MATH 223, Linear Algebra
Fall, 2007
Solutions to Assignment 3
1. Suppose that
A
and
B
are invertible matrices of the same size.
(a) Show that
B
(
A

1
+
B

1
) = (
A
+
B
)
A

1
.
Solution: Easy. By distributivity, the lefthand side is
BA

1
+
BB

1
=
BA

1
+
I
and the righthand side is
AA

1
+
BA

1
=
I
+
BA

1
. By commutativity of +, the sides are equal.
(b) Suppose that
A
+
B
is also invertible. Show that
A

1
+
B

1
is
invertible, and its inverse is
A
(
A
+
B
)

1
B
.
Solution: Being a product of invertible matrices,
A
(
A
+
B
)

1
B
is
invertible, and its inverse is
B

1
((
A
+
B
)

1
)

1
A

1
. This equals
B

1
(
A
+
B
)
A

1
= (
B

1
A
+
I
)
A

1
=
B

1
AA

1
+
IA

1
=
B

1
+
A

1
=
A

1
+
B

1
.
So
A

1
+
B

1
is invertible and its inverse is
A
(
A
+
B
)

1
B
.
2. (a) Let
i
=
0
1
0
0

1 0
0
0
0
0
0
1
0
0

1 0
, a matrix over the reals. Show that
i
2
=

I
, that the set
C
=
{
aI
+
bi
:
a,b
∈ R}
is a subspace of
M
4
(
R
)
(regarded as a vector space over
R
),
C
is closed under matrix multi
plication, and that every nonzero matrix in
C
has an inverse, which
is also in
C
. (We can identify
C
with the ﬁeld of complex numbers
in a pretty obvious manner. Reminder:
M
4
(
R
)) is the collection of
all 4
×
4 matrices with real entries.)
Solution: I’ll skip the direct and easy calculation that
i
2
=

I
. The
zero matrix is in
C
since we can take
a
=
b
= 0. Now if we have two
elements of
C
, say they are
a
1
I
+
b
1
i
and
a
2
I
+
b
2
i
(with
a
1
,
b
1
,
a
2
and
b
2
real); their sum is (
a
1
+
a
2
)
I
+ (
b
1
+
b
2
)
i
by basic properties
of matrix algebra, and this is in
C
. So
C
is closed under addition.
If
α
∈ R
and
aI
+
bi
∈
C
, then
α
(
aI
+
bi
) = (
αa
)
I
+ (
αb
)
i
, also an
element of
C
, so
C
is closed under scalar multiplication. So it’s a
subspace of
M
4
(
R
).
Next, suppose that
a
1
I
+
b
1
i
and
a
2
I
+
b
2
i
are elements of
C
. Using
the basic rules, we get
(
a
1
I
+
b
1
i
)(
a
2
I
+
b
2
i
) = (
a
1
I
+
b
1
i
)(
a
2
I
) + (
a
1
I
+
b
1
i
)(
b
2
i
)
= (
a
1
I
)(
a
2
I
)+(
b
1
i
)(
a
2
I
)+(
a
1
I
)(
b
2
i
)+(
b
1
i
)(
b
2
i
) =
a
1
a
2
I
+
b
1
a
2
i
+
a
1
b
2
i
+
b
1
b
2
i
2
= (
a
1
a
2

b
1
b
2
)
I
+ (
b
1
a
2
+
a
1
b
2
)
i,
1