Characterization of Normal Operators
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Theorem
Let $\GF \in \set {\R, \C}$.
Let $\HH$ be a Hilbert space over $\GF$.
Let $A$ be a bounded linear operator on $\HH$.
Then the following are equivalent:
- $(1): \quad A A^* = A^* A$, that is, $A$ is normal
- $(2): \quad \forall h \in H: \norm {A h}_\HH = \norm {A^*h}_\HH$
where:
- $A^*$ denotes the adjoint of $A$
- $\norm {\, \cdot\,}_\HH$ denotes the inner product norm of $\HH$
If $\GF = \C$, these are also equivalent to:
- $(3): \quad \map \Re A \map \Im A = \map \Im A \map \Re A$, that is, the real and imaginary parts of $A$ commute.
Proof
$(3)$ equivalent to $(1)$
Suppose $\GF = \C$.
We have:
\(\ds \map \Re A \map \Im A\) | \(=\) | \(\ds \paren {\frac 1 2 \paren {A + A^\ast} } \paren {\frac 1 {2 i} \paren {A - A^\ast} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {4 i} \paren {A + A^\ast} \paren {A - A^\ast}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {4 i} \paren {A^2 + A^\ast A - A A^\ast - \paren {A^\ast}^2}\) |
and:
\(\ds \map \Im A \map \Re A\) | \(=\) | \(\ds \paren {\frac 1 {2 i} \paren {A - A^\ast} } \paren {\frac 1 2 \paren {A + A^\ast} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {4 i} \paren {A - A^\ast} \paren {A + A^\ast}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {4 i} \paren {A^2 - A^\ast A + A A^\ast - \paren {A^\ast}^2}\) |
So $\map \Re A \map \Im A = \map \Im A \map \Re A$ if and only if
- $A^\ast A - A A^\ast = -A^\ast A + A A^\ast$
This is equivalent to:
- $A^\ast A = A A^\ast$
$\Box$
$(1)$ implies $(2)$
We have:
\(\ds \norm {A h}_\HH\) | \(=\) | \(\ds \sqrt {\innerprod {A h} {A h}_\HH}\) | Definition of Inner Product Norm | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\innerprod h {A^\ast A h}_\HH}\) | Definition of Adjoint Linear Transformation | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\innerprod h {A A^\ast h}_\HH}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\innerprod {A^\ast h} {A^\ast h}_\HH}\) | Adjoint is Involutive | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {A^\ast h}_\HH\) |
$\Box$
$(2)$ implies $(1)$
As above, we have:
- $\norm {A h}_\HH = \sqrt {\innerprod h {A^\ast A h}_\HH}$
So from Adjoint is Involutive, we have:
- $\norm {A h}_\HH = \sqrt {\innerprod h {A A^\ast h}_\HH}$
So from Inner Product is Sesquilinear, we have:
- $\innerprod h {\paren {A^\ast A - A A^\ast} h}_\HH = 0$
We have:
\(\ds \paren {A^\ast A - A A^\ast}^\ast\) | \(=\) | \(\ds \paren {A^\ast A}^\ast - \paren {A A^\ast}^\ast\) | Adjoining is Linear | |||||||||||
\(\ds \) | \(=\) | \(\ds A^\ast \paren {A^\ast}^\ast - \paren {A^\ast}^\ast A^\ast\) | Adjoint of Composition of Linear Transformations is Composition of Adjoints | |||||||||||
\(\ds \) | \(=\) | \(\ds A^\ast A - A A^\ast\) | Adjoint is Involutive |
So $A^\ast A - A A^\ast$ is Hermitian and:
- $\innerprod {\paren {A^\ast A - A A^\ast} h} h_\HH = 0$
for each $h \in \HH$.
So by Norm of Hermitian Operator: Corollary, we have:
- $A^\ast A = A A^\ast$
$\blacksquare$
Sources
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next) $II.2.16$