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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).

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up vote 3 down vote accepted

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 $w$.

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