Answer to Question #219721 in Linear Algebra for Kailash

Question

What do you mean by Norm of an Inner Product Space

Answer 1

Any inner product induces a norm given by

$$\|v\|=\sqrt{\langle v, v\rangle}$$

Proof. The axioms for norms mostly follow directly from those for inner products.

If $u, v ∈ V$ and $α ∈ F,$ then

(i)

$$\|v\|=\sqrt{\langle v, v\rangle}\geq0,$$

since $\langle v, v\rangle\geq0$ with equality if and only if $v = 0.$

(ii)

$$\|\alpha v\|=\sqrt{\langle \alpha v,\alpha v\rangle}=\sqrt{|\alpha|^2\langle v, v\rangle}$$

$$=|\alpha|\sqrt{\langle v, v\rangle}=|\alpha|\|a v\|$$

(iii) The triangle inequality

Cauchy-Schwarz inequality

If $V$ is an inner product space, then

$$|\langle u, u\rangle|\leq\|u\|\|v\|$$

for all $u, v ∈ V .$ Equality holds exactly when $u$ and $v$ are linearly dependent.

Using the Cauchy-Schwarz inequality,

$$\|u+ v\|^2=\langle u+v, u+v\rangle$$

$$=\langle u, u\rangle+\langle u, v\rangle+\langle v, u\rangle+\langle v, v\rangle$$

$$=\|u\|^2+\langle u, v\rangle+\overline{\langle u, v\rangle}+\|v\|^2$$

$$=\|u\|^2+2Re\langle u, v\rangle+\|v\|^2$$

$$\leq\|u\|^2+2|\langle u, v\rangle|+\|v\|^2$$

$$\leq\|u\|^2+2\|u\|\|v\|+\|v\|^2$$

$$=(\|u\|+\|v\|)^2$$

Taking square roots yields

$$\|u+ v\|\leq\|u\|+\|v\|,$$

since both sides are nonnegative.

Therefore $\|v\|=\sqrt{\langle v, v\rangle}$ is a Norm.