Semi-Inner Product induces Inner Product on Quotient

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Theorem

Let $\GF \in \set {\R, \C}$.

Let $\struct {V, \innerprod \cdot \cdot}$ be a semi-inner product space over $\GF$.

Let:

$N = \set {x \in V : \innerprod x x = 0}$

Let $X/N$ be the quotient vector space of $X$ modulo $N$.

Define:

$\innerprod {x + N} {y + N}_{X/N} = \innerprod x y$

for each $x, y \in X$.


Then $\innerprod \cdot \cdot_{X/N}$ is an inner product on $X/N$.


Proof

First note that Zero Set of Semi-Inner Product is Vector Subspace shows that $N$ is a vector subspace and hence $X/N$ is well-defined.

We show that $\innerprod \cdot \cdot_{X/N}$ is well-defined.

We need to show that if $x, y, x', y' \in X$ are such that $x - x' \in N$ and $y - y' \in N$, then $\innerprod x y = \innerprod {x'} {y'}$.

We have:

\(\ds \innerprod x y\) \(=\) \(\ds \innerprod {x - x' + x'} {y - y' + y'}\)
\(\ds \) \(=\) \(\ds \innerprod {x - x'} {y - y' + y'} + \innerprod {x'} {y - y' + y'}\)
\(\ds \) \(=\) \(\ds \innerprod {x - x'} {y - y'} + \innerprod {x - x'} {y'} + \innerprod {x'} {y - y'} + \innerprod {x'} {y'}\)

From Cauchy-Bunyakovsky-Schwarz Inequality: Inner Product Spaces, we have:

$\cmod {\innerprod {x - x'} {y - y'} }^2 \le \innerprod {x - x'} {x - x'} \innerprod {y - y'} {y - y'} = 0$

since $x - x' \in N$ and $y - y' \in N$.

We also have:

$\cmod {\innerprod {x - x'} {y'} }^2 \le \innerprod {x - x'} {x - x'} \innerprod {y'} {y'} = 0$

since $x - x' \in N$.

Finally:

$\cmod {\innerprod {x'} {y - y'} }^2 \le \innerprod {x'} {x'} \innerprod {y - y'} {y - y'} = 0$

since $y - y' \in N$.

So we have $\innerprod {x - x'} {y - y'} + \innerprod {x - x'} {y'} + \innerprod {x'} {y - y'} = 0$ and we obtain:

$\innerprod x y = \innerprod {x'} {y'}$

Hence $\innerprod \cdot \cdot_{X/N}$ is well-defined.

We just need to verify that $\innerprod \cdot \cdot_{X/N}$ is an inner product on $X/N$.


Proof of conjugate symmetry

Let $x, y \in X$.

We have:

\(\ds \overline {\innerprod {x + N} {y + N}_{X/N} }\) \(=\) \(\ds \overline {\innerprod x y}\)
\(\ds \) \(=\) \(\ds \innerprod y x\) conjugate symmetry of $\innerprod \cdot \cdot$
\(\ds \) \(=\) \(\ds \innerprod {y + N} {x + N}_{X/N}\)

$\Box$


Proof of linearity in first argument

Let $x, y, z \in X$, we have:

\(\ds \innerprod {\paren {x + N} + \paren {y + N} } {z + N}_{X/N}\) \(=\) \(\ds \innerprod {\paren {x + y} + N} {z + N}_{X/N}\) Definition of Quotient Vector Space
\(\ds \) \(=\) \(\ds \innerprod {x + y} z\)
\(\ds \) \(=\) \(\ds \innerprod x z + \innerprod y z\) linearity in first argument for $\innerprod \cdot \cdot$
\(\ds \) \(=\) \(\ds \innerprod {x + N} {z + N}_{X/N} + \innerprod {y + N} {z + N}_{X/N}\)

$\Box$


Proof of non-negative definiteness

Let $x \in X$.

We have:

$\innerprod {x + N} {x + N}_{X/N} = \innerprod x x \in \R_{\ge 0}$

From non-negative definiteness of $\innerprod \cdot \cdot$ .

$\Box$


Proof of positiveness

Let $x \in X$.

Again we have:

$\innerprod {x + N} {x + N}_{X/N} = \innerprod x x$

So $\innerprod {x + N} {x + N}_{X/N} = 0$ if and only if $\innerprod x x = 0$ if and only if $x \in N$.

This is, if and only if $x + N = {\mathbf 0}_{X/N}$.

$\blacksquare$

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