4.4. BLOCK MULTIPLICATION OF MATRICES 89
Next consider the question of multiplication of two block matrices. Let B be a blockmatrix of the form
B11 · · · B1p...
. . ....
Br1 · · · Brp
(4.17)
and A is a block matrix of the formA11 · · · A1m
.... . .
...Ap1 · · · Apm
(4.18)
and that for all i, j, it makes sense to multiply BisAs j for all s ∈ {1, · · · , p}. (That is the twomatrices, Bis and As j are conformable.) and that for fixed i j, it follows BisAs j is the samesize for each s so that it makes sense to write ∑s BisAs j.
The following theorem says essentially that when you take the product of two matrices,you can do it two ways. One way is to simply multiply them forming BA. The other way isto partition both matrices, formally multiply the blocks to get another block matrix and thisone will be BA partitioned. Before presenting this theorem, here is a simple lemma whichis really a special case of the theorem.
Lemma 4.4.1 Consider the following product. 0
I0
( 0 I 0)
where the first is n× r and the second is r×n. The small identity matrix I is an r× r matrixand there are l zero rows above I and l zero columns to the left of I in the right matrix.Then the product of these matrices is a block matrix of the form 0 0 0
0 I 0
0 0 0
Proof: From the definition of the way you multiply matrices, the product is
0
I0
0 · · ·
0
I0
0
0
I0
e1 · · ·
0
I0
er
0
I0
0 · · ·
0
I0
0
which yields the claimed result. In the formula e j refers to the column vector of length rwhich has a 1 in the jth position. ■
Theorem 4.4.2 Let B be a q× p block matrix as in 4.17 and let A be a p×n block matrixas in 4.18 such that Bis is conformable with As j and each product, BisAs j for s = 1, · · · , pis of the same size so they can be added. Then BA can be obtained as a block matrix suchthat the i jth block is of the form
∑s
BisAs j. (4.19)