Kenneth Kuttler

10 CHAPTER 1. REVIEW OF SOME LINEAR ALGEBRA

Ai j? Here is how for A an m×n matrix:

si×m︷︸︸︷(0 Isi×si 0

n×p j︷︸︸︷ 0Ip j×p j

. (1.2)

In the block column matrix on the right, you need to have c j− 1 rows of zeros above thesmall p j× p j identity matrix where the columns of A involved in Ai j are c j, · · · ,c j + p j−1and in the block row matrix on the left, you need to have ri− 1 columns of zeros to theleft of the si× si identity matrix where the rows of A involved in Ai j are ri, · · · ,ri + si. Animportant observation to make is that the matrix on the right specifies columns to use in theblock and the one on the left specifies the rows. Thus the block Ai j, in this case, is a matrixof size si× p j. There is no overlap between the blocks of A. Thus the identity n×n identitymatrix corresponding to multiplication on the right of A is of the form Ip1×p1 0

. . .0 Ipm×pm

 ,

where these little identity matrices don’t overlap. A similar conclusion follows from con-sideration of the matrices Isi×si . Note that in (1.2), the matrix on the right is a block columnmatrix for the above block diagonal matrix, and the matrix on the left in (1.2) is a blockrow matrix taken from a similar block diagonal matrix consisting of the Isi×si .

Next consider the question of multiplication of two block matrices. Let B be a blockmatrix of the form  B11 · · · B1p

.... . .

...Br1 · · · Brp

 (1.3)

and A is a block matrix of the form A11 · · · A1m...

. . ....

Ap1 · · · Apm

 (1.4)

such that for all i, j, it makes sense to multiply BisAs j for all s ∈ {1, · · · , p}. (That is thetwo matrices Bis and As j are conformable.) and that for fixed i j, it follows that BisAs j is thesame size 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 partition both matrices, formally multiply the blocks to get another block matrixand this one will be BA partitioned. Before presenting this theorem, here is a simple lemmawhich is really a special case of the theorem.

Lemma 1.2.1 Consider the following product. 0I0

( 0 I 0)

10 CHAPTER 1. REVIEW OF SOME LINEAR ALGEBRAAj;? Here is how for A an m Xx n matrix:NXPjsyxm 0(0 Isxs,; O )AL Ipjxp; ]- (1.2)0In the block column matrix on the right, you need to have c; — 1 rows of zeros above thesmall p; x p; identity matrix where the columns of A involved in Aj; are cj,--- ,cj;+pj—1and in the block row matrix on the left, you need to have r; — 1 columns of zeros to theleft of the s; x s; identity matrix where the rows of A involved in Aj; are r;,--- ,rj-++s;. Animportant observation to make is that the matrix on the right specifies columns to use in theblock and the one on the left specifies the rows. Thus the block A;;, in this case, is a matrixof size s; x p;. There is no overlap between the blocks of A. Thus the identity n x n identitymatrix corresponding to multiplication on the right of A is of the formIp, XPI 00 Lom Pmwhere these little identity matrices don’t overlap. A similar conclusion follows from con-sideration of the matrices J,,.5,. Note that in (1.2), the matrix on the right is a block columnmatrix for the above block diagonal matrix, and the matrix on the left in (1.2) is a blockrow matrix taken from a similar block diagonal matrix consisting of the Is,s,.Next consider the question of multiplication of two block matrices. Let B be a blockmatrix of the formBy Bip: (1.3)Br Brpand A is a block matrix of the formAi ++) Aim: 7 ; (1.4)Apt + Apmsuch that for all i, 7, it makes sense to multiply BjsAs; for all s € {1,---,p}. (That is thetwo matrices Bj, and A,; are conformable.) and that for fixed ij, it follows that BjsAs; is thesame size for each s so that it makes sense to write )°, BisAsj.The following theorem says essentially that when you take the product of two matrices,you can partition both matrices, formally multiply the blocks to get another block matrixand this one will be BA partitioned. Before presenting this theorem, here is a simple lemmawhich is really a special case of the theorem.Lemma 1.2.1 Consider the following product.01)(0 7 0)0