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?
 A: 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.
A: 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.
A: 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..
