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.

$(3)$ equivalent to $(1)$
Suppose $\GF = \C$.

We have:

and:

So $\map \Re A \map \Im A = \map \Im A \map \Re A$


 * $A^\ast A - A A^\ast = -A^\ast A + A A^\ast$

This is equivalent to:


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

$(1)$ implies $(2)$
We have:

$(2)$ 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:

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:


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