This preview shows pages 1–3. Sign up to view the full content.
MATH235 Fall 2007
Assignment 3. Solutions.
(32 marks total)
1. Let
A
be an nxn matrix. Suppose that
λ
1
,...,λ
k
are the (distinct) eigenvalues
of
A
and that their multiplicites are
m
λ
1
,...,m
λ
k
respectively. Then, prove that
det
(
A
) =
k
Y
i
=1
λ
m
λ
i
i
Solution.
(4 marks)
The characteristic polynomial of
A
is given by:
p
(
x
) =
det
(
A

xI
) = (
λ
1

x
)
m
λ
1
(
λ
2

x
)
m
λ
2
···
(
λ
k

x
)
m
λ
k
By letting
x
= 0 we get
det
(
A
) =
λ
m
λ
1
1
λ
m
λ
2
2
λ
m
λ
k
k
2. Let A be a 2x2 matrix. Prove that the characteristic polynomial of
A
is
λ
2

trace
(
A
)
λ
+
det
(
A
)
Solution.
(4 marks)
Let
A
=
±
a
11
a
12
a
21
a
22
¶
.
Then the characteristic polynomial of
A
is given by
p
(
λ
) =
det
(
A

λI
) = (
a
11

λ
)(
a
22

λ
)

a
21
a
12
=
λ
2

(
a
11
+
a
22
)
λ
+ (
a
11
a
22

a
21
a
12
) =
λ
2

trace
(
A
)
λ
+
det
(
A
)
3. Let
A
be a 5x5 matrix. Suppose we are given the following facts about
A
:
trace
(
A
) = 10, 2 is an eigenvalue of
A
with multiplicity 3, and
A
is not invertible.
Prove that this information completely determines the characteristic polynomial of
A
, and state the characteristic polynomial of
A
.
Solution.
(4 marks)
A
is not invertible if and only if it has determinant zero. It follows from question
1 that
det
(
A
) = 0 if and only if at least one of the eigenvalues is zero. Thus we
already know four out ﬁve eigenvalues of
A
:
λ
1
= 2
,λ
2
= 2
3
= 2
4
= 0
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentSo we need to ﬁnd the last one. Using the fact that the trace of
A
is equal to the
sum of the eigenvalues of
A
(counted with multiplicities), we get
λ
1
+
λ
2
+
λ
3
+
λ
4
+
λ
5
= 10
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 CELMIN
 Math, Linear Algebra, Algebra

Click to edit the document details