Diagonalization of Normal Operators
April 18, 2007
Here is a proof of the diagonalizability of normal operators which does
not use Schur’s theorem.
Lemma
Let
T
:
V
→
W
be a linear operator. Assume that its adjoint
T
*
exists. Then
Ker
(
T
) =
Im
(
T
*
)
⊥
. If
V
is ﬁnite dimensional, then
Ker
(
T
)
⊥
=
Im
(
T
*
).
Proof:
Recall that an element
w
of
W
is zero if and only if (
w

w
0
) = 0
for all
w
0
∈
W
. Thus
x
∈
Ker
(
T
) if and only if (
Tx

y
) = 0 for all
y
∈
W
,
i.e.
, if and only if (
x

T
*
y
) = 0 for all
y
∈
W
,
i.e.
, if and only if
x
is
orthogonal to the image of
T
*
. This proves the ﬁrst statement. It follows
that (
Ker
(
T
))
⊥
= (
Im
(
T
*
)
⊥
)
⊥
, and the latter is
Im
(
T
*
) when
V
is ﬁnite
dimensional.
Lemma
Let
A
and
B
be linear operators on
V
. If
A
and
B
commute,
then
Ker
(
A
) and
Im
(
A
) are
B
invariant.
Proof:
If
x
∈
Ker
(
A
), then
AB
(
x
) =
BA
(
x
) = 0, so
B
(
x
)
∈
Ker
(
A
).
Thus
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 Spring '08
 GUREVITCH
 Linear Algebra, Ker, Normal Operators

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