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

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