Download AB− BA = A12B21 − A21B12 A11B12 + A12B22 − A12B11

Selected Solutions
Math 420
Homework 2
1/20/12
1.5.8 Let
C11 C12
C=
C21 C22
be a 2 × 2 matrix. We inquire when it is possible to find 2 × 2 matrices A and B such that
C = AB − BA. Prove that such matrices can be found if and only if C11 + C22 = 0.
It is easy to see that if
A11 A12
A=
A21 A22
B11 B12
and B =
,
B21 B22
then
A12 B21 − A21 B12
A11 B12 + A12 B22 − A12 B11 − A22 B12
AB − BA =
.
A21 B11 + A22 B21 − A11 B21 − A21 B22
A21 B12 − A12 B21
If there exist 2 × 2 matrices A and B such that C = AB − BA, then we have
C11 + C22 = ( A12 B21 − A21 B12 ) + ( A21 B12 − A12 B21 ) = 0.
Conversely, suppose C11 + C22 = 0, so we have
C11 C12
C=
.
C21 −C11
If C21 6= 0, let
"
A=
1
1
− C11C+21C12
#
"
and B =
0
C11 +C12
C21
1
C12 −
1 − C21
0
#
.
If C21 = 0 but C12 6= 0, let
"
A=
If C21 = C12 = 0, let
1
− CC11
12
1
0
#
"
and B =
0
C11
A=
−C11 0
#
1 + C12
.
0
1
11
− CC12
0 1
and B =
.
1 0
In each case, it is easy to check that AB − BA = C.
1.6.6 Suppose A is a 2 × 1 matrix and that B is a 1 × 2 matrix. Prove that C = AB is not
invertible.
1
Let
Then, we have
a
A = 1 and B = b1 b2 .
a2
a1 b1 a1 b2
C = AB =
.
a2 b1 a2 b2
If a1 = 0, the first row of C is all zeroes, so C is not invertible. Otherwise, subtracting aa12
times the first row from the second row gives a second row of all zeroes, and again this
implies C is not invertible.
1.6.7 Let A be an n × n (square) matrix. Prove the following two statements:
(a) If A is invertible and AB = 0 for some n × n matrix B, then B = 0.
(b) If A is not invertible, then there exists an n × n matrix B such that AB = 0 but
B 6= 0.
(a) If A in invertible and AB = 0, then B = A−1 AB = A−1 0 = 0.
(b) If A is not invertible, the system AX = 0 has a nontrivial solution X0 . Let B be the
n × n matrix whose columns are all equal to X0 . Then B 6= 0 but AB = 0.
1.6.10 Prove the following generalization of Exercise 6. If A is an m × n matrix, B is an
n × m matrix and n < m, then AB is not invertible.
The homogeneous system of linear equations given by BX = 0 consists of n equations in
m unknowns. Since n < m, this system will always have a nontrivial solution. Let X0 be
a nontrivial solution. Then ( AB) X = A( BX ) = A(0) = 0, so X0 is a nontrivial solution to
( AB) X = 0. Since a nontrivial solution to this system exists, Theorem 13 implies that AB
is not invertible.
2.1.3 If C is the field of complex numbers, which vectors in C3 are linear combinations
of (1, 0, −1), (0, 1, 1), and (1, 1, 1)?
It is easy to check, by performing elementary row operations, that the matrix


1 0 1
A : =  0 1 1
−1 1 1
is invertible. Thus, the system AX = Y has a solution for each 3 × 1 matrix Y. Since the
column Y is a linear combination of the columns of A (Exercise 1.6.6), this means that any
vector in C3 is a linear combination of the given vectors.
2
2.1.4 Let V be the set of all pairs ( x, y) of real numbers, and let F be the field of real
numbers. Define
( x, y) + ( x1 , y1 ) = ( x + x1 , y + y1 )
c( x, y) = (cx, y).
Is V, with these operations, a vector space over the field of real numbers?
No, V is not a vector space with these operations, as the distributive proberty 4(d) fails:
(1 + 1)(0, 1) = 2(0, 1) = (2(0), 1) = (0, 1),
but
1(0, 1) + 1(0, 1) = (0, 1) + (0, 1) = (0 + 0, 1 + 1) = (0, 2).
2.1.5 On Rn , define two operations
α⊕β = α−β
c · α = −cα.
The operations on the right are the usual ones. Which of the axioms for a vector space
are satisfied by (Rn , ⊕, ·)?
Consider each of the axioms.
3(a) This axiom fails. Addition is not commutative, since α − β 6= β − α whenever α 6= β.
3(b) This axiom fails. Addition is not associative. For example, take α = β = 0 and
γ 6= 0. Then (α ⊕ β) ⊕ γ = −γ, but α ⊕ ( β ⊕ γ) = γ.
3(c) This axiom does hold. Taking 0 to be the usual zero vector in Rn , we have α ⊕ 0 = α
for all α ∈ Rn , and this vector is unique since α − β = α in Rn implies β = α − α = 0.
3(d) This axiom also holds. We define −α = α for all α ∈ Rn . Then α ⊕ (−α) = α − α = 0.
Uniqueness easily follows from uniqueness of additive inverses in Rn (with usual
operations).
4(a) This axiom fails, as 1 · α = −α 6= α for nonzero α.
4(b) This axiom fails, as (c1 c2 ) · α = −c1 c2 α, but c1 · (c2 · α) = c1 c2 α, and these are not
equal when c1 , c2 , and α are nonzero.
4(c) This axiom holds, since c · (α ⊕ β) = −c(α − β) = −cα − (−cβ) = c · α ⊕ c · β.
4(d) This axiom fails. For example, for α 6= 0, (1 + 1) · α = −2α 6= 0, but 1 · α ⊕ 1 · α =
−α − (−α) = 0.
3