Operator is Hermitian iff Inner Product is Real

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Theorem

Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ be a Hilbert space over $\C$.

Let $A : \HH \to \HH$ be a bounded linear operator.


Then $A$ is Hermitian if and only if:

$\forall h \in \HH: \innerprod {A h} h_\HH \mathop \in \R$


Proof

Necessary Condition

Suppose that $A$ is Hermitian.

Then:

$A^* = A$

where $A^*$ denotes the adjoint of $A$.

Let $x \in \HH$.

Then, by the definition of the adjoint, we have:

$\innerprod {A x} x_\HH = \innerprod x {A x}_\HH$

From the conjugate symmetry of the inner product, we have:

$\innerprod x {A x}_\HH = \overline {\innerprod {A x} x_\HH}$

So:

$\innerprod {A x} x_\HH = \overline {\innerprod {A x} x_\HH}$

So, from Complex Number equals Conjugate iff Wholly Real:

$\innerprod {A x} x_\HH$ is a real number.

$\Box$


Sufficient Condition

Suppose that:

$\innerprod {A x} x_\HH$ is a real number for each $x \in \HH$.

Let $\alpha \in \C$.

Let $x, y \in \HH$.

We have:

\(\ds \innerprod {\map A {x + \alpha y} } {x + \alpha y}_\HH\) \(=\) \(\ds \innerprod {A x + \alpha A y} {x + \alpha y}_\HH\) Definition of Linear Transformation
\(\ds \) \(=\) \(\ds \innerprod {A x} {x + \alpha y}_\HH + \alpha \innerprod {A y} {x + \alpha y}_\HH\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds \innerprod {A x} x_\HH + \overline \alpha \innerprod {A x} y_\HH + \alpha \innerprod {A y} x_\HH + \alpha \overline \alpha \innerprod {A y} y_\HH\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds \innerprod {A x} x_\HH + \overline \alpha \innerprod {A x} y_\HH + \alpha \innerprod {A y} x_\HH + \size \alpha^2 \innerprod {A y} y_\HH\) Product of Complex Number with Conjugate

We have that:

$\innerprod {\map A {x + \alpha y} } {x + \alpha y}_\HH$

is a real number.

Note that both:

$\innerprod {A x} x_\HH$

and:

$\size \alpha^2 \innerprod {A y} y_\HH$

are also real numbers.

So, we must have that:

$\overline \alpha \innerprod {A x} y_\HH + \alpha \innerprod {A y} x_\HH$

is a real number.

We therefore have:

\(\ds \overline \alpha \innerprod {A x} y_\HH + \alpha \innerprod {A y} x_\HH\) \(=\) \(\ds \overline {\overline \alpha \innerprod {A x} y_\HH + \alpha \innerprod {A y} x_\HH}\) Complex Number equals Conjugate iff Wholly Real
\(\ds \) \(=\) \(\ds \overline {\overline \alpha \innerprod {A x} y_\HH} + \overline {\alpha \innerprod {A y} x_\HH}\) Sum of Complex Conjugates
\(\ds \) \(=\) \(\ds \alpha \overline {\innerprod {A x} y_\HH} + \overline \alpha \overline {\innerprod {A y} x_\HH}\) Product of Complex Conjugates
\(\ds \) \(=\) \(\ds \alpha \innerprod y {A x}_\HH + \overline \alpha \innerprod x {A y}_\HH\) conjugate symmetry of inner product
\(\ds \) \(=\) \(\ds \alpha \innerprod {A^* y} x_\HH + \overline \alpha \innerprod {A^* x} y_\HH\) Definition of Adjoint Linear Transformation, Adjoint is Involutive

Setting $\alpha = 1$, we have:

$\innerprod {A x} y_\HH + \innerprod {A y} x_\HH = \innerprod {A^* y} x_\HH + \innerprod {A^* x} y_\HH$

Setting $\alpha = i$, we have:

$-i \innerprod {A x} y_\HH + i \innerprod {A y} x_\HH = i \innerprod {A^* y} x_\HH - i \innerprod {A^* y} x_\HH$

Dividing by $i$ we have:

$-\innerprod {A x} y_\HH + \innerprod {A y} x_\HH = \innerprod {A^* y} x_\HH - \innerprod {A^* y} x_\HH$

So:

$2 \innerprod {A y} x_\HH = 2 \innerprod {A^* y} x_\HH$

giving:

$\innerprod {A y} x_\HH = \innerprod {A^* y} x_\HH$

From Inner Product is Sesquilinear:

$\innerprod {A y - A^* y} x_\HH = 0$

Setting $x = A y - A^* y$ we have:

$\innerprod {A y - A^* y} {A y - A^* y}_\HH = 0$

From the positiveness of the inner product we have:

$A y - A^* y = 0$

so:

$A y = A^* y$

Since $y \in \HH$ was arbitrary, we have:

$A = A^*$

That is:

$A$ is Hermitian.

$\blacksquare$


Sources