The Case of Complex Eigenvalues

First let us convince ourselves that there exist matrices with complex eigenvalues.

Example. Consider the matrix

\begin{displaymath}A = \left(\begin{array}{rrr}

The characteristic equation is given by

\end{array}\right\vert = \lambda^2 - 2 \lambda + 5= 0.\end{displaymath}

This quadratic equation has complex roots given by

\begin{displaymath}\lambda = \frac{2 \pm i \sqrt{16}}{2} = 1 \pm 2i\end{displaymath}

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 $\lambda = 1+2i$. The associated eigenvectors are given by the linear system

A X = (1+2i) X

which may be rewritten as

(2-2i)x - 2y &=& 0\\
4x-(2+2i)y &=& 0\\

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

\begin{displaymath}X = \left(\begin{array}{rrr}
\end{array}\right) = \... = c \left(\begin{array}{c}

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 $\lambda$ is a complex eigenvalue, then its conjugate $\bar{\lambda}$ is also an eigenvalue. Moreover, if X is an eigenvector of A associated to $\lambda$, then the vector $\bar{X}$, obtained from X by taking the complex-conjugate of the entries of X, is an eigenvector associated to $\bar{\lambda}$. So the eigenvectors of the above matrix A associated to the eigenvalue (1-2i) are given by

\begin{displaymath}X = c \left(\begin{array}{c}

where c is an arbitrary number.

Let us summarize what we did in the above example.

Summary: Let A be a square matrix. Assume $\lambda$ is a complex eigenvalue of A. In order to find the associated eigenvectors, we do the following steps:

Write down the associated linear system

\begin{displaymath}A X = \lambda X\;\;\mbox{\bf or}\;\; \Big(A + \lambda I_n\Big) X = {\cal O}.\end{displaymath}

Solve the system. The entries of X will be complex numbers.
Rewrite the unknown vector X as a linear combination of known vectors with complex entries.
If A has real entries, then the conjugate $\bar{\lambda}$ 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

\begin{displaymath}A = \left(\begin{array}{rrr}

Its characteristic equation is given by

\begin{displaymath}\det(A - \lambda I_2) = \left\vert\begin{array}{cc}
a - \lamb...
...end{array}\right\vert = \lambda^2 - (a+c) \lambda + ac-b^2 = 0.\end{displaymath}

This is a quadratic equation. The nature of its roots (which are the eigenvalues of A) depends on the sign of the discriminant

\begin{displaymath}\Delta = (a+c)^2 - 4 (ac-b^2).\end{displaymath}

Using algebraic manipulations, we get

\begin{displaymath}\Delta = (a-c)^2 + 4 b^2.\end{displaymath}

Therefore, $\Delta$ 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 $\Delta = 0$. But this is possible only if a=c and b=0. In other words, we have

A = a I2.

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Author: M.A. Khamsi

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