Chapter 10
Canonical Forms
1. Show that if a triangular matrix is normal, then it must be diagonal.
Answer 10.1
We prove this by induction on
n
. Let
T
∈
C
n
×
n
be
normal and, without loss of generality, assume it is upper triangular.
For
n
=1
, the matrix
T
is simply a complex number, so it is diagonal.
Suppose
T
is diagonal for
1
≤
n < k
. Consider the case
n
=
k
. Since
T
is upper triangular, it has the form
±
α
v
H
0
S
²
for some
α
∈
C
,
v
∈
C
n

1
, and upper triangular
S
∈
C
(
n

1)
×
(
n

1)
. Since
T
is normal,
we have that
TT
H
=
±
α
v
H
0
S
²±
α
0
vS
H
²
=
±

α

2
+
v
H
vv
H
S
H
Sv
SS
H
²
=
T
H
T
=
±
α
0
H
α
v
H
0
S
²
=
±

α

2
αv
H
αv
vv
H
+
S
H
S
²
.
Thus

α

2
+
v
H
v
=

α

2
, which is true only if
v
=0
. Hence
H
=
S
H
S
, i.e.,
S
is normal. But
S
is also upper triangular, so by the induc
tion hypothesis,
S
is diagonal. Therefore,
T
=
±
α
0
0
S
²
is diagonal.
43
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CHAPTER 10. CANONICAL FORMS
Similarly, if
T
is lower triangular and normal, then it is diagonal.
Therefore for all positive integers
n
, if
T
∈
C
n
×
n
is triangular and
normal, then
T
is diagonal.
Note: This result can also be obtained directly by comparing diagonal
elements of
TT
H
and
T
H
T
.
2. Prove that if
A
∈
IR
n
×
n
is normal, then
N
(
A
)=
N
(
A
T
).
Answer 10.2
If
x
∈ N
(
A
)
, then
Ax
=0
. Hence,
A
T
Ax
and
x
T
A
T
Ax
. Since
A
is normal, we then have
x
T
AA
T
x
. But
this implies that
±
A
T
x
±
and hence
A
T
x
, i.e.,
x
(
A
T
)
.
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 Spring '11
 LAUB
 Determinant, Matrices, Normal matrix, positive definite

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