# Eigenvalue theorem

I was reading my differential equation book and there is a theorem I am having trouble understanding. What do they mean by this?

An $n\times n$ matrix $A$ has at least one and at most $n$ distinct complex eigenvalues.

If I were to have a matrix with an eigenvalue of distinct real roots how can that matrix also have at least one complex eigenvalue?

-
Real numbers are complex. But they are not imaginary. (Unless it's zero... maybe) – StuartHa Nov 27 '12 at 2:48
@Stuart but I thought complex numbers are imaginary numbers? – Q.matin Nov 27 '12 at 2:51
@Q.matin, not at all. If you use imaginary as in the common language, they're really not more imaginary than $\,3, 1/4, -16.35\,$ ,etc. If you're using the term imaginary as in the theory, a complex number $\,x+iy\,\,\,,\,\,x,y\in\Bbb R\,$ is imaginary iff $\,x=0\,\,,\,y\neq 0\,$ – DonAntonio Nov 27 '12 at 2:56

By a complex number, they mean any number of the form $x + iy$, where $x,y$ are real numbers. To understand why this theorem is true, we need to realize that we solve for the eigenvalues of a matrix $A$ by calculating the zeros of its characteristic polynomial, which is going to be an $n$th degree polynomial if $A$ is $n \times n$.

Every $n$th degree polynomial has at least one distinct zero and at most $n$ distinct zeros, if you allow the zeros to be complex, and hence the matrix $A$ has at least one complex eigenvalue and at most $n$ of them. Note that both real numbers and imaginary numbers are complex, so this theorem is not placing any restrictions on having real eigenvalues.

-
When you say "if you allow the zeros to be complex" does that mean that zero is a complex number which is why they said it has at least one complex number? – Q.matin Nov 27 '12 at 3:05
By a zero, I mean a point at which the polynomial is equal to zero. For example, if the characteristic polynomial of $A$ is $x^2 + 2x + 5$, then there no real zeros, but if you allow complex zeros, then the zeros are $-1 + 2i$ and $-1 - 2i$. – Christopher A. Wong Nov 27 '12 at 3:12

The characteristic polynomial for a matrix $A$ is given by $$\operatorname{det}(\lambda I-A).$$ The roots of this polynomial are the eigenvalues. This polynomial has degree $n$, which implies by the fundamental theorem of algebra that there are exactly $n$ eigenvalues, including repetition.

If all eigenvalues are distinct, then there are $n$ distinct values. If all eigenvalues are equal, then there is only one eigenvalue.

-
Thanks! I am going to read that wiki article you provided now to understand it more fully. – Q.matin Nov 27 '12 at 3:08

The theorem follows if we take the characteristic polynomial of the matrix $\,A\,$ of order $\,n\times n\,$: this is a complex polynomial and from the Fundamental Theorem of Algebra we know this polynomial has exactly $\,n\,$ roots, counting multiplicites, and this means exactly that the polynomial has at least one complex root = at least one eigenvalue, and at most $\,n\,$ different ones=at most $\,n\,$ different eigenvalues..

-
What do you mean by "counting multiplicites, and this means exactly the pol"? – Q.matin Nov 27 '12 at 3:06
Changed slightly the wording so that perhaps this time is clearer. – DonAntonio Nov 27 '12 at 3:19