Operator is Hermitian iff Inner Product is Real
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
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next) $\text {II}.2.12$