R}^n \rightarrow \mathbb{R}^n$ can be written as a sum of four linear transformations $\mathbb{R}^{n/2} \rightarrow \mathbb{R}^{n/2}$:
$$
X: (v, w) \mapsto (Av + Bw, Cv + Dw).
$$
And similarly
$$
Y: (v, w) \mapsto (Ev + Fw, Gv + Hw).
$$
Now write out the composition $X \circ Y$ and see what you get in terms of $v$ and $w$.