# Perpendicularity on complex vector space

1. Dec 28, 2009

### mnb96

Hi,
given a complex vector space with a hermitian inner product, how is the cosine of the angle between two vectors defined?
I tried to follow a similar reasoning as in the real case and I got the following:

$$cos(\theta)=\mathcal{R}e \frac{ \left\langle u,v\right\rangle}{\left\|u\right\| \left\|v\right\|}$$

Does this make any sense?
If that is correct it means two vectors are perpendicular whenever the real part of their hermitian inner product is zero.

Again, if that is correct, how can we compute the projection of one vector onto another?
Thanks!

2. Dec 28, 2009

### jasonRF

The problem with your definition is that orthogonality would only involve the real part of the inner product. A more useful definition of angle would be

$$cos(\theta)= \frac{ \left|\left\langle u,v\right\rangle\right|}{\left\|u\right\| \left\|v\right\|}$$

Last edited: Dec 28, 2009
3. Dec 28, 2009

### JSuarez

One way to define it is as the smallest of the angles $$\theta$$ that makes the expression:

$$\frac{\left\langle U,V \right\rangle}{\left\| U\right\| \left\| V \right\|}e^{i\theta}$$

A real number.

4. Dec 29, 2009

### mnb96

I see.
it seems the definition of angle for complex vector spaces is quite arbitrary!
Aren't you aware of any of these definitions which is more "meaningful" than others? For example one that has been used in physics or has found some application?

I came up to that definition by starting from the law of cosines for triangles which says:

$$c^2=a^2+b^2-2abcos(\theta)$$

so ones has:

$$|c|^2=<a-b,a-b>=|a|^2+|b|^2-(<a,b>+<b,a>)$$

which implies that:

$$\frac{1}{2}(<a,b>+<b,a>)=\mathcal{R}e<a,b>=abcos(\theta)$$

What is the geometric reasoning behind the other definitions you are suggesting?

5. Dec 29, 2009

### Fredrik

Staff Emeritus
The standard definition for real vector spaces is

$$\cos\theta=\frac{\langle x,y\rangle}{\|x\|\|y\|}$$

(No absolute value in the numerator). I don't think there is a useful definition for complex vector spaces. But x and y are still said to be orthogonal if <x,y>=0.

6. Dec 29, 2009

### JSuarez

The geometric reasoning behind the definition I gave (and that is sometimes used in signal processing), is a generalization of the situation in C (considered as a vector space): if you have two complex numbers with unit modulus, then $$\theta$$ is the rotation angle that takes their product to the real axis.
But Fredrik is also correct: in complex vector spaces (with inner product, of course), orthogonality can always be defined, while angle is somewhat arbitrary.