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My textbook says that multiplying block matrices works the same as regular matrix multiplication (granted the dimensions of the sub matrices are appropriate). Wikipedia also has an example saying so.

It seems like proving it is purely technical and yet I'm having trouble putting it into words. What would be a good way to go?

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Doing a small example will probably convince you. If you want to do it rigorously, use the formula $\sum_k a_{ik}b_{kj}=c_{ij}$ for the $(i,j)$th entry of the product $C=AB$ of two matrices $A$ and $B$, and split it cases: one for each block that $(i,j)$ can belong to. –  Samuel Mar 31 '11 at 14:55
    
definitely work out some easier/smaller examples to convince yourself, or (depending on your disposition) work out the general say $4\times4$ with $2\times2$ blocks –  yoyo Mar 31 '11 at 15:55

1 Answer 1

up vote 5 down vote accepted

It is the same as regular multiplication except that matrix multiplication is not usually commutative.

That said I think you can develop the notation and proof by bootstrapping the 2x2 case. Suppose A is a 2x2 block matrix, say having I+J rows and K+L columns, so that the block in the upper left corner is IxK, etc. Then for B a 2x2 block matrix having K+L rows and M+N columns, the block multiplication would be compatible:

$$ A = \left( \begin{array} {c,c} A_{11} A_{12} \\ A_{21} A_{22} \end{array} \right) $$

$$ B = \left( \begin{array} {c,c} B_{11} B_{12} \\ B_{21} B_{22} \end{array} \right) $$

$$ AB = \left( \begin{array} {c,c,c} {A_{11}*B_{11}+A_{12}*B_{21}} {\; \;}{A_{11}*B_{12}+A_{12}*B_{22}} \\ {A_{21}*B_{11}+A_{22}*B_{21}} {\; \;}{A_{21}*B_{12}+A_{22}*B_{22}} \end{array} \right) $$

Cases with more than two blocks per row or column can then be reduced to this simple case by lumping blocks together and applying the multiplication recursively.

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