4.21. Ex 2
Prove of each of the following statements about nn matrices, or exhibit a counter
example.
(a)
(b)
(c)
(d)
(e)
If AB  BA  0 , then A2 B3  B3 A2 .
If A and B are nonsingular, then A  B is nonsingular.
If A and B are nonsingular, then AB is nonsingular.
If the product AB is nonsingular, then A and B are nonsingular.
If A, B and A  B are nonsingular, then A  B is nonsingular.
(f) If A2  O , then A  O .
(g) If A2  O , then A  I is nonsingular.
(h) If A2  O , then A  I is nonsingular.
 A  2I 
2
 O , then A is nonsingular.
(i)
If
(j)
If AB  BA , then ABk  Bk A for every integer k  1 .
Solution
(a)
If AB  BA  0 , then AB   BA . Hence,
A2 B  AAB  A  BA   ABA    BA A  BA2
Assuming A2 Bn1  Bn1 A2 , we can right multiply by B to get
A2 Bn  Bn1 A2 B  Bn1BA2  B n A2
Hence, by induction, A2 B n  B n A2 for all n.
In particular, setting n  3 , we have
AB B A
2
3
3
2
(b)
Not true.
Counter example:
Let B   A . Obviously, B is nonsingular if A is.
However, A  B  A  A  O is
singular.
(c)
True.
Proof:
If A and B are nonsingular, then both A1 and B 1 exist. Consider then
B
1
A1   AB   B 1 A1 AB  BB 1  I
Hence, B 1 A1 is the left inverse of AB.
AB. Hence AB is nonsingular.
By theorem 4.20, it is also the inverse of
(d)
True.
Proof:
If the product AB is nonsingular, then
I   AB 
Hence,
 AB 
1
1
 AB    AB 
1
 AB 
1
exists so that
A B

A is the inverse of B so that B is nonsingular.
argument to I   AB  AB  , we see that B  AB 
1
1
Appying the same
is the inverse of A so that A is
nonsingular.
(e)
Statement is false.
Counter example:
Let A  B be nonsingular. Thus A1  B 1 exists.
Thus, the inverse of A  B  2 A exists and equal to
1 1
A .
2
Hence A  B is
nonsingular.
But A  B  A  A  O is obviously singular.
(f)
Statement is false.
Counter example:
a b
Consider a 2  2 matrix A  
. The condition A2  O requires

c d
a 2  bc  0
ab  bd  0
bc  d 2  0
ca  dc  0
Only 2 of these equations are independent:
bc  a 2
and
(1)
a  d
Thus, any A  O with elements satisfying (1) will satisfy A2  O . One example
is
4 
 10
A

 25 10 
Note that A  O must be singular since if A1 exists, then A2  O implies
A1 A2  A1O

AO
(g)
Consider the identity
 I  A A  I   I  A2
Hence, if A2  O , then I  A is the left inverse of A  I . By theorem 4.20,
A  I is nonsingular.
(h)
Consider the identity
A
2
 A  I   A  I   A3  I
Hence, if A3  O , then   A2  A  I  is the left inverse of A  I . By theorem
4.20, A  I is nonsingular.
(i)
 A  2I 
2
 O implies
A2  4 A  4 I  O
(1)
1
Assuming A exists, then A1  (1) gives
A  4 I  4 A1  O
or
A1  
1
 A  4I 
4
(2)
Thus, by construction according to (2), A1 indeed exists so that A is nonsingular.
(j)
We shall prove it by induction. The case n  1, or AB  BA , is given. Assuming
AB k 1  B k 1 A , we can right multiply by B to get
ABk  Bk 1 AB  Bk 1BA  Bk A
QED.