Inner product

Inner product
$\langle\cdot,\cdot\rangle:V\times V\rightarrow\mathbb{F}$ Where $V$ is a vector space over the field $\mathbb{F}$ $\mathbb{F}$ may be $\mathbb{R}$ or $\mathbb{C}$.
relation to other topological spaces
is a	topological space metric space normed space
contains all	(none)
Related objects
Induced norm	$\Vert\cdot\Vert_{\langle\cdot,\cdot\rangle}:V\rightarrow\mathbb{R}_{\ge 0}$ $\Vert\cdot\Vert_{\langle\cdot,\cdot\rangle}:x\mapsto\sqrt{\langle x,x\rangle}$ For $V$ a vector space over $\mathbb{R}$ or $\mathbb{C}$
Induced metric	$d_{\langle\cdot,\cdot\rangle}:V\times V\rightarrow\mathbb{R}_{\ge 0}$ $d_{\langle\cdot,\cdot\rangle}:(x,y)\mapsto\sqrt{\langle x-y,x-y\rangle}$ (As every metric induces a norm) For $V$ considered as a set

Definition

Given a vector space, $(V,F)$ (where $F$ is either $\mathbb{R}$ or $\mathbb{C}$), an inner product is a map:

• $$\langle\cdot,\cdot\rangle:V\times V\rightarrow\mathbb{R}$$ (or sometimes $$\langle\cdot,\cdot\rangle:V\times V\rightarrow\mathbb{C}$$ )

Such that:

• $$\langle x,y\rangle = \overline{\langle y, x\rangle}$$ (where the bar denotes Complex conjugate)
  • Or just $$\langle x,y\rangle = \langle y,x\rangle$$ if the inner product is into $\mathbb{R}$
• $$\langle\lambda x+\mu y,z\rangle = \lambda\langle y,z\rangle + \mu\langle x,z\rangle$$ ( linearity in first argument )

  This may be alternatively stated as:

  • $$\langle\lambda x,y\rangle=\lambda\langle x,y\rangle$$ and $$\langle x+y,z\rangle = \langle x,z\rangle + \langle y,z\rangle$$

• $$\langle x,x\rangle \ge 0$$ but specifically:
  • $$\langle x,x\rangle=0\iff x=0$$

Terminology

Given a vector space $X$ over either $\mathbb{R}$ or $\mathbb{C}$, and an inner product $\langle\cdot,\cdot\rangle:X\times X\rightarrow F$ we call the space $(X,\langle\cdot,\cdot\rangle)$ an:

• Inner product space (or i.p.s for short) or sometimes a
• pre-hilbert space

Properties

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• The most important property by far is that: $\forall x\in X[\langle x,x\rangle\in\mathbb{R}_{\ge 0}]$ - that is $\langle x,x\rangle$ is real

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Proof:

Notice that we (by definition) have $\langle x,x\rangle=\overline{\langle x,x\rangle}$, so we must have:

• $a+bj=a-bj$ where $a+bj:=\langle x,x\rangle$, and by equating the real and imaginary parts we see immediately that we have:
  • $b=-b$ and conclude $b=0$, that is there is no imaginary component.

To complete the proof note that by definition $\langle x,x\rangle\ge 0$.

Thus $\langle x,x\rangle\in\mathbb{R}_{\ge 0}$ - as I claimed.

Notice that

$$\langle\cdot,\cdot\rangle$$ is also linear (ish) in its second argument as:

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• $$\langle x,\lambda y+\mu z\rangle =\bar{\lambda}\langle x,y\rangle+\bar{\mu}\langle x,z\rangle$$

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$$\langle x,\lambda y+\mu z\rangle$$

$$=\overline{\langle \lambda y+\mu z, x\rangle}$$

$$=\overline{\lambda\langle y,x\rangle + \mu\langle z,x\rangle}$$

$$=\bar{\lambda}\overline{\langle y,x\rangle}+\bar{\mu}\overline{\langle z,x\rangle}$$

$$=\bar{\lambda}\langle x,y\rangle+\bar{\mu}\langle x,z\rangle$$

As required.

From this we may conclude the following:

• $$\langle x,\lambda y\rangle = \bar{\lambda}\langle x,y\rangle$$ and
• $$\langle x,y+z\rangle = \langle x,y\rangle + \langle x,z\rangle$$

This leads to the most general form:

[Expand]

• $\langle au+bv,cx+dy\rangle=a\overline{c}\langle u,x\rangle+a\overline{d}\langle u,y\rangle+b\overline{c}\langle v,x\rangle+b\overline{d}\langle v,y\rangle$ - which isn't worth remembering!

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Proof:

$\langle au+bv,cx+dy\rangle$

$=a\langle u,cx+dy\rangle+b\langle v,cx+dy\rangle$

$=a\overline{\langle cx+dy,u\rangle}+b\overline{\langle cx+dy,v\rangle}$

$=a(\overline{c\langle x,u\rangle} + \overline{d\langle y,u\rangle})+b(\overline{c\langle x,v\rangle}+\overline{d\langle y,v\rangle})$

$=a\overline{c}\langle u,x\rangle+a\overline{d}\langle u,y\rangle+b\overline{c}\langle v,x\rangle+b\overline{d}\langle v,y\rangle$

As required

Notation

Typically, $\langle\cdot,\cdot\rangle$ is the notation for inner products, however I have seen some authors use $\langle a,b\rangle$ to denote the ordered pair containing $a$ and $b$. Also, notably use $(\cdot,\cdot)$ for an inner product (and $\langle\cdot,\cdot\rangle$ for an ordered pair!)

Immediate theorems

Here $\langle\cdot,\cdot\rangle:X\times X\rightarrow \mathbb{C}$ is an inner product

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Theorem: if $\forall x\in X[\langle x,y\rangle=0]$ then $y=0$

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Suppose that $y\ne 0$, then by hypothesis:

• $\forall x\in X[\langle x,y\rangle =0]$

Specifically that means for $y\in X$ we have $\langle y,y\rangle=0$

• Of course by definition, $\langle y,y\rangle\ge 0$ for $\forall y\in X$, and specifically
  • $\langle x,x\rangle = 0\iff x=0$

So we have $\langle y,y\rangle =0$ contradicting that $y\ne 0$

• We conclude that if $\forall x\in X[\langle x,y\rangle=0]$ then we must have $y=0$

  (As required)

Norm induced by

• Given an inner product space $(X,\langle\cdot,\cdot\rangle)$ we can define a norm as follows:
  • $\forall x\in X$ the inner product induces the norm $\Vert x\Vert:=\sqrt{\langle x,x\rangle}$

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TODO: Find out what this is called, eg compared to the metric induced by a norm

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Prominent examples

• Vector dot product

See also

• Hilbert space
• Inner product examples
• Inequalities for inner products
• Perpendicular