Partial Solution Set, Leon
§
6.3
6.3.1a
We are to ﬁnd a matrix
X
that diagonalizes
A
=
±
0 1
1 0
²
.
The eigenvalues of
A
in
this case are
λ
1
= 1 and
λ
2
=

1. By Theorem 6.3.2, such a matrix
X
exists, and
the columns of
X
are eigenvectors of
A
corresponding to the eigenvalues. In the usual
fashion, we ﬁnd eigenvectors
x
1
= (1
,
1)
T
and
x
2
= (1
,

1)
T
. So
X
=
±
1
1
1

1
²
, and
X

1
=
1
2
±
1
1
1

1
²
. It is easy to verify that
X

1
AX
=
D
=
±
1
0
0

1
²
, so
A
=
XDX

1
.
6.3.1b
As in (a), but with
A
=
±
5
6

2

2
²
. The eigenvalues are
λ
1
= 1 and
λ
2
= 2.
An eigenvector associated with
λ
1
is
x
1
= (

3
,
2)
T
, and an eigenvector associated with
λ
2
is
x
2
= (

2
,
1)
T
. So
X
=
±

3

2
2
1
²
, and
X

1
=
±
1
2

2 13
²
. As expected,
D
=
X

1
AX
=
±
1 0
0 2
²
.
6.3.1d
As in (a) and (b), but now
A
=
2 2
1
0 1
2
0 0

1
. The eigenvalues are
λ
1
= 2
, λ
2
= 1,
and and
λ
3
=

1. The associated eigenvectors are (any scalar multiples of)
x
1
= (1
,
0
,
0)
T
,
x
2
= (

2
,
1
,
0)
T
, and
x
3
= (1
,

3
,
3)
T
. So
X
=
1

2
1
0
1

3
0
0
3
, and
X

1
=
1 2
5
3
0 1 1
0 0
1
3
.
6.3.4a
Given
A
=
±
2
1

2

1
²
, ﬁnd a matrix
B
such that
B
2
=
A
. In other words, ﬁnd a
square root for
A
. This is easily done using diagonalization, i.e., once we factor