Now what if instead of numbers, the entries, A,B,C,D,E,F,G are matrices of a size such that
the multiplications and additions needed in the above formula all make sense. Would the formula
be true in this case? I will show below that this is true.
Suppose A is a matrix of the form
( )
| A11 ⋅⋅⋅ A1m |
A = |( ... ... ... |) (3.14)
A ⋅⋅⋅ A
r1 rm
(3.14)
where A_{ij} is a s_{i}× p_{j} matrix where s_{i} is constant for j = 1,
⋅⋅⋅
,m for each i = 1,
⋅⋅⋅
,r. Such a
matrix is called ablock matrix, also a partitioned matrix. How do you get the block A_{ij}?
Here is how for A an m × n matrix:
In the block column matrix on the right, you need to have c_{j}− 1 rows of zeros above the small
p_{j}× p_{j} identity matrix where the columns of A involved in A_{ij} are c_{j},
⋅⋅⋅
,c_{j} + p_{j}− 1 and in the
block row matrix on the left, you need to have r_{i}− 1 columns of zeros to the left of
the s_{i}× s_{i} identity matrix where the rows of A involved in A_{ij} are r_{i},
⋅⋅⋅
,r_{i} + s_{i}. An
important observation to make is that the matrix on the right specifies columns to use in
the block and the one on the left specifies the rows used. Thus the block A_{ij} in this
case is a matrix of size s_{i}× p_{j}. There is no overlap between the blocks of A. Thus the
identity n × n identity matrix corresponding to multiplication on the right of A is of the
form
( )
| Ip1×p1 0 |
|( ... |)
0 I
pm×pm
where these little identity matrices don’t overlap. A similar conclusion follows from
consideration of the matrices I_{si×si}. Note that in 3.15 the matrix on the right is a
block column matrix for the above block diagonal matrix and the matrix on the left in
3.15 is a block row matrix taken from a similar block diagonal matrix consisting of the
I_{si×si}.
Next consider the question of multiplication of two block matrices. Let B,A be block matrices
of the form
( B ⋅⋅⋅ B ) ( A ⋅⋅⋅ A )
| .11 . 1.p | | 1.1 . 1m. |
|( .. .. .. |) ,|( .. .. .. |) (3.16)
Br1 ⋅⋅⋅ Brp Ap1 ⋅⋅⋅ Apm
(3.16)
and that for all i,j, it makes sense to multiply B_{is}A_{sj} for all s ∈
{1,⋅⋅⋅,p}
. (That is the two
matrices, B_{is} and A_{sj} are conformable.) and that for fixed ij, it follows B_{is}A_{sj} is the same size for
each s so that it makes sense to write ∑_{s}B_{is}A_{sj}.
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 is to
partition both matrices, formally multiply the blocks to get another block matrix and this one will
be BA partitioned. Before presenting this theorem, here is a simple lemma which is really a
special case of the theorem.
Lemma 3.5.1Consider the following product.
( )
0 ( )
|( I |) 0 I 0
0
where the first is n×r and the second is r ×n. The small identity matrix I is an r ×r matrix andthere are l zero rows above I and l zero columns to the left of I in the right matrix. Then theproduct 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 0 0 0 0 0
|( |( I |) 0 ⋅⋅⋅|( I |) 0 |( I |)e1 ⋅⋅⋅ |( I |) er |( I |) 0 ⋅⋅⋅ |( I |) 0 |)
0 0 0 0 0 0
which yields the claimed result. In the formula e_{j} refers to the column vector of length r which
has a 1 in the j^{th} position. ■
Theorem 3.5.2Let B be a q ×p blockmatrix as in 3.16and let A be a p×n block matrix as in3.16such that B_{is}is conformable with A_{sj}and each product, B_{is}A_{sj}for s = 1,
⋅⋅⋅
,p is of thesame size so they can be added. Then BA can be obtained as a block matrix such that the ij^{th}block is of the form
( ) ( )
( ) 0 ( ) 0
BisAsj = 0 Ir ×r 0 B |( Ip ×p |) 0 Ip ×p 0 A |( Iq ×q |)
i i s0 s s s j0 j
where here it is assumed B_{is} is r_{i}× p_{s} and A_{sj} is p_{s}× q_{j}. The product involves the s^{th} block in
the i^{th} row of blocks for B and the s^{th} block in the j^{th} column of A. Thus there are the same
number of rows above the I_{ps×ps} as there are columns to the left of I_{ps×ps} in those two inside
matrices. Then from Lemma 3.5.1
which equals the ij^{th} block of BA. Hence the ij^{th} block of BA equals the formal multiplication
according to matrix multiplication, ∑_{s}B_{is}A_{sj}. ■
Example 3.5.3Let an n×n matrix have the form A =
( )
a b
c P
where P is n−1×n−1.Multiply it by B =
( p q )
r Q
where B is also an n × n matrix and Q is n − 1 × n − 1.
You use block multiplication
( ) ( ) ( )
a b p q = ap + br aq+ bQ
c P r Q pc +P r cq+ P Q
Note that this all makes sense. For example, b = 1 × n − 1 and r = n − 1 × 1 so br is a 1 × 1.
Similar considerations apply to the other blocks.
Here is an interesting and significant application of block multiplication. In this
theorem, q_{M}
(t)
denotes the characteristic polynomial, det
(tI − M )
. The zeros of this
polynomial will be shown later to be eigenvalues of the matrix M. First note that from block
multiplication, for the following block matrices consisting of square blocks of an appropriate
size,
( ) ( ) ( )
A 0 A 0 I 0
B C = B I 0 C so
( ) ( ) ( )
A 0 A 0 I 0
det B C = det B I det 0 C = det(A)det(C)
Theorem 3.5.4Let A be an m × n matrix and let B be an n × m matrix for m ≤ n.Then
qBA (t) = tn−mqAB (t),
so the eigenvalues of BA and AB arethe same including multiplicities except that BA has n−mextra zero eigenvalues. Here q_{A}
(t)
denotes the characteristic polynomial of the matrixA.
Proof:Use block multiplication to write
( ) ( ) ( )
AB 0 I A AB ABA
B 0 0 I = B BA
( I A ) ( 0 0 ) ( AB ABA )
= .
0 I B BA B BA
( ) ( ) ( ) ( )
I A 0 0 = AB 0 I A
0 I B BA B 0 0 I
Therefore,
( ) −1( ) ( ) ( )
I A AB 0 I A = 0 0
0 I B 0 0 I B BA
Since the two matrices above are similar, it follows that
( ) ( )
0m×m 0 AB 0
B BA , B 0n×n
have the same characteristic polynomials. See Problem 8 on Page 252. Thus
( tI 0 ) ( tI − AB 0 )
det m ×m = det (3.18)
− B tI − BA − B tIn×n