EE 221a Homework 5 Solutions
1
Fall 2007
Problem 1.
Since
f
(
s
) =
e
s
is an analytic function, the eigenvalues of
e
A
are
e
λ
i
, i
= 1
,
2
, . . . , n
, where
λ
i
’s are eigenvalues of
A
. So,
det(
e
A
) =
n
i
=1
e
λ
i
= 0
Problem 2.
∀
v
∈
R
n
, Av
∈
N
(
A
), that is,
A
2
v
=
θ,
∀
v
. So,
A
2
= 0.
The statement is wrong.
A
1
= 0 is a solution of
A
2
= 0, and
A
2
=
0
1
. . .
0
0
0
. . .
0
. . .
. . .
. . .
. . .
0
0
. . .
0
is also a solution. Even with change of basis,
A
1
is not
A
2
.
Problem 3.
The minimum polynomial is
s
3
, so the eigenvalues of
A
are all
zeros, and the largest Jordan block has size 3. Write
A
=
V JV

1
, where
J
is
in Jordan form. Clearly
rank
(
A
) =
rank
(
J
). The
J
which has the lowest rank
is
J
=
0
1
0
1
0
0
0
0
The
J
which has the highest rank is
J
=
0
1
0
1
0
0
1
0
1
0
So, 2
≤
rank
(
J
)
≤
4.
1
Thanks to Onureena Banerjee for providing part of the solution
1
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Problem 4.
f
(
s
) =
cos
(
e
s
).
So
f
(
s
) =

sin
(
e
s
)
e
s
, f
(
s
) =

cos
(
e
s
)
e
2
s

sin
(
e
s
)
e
s
.
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 Fall '10
 ClaireTomlin
 Linear Algebra, Characteristic polynomial, Zagreb, Highways in Croatia, Eλi

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