Course 18.06, Fall 2002: Quiz 3, Solutions
1
(a) One eigenvalue of
A
=
ones
(5) is
λ
1
= 5, corresponding to the eigenvector
x
1
= (1
,
1
,
1
,
1
,
1).
Since the rank of
A
is 1, all the other eigenvalues
λ
2
, . . . , λ
5
are zero. Check: The trace
of
A
is 5.
(b) The initial condition
u
(0) can be written as a sum of the two eigenvectors
x
1
= (1
,
1
,
1
,
1
,
1)
and
x
2
= (

1
,
0
,
0
,
0
,
1), corresponding to the eigenvalues
λ
1
= 5 and
λ
2
= 0:
u
(0) = (0
,
1
,
1
,
1
,
2) = (1
,
1
,
1
,
1
,
1) + (

1
,
0
,
0
,
0
,
1) =
x
1
+
x
2
.
The solution to
d
u
dt
=
A
u
is then
u
(
t
) =
c
1
e
λ
1
t
x
1
+
c
2
e
λ
2
t
x
2
= (1
,
1
,
1
,
1
,
1)
e
5
t
+ (

1
,
0
,
0
,
0
,
1)
.
(c) The eigenvectors of
B
=
A

I
are the same as for
A
, and the eigenvalues are smaller by
1:
B
x
= (
A

I
)
x
=
A
x

x
=
λ
x

x
= (
λ

1)
x
,
where
x
, λ
are an eigenvector and an eigenvalue of
A
.
The eigenvalues of
B
are then
4
,

1
,

1
,

1
,

1, the trace is
∑
i
λ
i
= 0, and the determinant is
i
λ
i
= 4.
2
(a)
B
is similar to
A
when
B
=
M

1
AM
, with
M
invertible. The exponential of
A
is
e
A
=
I
+
A
+
1
2
A
2
+
1
6
A
3
+
· · ·
.
Every power
B
k
of
B
is similar to the same power
A
k
of
A
:
B
k
=
M

1
AMM

1
AM
· · ·
M

1
AM
=
M

1
A
k
M.