Editing Inner product space (section)

===Real and complex parts of inner products===

Suppose that <math>\langle \cdot, \cdot \rangle</math> is an inner product on <math>V</math> (so it is antilinear in its second argument). The [[polarization identity]] shows that the [[real part]] of the inner product is
<math display=block>\operatorname{Re} \langle x, y \rangle = \frac{1}{4} \left(\|x + y\|^2 - \|x - y\|^2\right).</math>

If <math>V</math> is a real vector space then
<math display=block>\langle x, y \rangle
= \operatorname{Re} \langle x, y \rangle
= \frac{1}{4} \left(\|x + y\|^2 - \|x - y\|^2\right)</math>
and the [[imaginary part]] (also called the {{em|complex part}}) of <math>\langle \cdot, \cdot \rangle</math> is always <math>0.</math>

Assume for the rest of this section that <math>V</math> is a complex vector space.
The [[polarization identity]] for complex vector spaces shows that
<math display="block">\begin{alignat}{4}
\langle x, \ y \rangle
&= \frac{1}{4} \left(\|x + y\|^2 - \|x - y\|^2 + i\|x + iy\|^2 - i\|x - iy\|^2 \right) \\
&= \operatorname{Re} \langle x, y \rangle + i \operatorname{Re} \langle x, i y \rangle. \\
\end{alignat}</math>

The map defined by <math>\langle x \mid y \rangle = \langle y, x \rangle</math> for all <math>x, y \in V</math> satisfies the axioms of the inner product except that it is antilinear in its {{em|first}}, rather than its second, argument. The real part of both <math>\langle x \mid y \rangle</math> and <math>\langle x, y \rangle</math> are equal to <math>\operatorname{Re} \langle x, y \rangle</math> but the inner products differ in their complex part:
<math display="block">\begin{alignat}{4}
\langle x \mid y \rangle
&= \frac{1}{4} \left(\|x + y\|^2 - \|x - y\|^2 - i\|x + iy\|^2 + i\|x - iy\|^2 \right) \\
&= \operatorname{Re} \langle x, y \rangle - i \operatorname{Re} \langle x, i y \rangle. \\
\end{alignat}</math>

The last equality is similar to the formula [[Real and imaginary parts of a linear functional|expressing a linear functional]] in terms of its real part. 

These formulas show that every complex inner product is completely determined by its real part. Moreover, this real part defines an inner product on <math>V,</math> considered as a real vector space. There is thus a one-to-one correspondence between complex inner products on a complex vector space <math>V,</math> and real inner products on <math>V.</math> 

For example, suppose that <math>V = \Complex^n </math> for some integer <math>n > 0.</math> When <math>V</math> is considered as a real vector space in the usual way (meaning that it is identified with the <math>2 n-</math>dimensional real vector space <math>\R^{2n},</math> with each <math>\left(a_1 + i b_1, \ldots, a_n + i b_n\right) \in \Complex^n</math> identified with <math>\left(a_1, b_1, \ldots, a_n, b_n\right) \in \R^{2n}</math>), then the [[dot product]] <math>x \,\cdot\, y = \left(x_1, \ldots, x_{2n}\right) \, \cdot \, \left(y_1, \ldots, y_{2n}\right) := x_1 y_1 + \cdots + x_{2n} y_{2n}</math> defines a real inner product on this space. The unique complex inner product <math>\langle \,\cdot, \cdot\, \rangle</math> on <math>V = \C^n</math> induced by the dot product is the map that sends <math>c = \left(c_1, \ldots, c_n\right), d = \left(d_1, \ldots, d_n\right) \in \Complex^n</math> to <math>\langle c, d \rangle := c_1 \overline{d_1} + \cdots + c_n \overline{d_n}</math> (because the real part of this map <math>\langle \,\cdot, \cdot\, \rangle</math> is equal to the dot product). 

