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I'm not sure that this will make a lot of sense, but I guess that I'm trying to decide which course of study or action to persue.

I have a system of equations that seems to be linear, if the complex numbers are treated as scalars.

In other words, I have a system of equations with complex coefficients. I'm wondering how to go about solving these, and if linear algebra may somehow suffice. The variables are complex themselves, but I wonder whether scalar variables (instead of complex variables) would change anything.

The question/Summary

To make more sense of this question and to sum up what I'm looking for, I'd like to know the best way to solve a system of equations with complex coefficients and variables. Which branch(es) of math/science does this involve? An example showing how to solve a system of equations of this form would be particularly helpful, but not necessary.

An Example I'm interested In I have a set of values, and I multiply them by complex values taken from the unit circle. So I may have a set${a,b,c}$ of complex values and I take this set and multiply it by $e^{i\pi d}$, which gives me three coefficients to use for three variables, say $x_1$, $x_2$, and $x_3$. I repeat this process several times (for different $d$ to create PART of several equations. So far, I can create euqations with these values, but they are complex multiples of one another.

These values (of the $x$'s) are added to another set of values that come from powers of complex numbers - this essentially ensures that the system of equations aren't "scalar" multiples of one another. I calculate corresponding sums of the variables. Then I have a system of equations.

What I'm trying to say is that I get a system of equations that aren't complex multiples of one another. I'm really thrilled at the possibility that I can solve this by linear algebra methods.

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To make more sense of this question and to sum up what [you are] looking for, you could show us an example of a system of equations that seems to be linear, if the complex numbers are treated as scalars. – Did Mar 24 '11 at 21:11
Linear algebra and specifically the theory of vector spaces explains why it works. – Raskolnikov Mar 24 '11 at 21:12
scalar variables (instead of complex variables) : "complex" numbers are usually considered "scalar" (though in some contexts -programming languages- they are implemented as a "vector" of two reals). The usual dichotomies are "real vs complex" "scalar vs vectorial" – leonbloy Mar 24 '11 at 21:21

2 Answers 2

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If you review your course in linear algebra, you will see that the only properties of the scalars that are invoked is that the obey the familiar arithmetic laws for addition, multiplication and inverses, i.e. they have the algebraic structure known as a field. Thus, for example, we can carry out Gaussian elimination over any field, e.g. $\rm\ \mathbb Q,\: \mathbb R,\: \mathbb C,\: \mathbb Q(x),\: \mathbb F_p\:$ etc. In fact the theory of linear algebra over $\rm\:\mathbb R\:$ can easily be abstracted to yield a theory of linear algebra over an arbitrary field of scalars.

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If you had attended the University of Chicago at the right time, you might have been lucky enough to learn linear algebra over the quaternions in your first semester undergrad linear algebra class! – Bill Dubuque Mar 24 '11 at 23:48
Or maybe you should have said "learn quaternions over linear algebra"... – Matt Groff Mar 25 '11 at 2:57

Yes: linear algebra works over any field, in particular over the complex numbers.

If your variables and coefficients are complex numbers, but the equations are linear on the unknowns, then the usual methods of complex numbers work, exactly as they do over the real numbers.

Rather than me invent a system of equations that may not match what you are looking for, perhaps you can post an example yourself, both in the "complex variable" and "scalar variable" case (I confess to being somewhat confused about what you mean by the latter).

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I'm working on a simple example... – Matt Groff Mar 24 '11 at 21:22
@Matt Groff: Okay; I'm about to leave the office, though. – Arturo Magidin Mar 24 '11 at 21:23
It was a bit hastily constructed, but I think that I've been reassured that I can do what I want by the answers submitted - so Thanks! – Matt Groff Mar 24 '11 at 21:40
@Matt Groff: Your description is rather vague. Better would have been to produce the specific/explicit problem you are trying to solve. But if you have linear equations, you can solve them exactly like you solve systems of linear equations over $\mathbb{R}$, only over $\mathbb{C}$. – Arturo Magidin Mar 25 '11 at 2:43
Unfortunately, I realized that I can't exploit the problem, even with linear algebra. I was naive enough to think I had a chance of exploiting P vs. NP using this. I guess I'm in bliss being ignorant - yet I still want to learn. – Matt Groff Mar 25 '11 at 3:06

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