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EE263
S. Lall
2009.10.08.01
Homework 3 Solutions
Due Thursday 10/15.
1.
Some basic properties of eigenvalues.
50045
Show that
(a) the eigenvalues of
A
and
A
T
are the same
(b)
A
is invertible if and only if
A
does not have a zero eigenvalue
(c) if the eigenvalues of
A
are
λ
1
,...,λ
n
and
A
is invertible, then the eigenvalues of
A

1
are 1
/λ
1
,...,
1
/λ
n
,
(d) the eigenvalues of
A
and
T

1
AT
are the same.
Hint:
you’ll need to use the facts that det
A
= det(
A
T
), det(
AB
) = det
A
det
B
, and, if
A
is invertible, det
A

1
= 1
/
det
A
.
Solution.
(a) The eigenvalues of a matrix
A
are given by the roots of the polynomial det(
sI
−
A
).
From determinant properties we know that det(
sI
−
A
) = det(
sI
−
A
)
T
= det(
sI
−
A
T
). We conclude that the eigenvalues of
A
and
A
T
are the same.
(b) First we recall that
A
is invertible if and only if det(
A
)
n
= 0. But det(
A
)
n
= 0
⇐⇒
det(
−
A
)
n
= 0.
i. If 0 is an eigenvalue of
A
, then det(
sI
−
A
) = 0 when
s
= 0. It follows that
det(
−
A
) = 0 and thus det(
A
) = 0, and
A
is not invertible. From this fact we
conclude that if
A
is invertible, then 0 is not an eigenvalue of
A
.
ii. If
A
is not invertible, then det(
A
) = det(
−
A
) = 0. This means that, for
s
= 0,
det(
sI
−
A
) = 0, and we conclude that in this case 0 must be an eigenvalue of
A
.
From this fact it follows that if 0 is not an eigenvalue of
A
, then
A
is invertible.
(c) From the results of the last item we see that 0 is not an eigenvalue of
A
. Now consider
the eigenvalue/eigenvector pair (
λ
i
,x
i
) of
A
. This pair satis±es
Ax
i
=
λ
i
x
i
. Now,
since
A
is invertible,
λ
i
is invertible. Multiplying both sides by
A

1
and
λ

1
i
we
have
λ

1
i
x
i
=
A

1
x
i
, and from this we conclude that the eigenvalues of the inverse
are the inverse of the eigenvalues.
(d) First we note that det(
sI
−
A
) = det(
I
(
sI
−
A
)) = det(
T

1
T
(
sI
−
A
)). Now, from
determinant properties, we have det(
T

1
T
(
sI
−
A
)) = det(
T

1
(
sI
−
A
)
T
). But this
is also equal to det(
sI
−
T

1
AT
), and the conclusion is that the eigenvalues of
A
and
T

1
AT
are the same.
2.
Tridiagonal Toeplitz matrices
0660
Some matrices have simple formulae for eigenvalues and eigenvectors. An example we
have seen is the circulant matrices. Another example is given by tridiagonal Toeplitz
matrices, as follows. Suppose
G
=
b a
c
b
a
c
b
a
.
.
.
a
c
b
is an
n
×
n
real matrix, and
a
n
= 0 and
c
n
= 0.
1
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View Full DocumentEE263
S. Lall
2009.10.08.01
(a) Suppose
x
=
x
1
x
2
.
.
.
x
n
is an eigenvector of
G
. Show that
cx
k

1
+
bx
k
+
ax
k
+1
=
λx
k
for all
k
= 1
,...,n
, where we deFne for convenience
x
0
= 0 and
x
n
+1
= 0. Hence
show that the solution
x
must be of the form
x
k
=
αr
k
1
+
βr
k
2
for some choice of
α,β,r
1
and
r
2
.
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