====Real vs. complex inner products====

Let <math>V_{\R}</math> denote <math>V</math> considered as a vector space over the real numbers rather than complex numbers.
The [[real part]] of the complex inner product <math>\langle x, y \rangle</math> is the map <math>\langle x, y \rangle_{\R} = \operatorname{Re} \langle x, y \rangle ~:~ V_{\R} \times V_{\R} \to \R,</math> which necessarily forms a real inner product on the real vector space <math>V_{\R}.</math> Every inner product on a real vector space is a [[Bilinear map|bilinear]] and [[symmetric map]]. 

For example, if <math>V = \Complex</math> with inner product <math>\langle x, y \rangle = x \overline{y},</math> where <math>V</math> is a vector space over the field <math>\Complex,</math> then <math>V_{\R} = \R^2</math> is a vector space over <math>\R</math> and <math>\langle x, y \rangle_{\R}</math> is the [[dot product]] <math>x \cdot y,</math> where <math>x = a + i b \in V = \Complex</math> is identified with the point <math>(a, b) \in V_{\R} = \R^2</math> (and similarly for <math>y</math>); thus the standard inner product <math>\langle x, y \rangle = x \overline{y},</math> on <math>\Complex</math> is an "extension" the dot product . Also, had <math>\langle x, y \rangle</math> been instead defined to be the {{EquationNote|Symmetry|symmetric map}} <math>\langle x, y \rangle = x y</math>  (rather than the usual {{EquationNote|Conjugate symmetry|conjugate symmetric map}} <math>\langle x, y \rangle = x \overline{y}</math>) then its real part <math>\langle x, y \rangle_{\R}</math> would {{em|not}} be the dot product; furthermore, without the complex conjugate, if <math>x \in \C</math> but <math>x \not\in \R</math> then <math>\langle x, x \rangle = x x = x^2 \not\in [0, \infty)</math> so the assignment <math display="inline">x \mapsto \sqrt{\langle x, x \rangle}</math> would not define a norm.

The next examples show that although real and complex inner products have many properties and results in common, they are not entirely interchangeable.
For instance, if <math>\langle x, y \rangle = 0</math> then <math>\langle x, y \rangle_{\R} = 0,</math> but the next example shows that the converse is in general {{em|not}} true.
Given any <math>x \in V,</math> the vector <math>i x</math> (which is the vector <math>x</math> rotated by 90°) belongs to <math>V</math> and so also belongs to <math>V_{\R}</math> (although scalar multiplication of <math>x</math> by <math>i = \sqrt{-1}</math> is not defined in <math>V_{\R},</math> the vector in <math>V</math> denoted by <math>i x</math> is nevertheless still also an element of <math>V_{\R}</math>). For the complex inner product, <math>\langle x, ix \rangle = -i \|x\|^2,</math> whereas for the real inner product the value is always <math>\langle x, ix \rangle_{\R} = 0.</math>

If <math>\langle \,\cdot, \cdot\, \rangle</math> is a complex inner product and <math>A : V \to V</math> is a continuous linear operator that satisfies <math>\langle x, A x \rangle = 0</math> for all <math>x \in V,</math> then <math>A = 0.</math> This statement is no longer true if <math>\langle \,\cdot, \cdot\, \rangle</math> is instead a real inner product, as this next example shows. 
Suppose that <math>V = \Complex</math> has the inner product <math>\langle x, y \rangle := x \overline{y}</math> mentioned above. Then the map <math>A : V \to V</math> defined by <math>A x = ix</math> is a linear map (linear for both <math>V</math> and <math>V_{\R}</math>) that denotes rotation by <math>90^{\circ}</math> in the plane. Because <math>x</math> and <math>A x</math> are perpendicular vectors and <math>\langle x, Ax \rangle_{\R}</math> is just the dot product, <math>\langle x, Ax \rangle_{\R} = 0</math> for all vectors <math>x;</math> nevertheless, this rotation map <math>A</math> is certainly not identically <math>0.</math> In contrast, using the complex inner product gives <math>\langle x, Ax \rangle = -i \|x\|^2,</math> which (as expected) is not identically zero.