1.4 Properties of Matrix Operations
THM 1 (Properties of Matrix Addition) Let A, B, C and D be
m × n matrices.
(a) A + B = B + A.
(b) A + (B + C) = (A + B) + C.
(c) There is a unique m × n matrix O such that
A+O =A
for any A. The matrix O is called the additive identity or zero
matrix.
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(d) For each A, there is a unique matrix D such that
A + D = O.
Write D as (−A) and it is called the additive inverse or negative of A.
• Note: It can be shown that
−A = (−1)A.
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Proof.
(a) Since for 1 ≤ i ≤ m, 1 ≤ j ≤ n,
(A + B)ij = aij + bij = bij + aij = (B + A)ij ,
thus A + B = B + A.
(c) Let U = [uij ]m×n which satisfies A + U = A. Since for all
1 ≤ i ≤ m, 1 ≤ j ≤ n,
(A + U )ij = aij + uij = aij then uij = 0.
Thus, U = [0]m×n. Denote U by O.
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EX 1 (Ex. 2) The 2 × 3 zero matrix is
Ã
O=
If
Ã
A=
2 3 4
−4 5 −2
0 0 0
0 0 0
!
.
Ã
!
then (−A) =
−2 −3 −4
4 −5 2
!
,
and A + (−A) = O.
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THM 2 (Properties of Matrix Multiplication) A, B, C are matrices of appropriate sizes.
(a) A(BC) = (AB)C.
(b) A(B + C) = AB + AC.
(c) (A + B)C = AC + BC.
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Proof.
(a) Let A = [aij ]m×n, B = [bij ]n×p, C = [cij ]p×k . Then for 1 ≤ i ≤
m, 1 ≤ j ≤ k,
(A(BC))ij = rowi(A) · colj (BC).
Since
colj (BC) = Bcolj (C), thus (A(BC))ij = rowi(A)Bcolj (C).
Similarly, since
((AB)C)ij = rowi(AB) · colj (C),
rowi(AB) = rowi(A)B,
and
((AB)C)ij = rowi(A)Bcolj (C).
Thus A(BC) = (AB)C.
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EX 2 (Ex. 5) Let
Ã
A=
2 2 3
3 −1 2

!
Ã
and
Ã
AB + AC =
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

−1 2
0



0 .
2 , C =  1
2 −2
3 −1
1

, B= 2
Then
A(B + C) =

2 2 3
3 −1 2
15 1
7 −4
!
!
+


Ã
!
!
Ã
0 2
21 −1


 3 2 =
7 −2
5 −3
6 −2
0 2
Ã
=
21 −1
7 −2
!
.
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DEF 1 The scalar matrix



In = 


1 0 ··· 0
0 1 ··· 0 

...
. . . ... 

0 0 ··· 1
is called the identity matrix of order n.
• For any Am×n,
ImAm×n = Am×nIn = Am×n.
• Every scalar matrix Sn = rIn, for some r.
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• If A is a square matrix, p is a positive integer, then define
the powers of a matrix as
Ap = AA · · · A.
Moreover, define A0 = I.
• For nonnegative integers p and q, square matrix A,
ApAq = Ap+q , (Ap)q = Apq .
• Note that
(AB)p 6= ApB p, unless AB = BA.
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• There exist A, B 6= O but AB = O.
EX 3 (Ex. 7) If
Ã
A=
1 2
2 4
!
Ã
4 −6
−2 3
, B=
!
,
then A, B are not zero matrix, but
Ã
AB =
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0 0
0 0
!
.
10
• There exist A, B, C such that AB = AC but B 6= C.
EX 4 (Ex. 8) If
Ã
A=
1 2
2 4
!
Ã
, B=
2 1
3 2
then
Ã
AB = AC =
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!
Ã
, C=
8 5
16 10
−2 7
5 −1
!
!
.
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EX 5 (Ex. 9) Suppose R and S are manufacturers of a product.
Each year, R keeps (1/4) of its customers and (3/4) switch to
S. S keeps (2/3) of its customers and (1/3) switch to R. Let R
be the company 1 and S be the company 2. Define the matrix A
with entry (i,j), aij = the chance of the customers of company
j switch to company i. Then
Ã
A=
1/4 1/3
3/4 2/3
!
.
If the initial distribution of the market is
Ã
x0 =
3
5
2
5
!
,
then one year later, the distribution of the market is
Ã
x1 = Ax0 =
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1
4
3
4
1
3
2
3
!Ã
3
5
2
5
!
Ã
=
3 ) + ( 1 )( 2 )
)(
(1
3 5
4 5
2)
3
3
)(
( 4 )( 5 ) + ( 2
3 5
!
Ã
=
17
60
43
60
!
.
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Similarly, at the end of 2 years, the distribution of the market
will be
x2 = Ax1 = A(Ax0) = A2x0.
If
Ã
x0 =
a
b
!
, where a + b = 1.
Find a, b such that the distribution will be the same from year to
year. That is,
x0 = Ax0 = A2x0 = A3x0 = · · · .
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x0 = Ax0 implies x0 = Apx0 for any integer p > 0, it suffices to
find the solution of Ax0 = x0.
Since
Ã
Ax0 = x0 ⇔
1
4
3
4
1
3
2
3
!Ã
a
b
!
Ã
=
a
b
!
.
Then
2
3
1
1
a + b = b.
a + b = a,
3
4
3
4
Subsequently, it can be found that
9
4
.
, b=
13
13
Note : This is an example of a Markov Chain.
a=
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THM 3 (Properties of Scalar Multiplication) If r, s are real numbers and A, B are matrices, then
(a) r(sA) = (rs)A.
(b) (r + s)A = rA + sA.
(c) r(A + B) = rA + rB.
(d) A(rB) = r(AB) = (rA)B.
Recall that by definition (rA)ij = raij .
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THM 4 (Properties of Transpose) If r is a real number and A, B
are matrices, then
(a) (AT )T = A.
(b) (A + B)T = AT + B T .
(c) (AB)T = B T AT .
(d) (rA)T = rAT .
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Proof.
(c) Let A = [aij ]m×p, B = [bij ]p×n, C = AB = [cij ]m×n and
(AB)T = [cT
ij ]n×m . Since
cT
ij = cji = rowj (A) · coli (B)
= aj1b1i + aj2b2i + · · · + ajpbpi
T + aT bT + · · · + aT bT
b
= aT
pj ip
2j i2
1j i1
T + bT aT + · · · + bT aT
a
= bT
ip pj
i2 2j
i1 1j
= rowi(B T ) · colj (AT ) = (B T AT )ij .
Thus, (AB)T = B T AT .
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EX 6 (Ex. 11) Let
Ã
A=
1 3 2
2 −1 3
Then
Ã
(AB)T =
and
Ã
B T AT =
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!


0 1


and B =  2 2  .
3 −1
12 7
5 −3
!
0 2 3
1 2 −1
,
!


!
Ã
1 2
12 7


.
 3 −1  =
5 −3
2 3
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DEF 2 A = [aij ] is symmetric if AT = A.
• A is symmetric if it is a square matrix with aij = aji.
• If A is symmetric, the elements of A are symmetric with
respect to the main diagonal of A.
EX 7 (Ex. 12) The matrices




1 0 0
1 2 3




A =  2 4 5  and I3 =  0 1 0 
0 0 1
3 5 6
are symmetric.
EXERCISES. 9, 16, 19, T.1, T.7, T.8, T.23, T.32
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