Why is the complex dot product defined to only be symmetric up to a conjugate?

Question

We all have a good interpretation of orthogonal vectors in Euclidean Space, how does this extend to Complex vectors? Algebraically, the thing that bugs me, is that the dot product is only symmetric up to a conjugate-which seems strange to me.

I considered making a system of equations to solve for an example of two orthogonal complex vectors, but I wasn't sure if it was going anywhere. I said $[a+bi,c+di]*[e+fi,g+hi]=0$, which got me to $eb-af+dg-hc=ae+fb+gc+hd$ which I think is right. Does anyone have an intuitive way of thinking about orthogonality in a complex space?

Answer 1

Definitions are set up to produce useful results. In the complex field there are probably two motivators for definitions. One is to have analogues to the real spaces. In particular, if you use a complex operation or function, when confined to real numbers it should give the same answer as the analogous operation for the reals. The other is that complex functions and vectors often result from the study of problems in physics. If the definitions and notation are set up poorly it would complicate the mathematics or even make it impossible.

Consider, for example, the length of a vector u. For real vectors it is $(u \cdot u)^{1/2}$. In the complex case if you don't use a conjugate for the inner product, $(u \cdot u)$ won't even be a real number. This won't work as a length. The length of a vector is pretty fundamental. You can hardly do any complex mathematics without it. By itself it is sufficient reason for the way the complex inner product is defined.

In turn all sorts of complex-valued matrix manipulations like multiplication then depend on that inner product, which has to be consistent; the operations will fail without a properly defined inner product.

Likewise, if you are trying to work with heat equations, or wave equations, etc. you are going to wind up with functions of complex variables. You need the arithmetic to work correctly or you will get not so much wrong answers as nonsense.

Now as to why this inner product seems unintuitive. We are used to thinking of vectors as arrows in the plane, or maybe in 3 dimensions. This is not a good geometric model for complex vectors. Better is to think of them as describing directionality within ongoing physical processes that involve two or three dimensional time dependent behavior. So the complex vectors are describing the directional relationships between various aspects of the process; and quite commonly these directions are orthogonal in the geometric sense.

Here's a quote. "In an electromagnetic plane wave, E and B are always perpendicular to each other and the direction of propagation" (where E is the electric field and B is the magnetic flux density.) I'm sure you can find whole books full of observations like this.

So that is the kind of thing orthogonality means for complex vectors. There are some very nice pictures on this website: http://en.wikipedia.org/wiki/Poynting_vector.

Answer 2

The main reason in my opinion for defining dot products is so we can get a "natural" norm to describe our vectors. If the components are real, there is no problem with the usual definition as $aa$ can never be negative whenever $a$ is real, so that we may take square roots to define the length.

But if $a$ is complex this fails. The corresponding property in the complex field, however, is that $a\bar{a}$ can never be negative. Thus, from the point of view of euclidean norms this is the natural way to extend the dot product to complex vector spaces.

Furthermore, if $a$ is real, then $a=\bar a,$ so that the complex definition smoothly generalises the real one, so that we then define the inner product that way.

Answer 3

$i^T i = -1$, whereas $i^H i = (-i)i=1$.

More generally, if $x,y\in\mathbb R^n$, then $(x+iy)^H(x+iy)=\sum_{j=1}^n (x_j-iy_j)(x_j+iy_j)=\sum_{j=1}^n (x_j^2-(iy_j)^2)=\sum_{j=1}^n (x_j^2+y_j^2)=\sum_{j=1}^n |x_j^2+y_j^2|=|x+iy|^2,$

as required for an inner product.

For mathematicians, usually dot product = inner product, unlike for physicists, whose dot product does not conjugate! Moreover, old-school mathematicians conjugate and transpose the second argument instead of the first one. See https://math.stackexchange.com/a/3567684/ for more.

Answer 4

(Disclaimer: this area is far from my domain of knowledge.)

You made a mistake with your inner product. You should get two equations, one for the complex component, and one for the real component.

If you're looking at the inner product of two vectors, you can choose a basis such that one of the vectors is of the form $[a+bi,0]$. Your inner product then becomes $[a+bi,0]∗[e+fi,g+hi]=0$. Then it's clear that $g=0, h=0$. Orthogonality doesn't change much in a complex vector space compared to a real one. The inner product of orthogonal vectors is symmetric, since the complex conjugate of zero is itself.

What's trickier to understand is the dot product of parallel vectors. Personally, I think of complex vectors more in the form $[R_ae^{i\theta_a},R_be^{i\theta_b}]$. If we imagine the dot product of two parallel vectors (again choosing a convenient basis):

$[R_1e^{i\theta_1},0]*[R_2e^{i\theta_2},0]=R_1R_2e^{i(\theta_1-\theta_2)}$

It's clear that their magnitudes multiply, just like with vectors over the reals. The key difference is that the inner product also has a phase component. Thus the inner product tells you to what degree two vectors are parallel, and also to what degree the parallel components are in phase.