Inner Product is Sesquilinear
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Theorem
Let $\mathbb F$ be a subfield of $\C$.
Let $V$ be a inner product space over $V$ with inner product $\innerprod \cdot \cdot$.
Define the $f : V \times V \to \mathbb F$ by:
- $\map f {x, y} = \innerprod x y$
for each $x, y \in V$.
Then $f$ is sesquilinear.
Proof
Let $\alpha \in \mathbb F$.
Let $x_1, x_2, y \in V$.
By the definition of the inner product, $f$ is linear in its first argument.
So, we have:
- $\innerprod {\alpha x_1 + x_2} y = \alpha \innerprod {x_1} y + \innerprod {x_2} y$
From the definition of the inner product, we also have that $f$ is conjugate symmetric, so:
\(\ds \innerprod y {\alpha x_1 + x_2}\) | \(=\) | \(\ds \overline {\innerprod {\alpha x_1 + x_2} y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \overline {\alpha \innerprod {x_1} y + \innerprod {x_2} y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \overline {\alpha \innerprod {x_1} y} + \overline {\innerprod {x_2} y}\) | Sum of Complex Conjugates | |||||||||||
\(\ds \) | \(=\) | \(\ds \overline \alpha \overline {\innerprod {x_1} y} + \overline {\innerprod {x_2} y}\) | Product of Complex Conjugates | |||||||||||
\(\ds \) | \(=\) | \(\ds \overline \alpha \innerprod y {x_1} + \innerprod y {x_2}\) | using the conjugate symmetry of the inner product |
So we have:
- $\innerprod {\alpha x_1 + x_2} y = \alpha \innerprod {x_1} y + \innerprod {x_2} y$
and:
- $\innerprod y {\alpha x_1 + x_2} = \overline \alpha \innerprod y {x_1} + \innerprod y {x_2}$
for $\alpha \in \mathbb F$ and $x_1, x_2, y \in V$.
So $f$ is sesquilinear.
$\blacksquare$