The Case of Complex Eigenvalues
First let us convince ourselves that there exist matrices with complex
eigenvalues.
Example. Consider the matrix
The characteristic equation is given by
This quadratic equation has complex roots given by
Therefore the matrix A has only complex eigenvalues.
The trick is to treat the complex eigenvalue as a real one. Meaning we deal
with it as a number and do the normal calculations for the eigenvectors. Let us
see how it works on the above example.
We will do the calculations for
 .
The associated eigenvectors are given by the linear system
A X = (1+2i) X
which may be rewritten as
In fact the two equations are identical since
(2+2i)(2-2i) = 8. So the system reduces to one equation
(1-i)x - y = 0.
Set x=c, then
y = (1-i)c. Therefore, we have
where c is an arbitrary number.
Remark. It is clear that one should expect to have complex entries in
the eigenvectors.
We have seen that (1-2i) is also an eigenvalue of the above matrix.
Since the entries of the matrix A are real, then one may easily show that
if
is a complex eigenvalue, then its conjugate
is also an eigenvalue. Moreover, if X is an eigenvector of A
associated to
,
then the vector
 ,
obtained from X by taking the complex-conjugate of the entries of X,
is an eigenvector associated to
.
So the eigenvectors of the above matrix A associated to the eigenvalue
(1-2i) are given by
where c is an arbitrary number.
Let us summarize what we did in the above example.
Summary: Let A be a square matrix. Assume
is a complex eigenvalue of A. In order to find the associated
eigenvectors, we do the following steps:
1.
Write down the associated linear system
2.
Solve the system. The entries of X will be complex numbers.
3.
Rewrite the unknown vector X as a linear combination of known
vectors with complex entries.
4.
If A has real entries, then the conjugate
is
also an eigenvalue. The associated eigenvectors are given by the same
equation found in 3, except that we should take the conjugate of the entries
of the vectors involved in the linear combination.
In general, it is normal to expect that a square matrix with real entries may
still have complex eigenvalues. One may wonder if there exists a class of
matrices with only real eigenvalues. This is the case for symmetric matrices.
The proof is very technical and will be discussed in another page. But for
square matrices of order 2, the proof is quite easy. Let us give it here for the
sake of being little complete.
Consider the symmetric square matrix
Its characteristic equation is given by
This is a quadratic equation. The nature of its roots (which are the eigenvalues
of A) depends on the sign of the discriminant
Using algebraic manipulations, we get
Therefore,
is a positive number which implies that the eigenvalues of A are real
numbers.
Remark. Note that the matrix A will have one eigenvalue, i.e.
one double root, if and only if
 .
But this is possible only if a=c and b=0. In other words,
we have
A = a I2.
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