Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer
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..

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.