MATH 223, Linear Algebra
Winter, 2008
Solutions to Assignment 6
1. Suppose that
A
is an
m
×
n
matrix and
B
is an
n
×
m
matrix. [Hence,
both
AB
and
BA
are defined and square, but not necessarily of the same
size.] Show that
tr
(
AB
) =
tr
(
BA
).
If
C
and
D
are square matrices of the same size which are similar, show
that
tr
(
C
) =
tr
(
D
). [
tr
(
C
) is the trace of
C
, that is, the sum of the entries
on the diagonal.]
Solution: If the (
j, k
)entry of
A
is
a
j,k
and the (
k, j
)entry of
B
is
b
k,j
,
then the (
j, j
)entry of
AB
is
∑
n
k
=1
a
j,k
b
k,j
.
The trace of
AB
is then
∑
m
j
=1
∑
n
k
=1
a
j,k
b
k,j
.
Similarly, the (
k, k
)entry of
BA
is
∑
m
j
=1
b
k,j
a
k,j
and the trace of
BA
is
∑
n
k
=1
∑
m
j
=1
b
k,j
a
j,k
. Multiplication in the field is
commutative, so these two sums are the same (in a different order). That
is,
tr
(
AB
) =
tr
(
BA
).
Now if
C
and
D
are similar, say
D
=
P

1
CP
.
Taking
A
=
CP
and
B
=
P

1
,
AB
=
C
and
BA
=
D
. By the last paragraph,
C
and
D
have
the same trace.
2. For each of the following matrices
A
j
(
j
= 1
,
2
,
3) over the real numbers,
(a) find all the eigenvalues of
A
j
;
(b) find a basis for each eigenspace;
(c) find the minimal polynomial
min
A
j
and the characteristic polynomial
χ
A
j
;
(d) find — if possible — an invertible matrix
P
j
such that
P

1
j
A
j
P
j
is
invertible. If this is not possible, explain why.
(e) Find
A
11
1
. (2
11
= 2048 and 3
11
= 177147.)
A
1
=

7

5
5
4
2

1

6

6
7
, A
2
=
3

2
3
10

3
5
5

4
7
, A
3
=
6
4

1
4
6

1

4

4
3
.
Solution: We start with
A
1
for some reason.
det
(
λI

A
1
) =
det
λ
+ 7
5

5

4
λ

2
1
6
6
λ

7
=
(
λ
+7)
det
λ

2
1
6
λ

7
¶

5
det

4
1
6
λ

7
¶

5
det

4
λ

2
6
6
¶
=
(
λ
+ 7)[(
λ

2)(
λ

7)

6]

5[

4(
λ

7)

6]

5[

24

6(
λ

2)] =
(
λ
+ 7)(
λ
2

9
λ
+ 8)

5(

4
λ
+ 22)

5(

6
λ

12) =
λ
3

2
λ
2

5
λ
+ 6
.
1