# Characterization of Normal Operators

## 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:

$A^\ast A = A A^\ast$

$\blacksquare$