Prof. Lydia Bieri
Solution Set 9
Posted: January 9th
Written by: Luca Candelori
Exercise 1
(7.1/2b)
.
The eigenvalues are
λ
= 4
,

1, each with multiplicity 1, with corresponding eigenvectors
v
1
=
(2
,
3) and
v
2
= (1
,

1) respectively. Each vector forms a basis for its generalized eigenspace. The
matrix is diagonalizable with Jordan Form:
J
=
±
4
0
0

1
²
Exercise 2
(7.1/3c)
.
Answers in the back of the book.
Exercise 3
(7.2/4b)
.
The eigenvalues are
λ
= 2 with multiplicity 2 and
λ
= 1 with multiplicity 1. A basis for the
generalized eigenspace of
λ
= 2 is given by the vectors
v
1
= (

1
,
0
,
2) and
v
2
= (1
,
2
,
0). A basis for
the generalized eigenspace of
λ
= 1 is given by the vector
v
3
= (1
,
2
,
1). In both cases the geometric
multiplicity equals the algebraic multiplicity, therefore the matrix is diagonalizable with
Q
=

1 1 1
0
2 2
2
0 1
and
J
=
2 0 0
0 2 0
0 0 1
Exercise 4
(7.2/5a)
.
Answers in the back of the book
Exercise 5
(6.1/20, 6.2/4)
.
(a) Note ﬁrst that
k
x
+
y
k
2
=
h
x
+
y,x
+
y
i
. By linearity of the inner product, we get
1
4
k
x
+
y
k
2

1
4
k
x

y
k
2
=
1
4
(
h
x
+
y,x
+
y
i  h
x

y,x

y
i
)
=
1
4
(
h
x,x
i
+ 2
h
x,y
i
+
h
y,y
i  h
x,x
i
+ 2
h
x,y
i  h
y,y
i
)
=
1
4
(4
h
x,y
i
) =
h
x,y
i
An entirely analogous process, though a little longer, will prove the second identity.
1