Complex eigenvalues and eigenvectors of a matrix

(Last Updated On: September 16, 2018)

Complex eigenvalues and eigenvectors of a matrix. Dear friends, today it’s all about the complex eigenvalues and eigenvectors of a matrix. Have a look!!

Complex eigenvalues and eigenvectors of a matrix

In my earlier posts, I have already shown how to find out eigenvalues and the corresponding eigenvectors of a matrix.

Now with eigenvalues of any matrix, three things can happen.

First one is a simple one – like all eigenvalues are real and different.

Next one is at least one eigenvalue is repeated, can be twice or even more.

And finally, the last one is that the eigenvalues are the complex conjugate numbers, not real anymore.

Today I’ll talk about only the complex eigenvalues of a matrix with real numbers.

Then I’ll also try to figure out the corresponding eigenvectors.

So, I’ll start with some examples.

Some examples of Complex eigenvalues and eigenvectors of a matrix

Disclaimer: None of these examples is mine. I have chosen these from some book or books. I have also given the due reference at the end of the post.

So here are the examples.

Example 1

According to Kreyszig (2005) “Find the eigenvalues and eigenvectors of the following matrices.

$\begin{pmatrix} 0.8&-0.6\\ 0.6 &0.8 \end{pmatrix}.$

”

Solution

Here the given matrix is

$\begin{pmatrix} 0.8&-0.6\\ 0.6 &0.8 \end{pmatrix}.$

First of all, I’ll give it a name, say, $A$ .

Therefore it will be

$\begin{pmatrix} 0.8&-0.6\\ 0.6 &0.8 \end{pmatrix}.$

Now I’ll start with the eigenvalues of the matrix $A$ .

Step 1

In the beginning, comes the characteristic equation of the matrix $A$ .

So it will be

$|A - \lambda I| = 0.$

Here $\lambda$ is the eigenvalue of the matrix $A$ .

Thus $|A - \lambda I|$ will be

$|A - \lambda I| = \left|\begin{array}{cc} 0.8 - \lambda &-0.6\\ 0.6 &0.8 - \lambda \end{array}\right|.$

Next, I’ll evaluate the determinant.

So it will be

$|A - \lambda I| = (0.8- \lambda)(0.8- \lambda) - (-0.6)(0.6).$

Now this gives

$\begin{eqnarray*} |A - \lambda I| &=& (0.8- \lambda)^2 +0.36\\&=& 0.64 + \lambda^2 - 1.6 \lambda + 0.36\\&=& \lambda^2 - 1.6 \lambda +1.\end{eqnarray*}$

Since $|A - \lambda I| = 0$ , this means $\lambda^2 - 1.6 \lambda +1 = 0$ .

Now I’ll solve this equation to get the values of $\lambda$ .

Thus it will be

$\lambda = \frac{1.6 \pm \sqrt{(1.6)^2 - 4(1)(1)}}{2}.$

So I’ll simplify it to get the values of $\lambda$ as

$\begin{eqnarray*} \lambda &=& \frac{1.6 \pm \sqrt{2.56 - 4}}{2}\\&=& \frac{1.6 \pm \sqrt{-1.44}}{2}\\&=& \frac{1.6 \pm 1.2 j}{2}\\&=& 0.8 \pm 0.6 j.\end{eqnarray*}$

Now I have two values of $\lambda$ – one is $\lambda_1 = 0.8 + 0.6 j$ and the other one is $\lambda_2 = 0.8 - 0.6 j$ .

So both are the complex conjugate numbers.

Hence these are the complex eigenvalues of a matrix with real numbers.

Now I’ll find out the eigenvectors corresponding to each eigenvalue.

Step 2

First of all, I’ll get the eigenvector corresponding to $\lambda_1 = 0.8 + 0.6 j$ .

For $\lambda_1 = 0.8 + 0.6 j$ , the matrix $(A - \lambda I)$ is

$A - \lambda I = \begin{pmatrix} 0.8 - 0.8 - 0.6 j&-0.6\\ 0.6 &0.8- 0.8 - 0.6 j \end{pmatrix}.$

This means

$A - \lambda I = \begin{pmatrix} - 0.6 j&-0.6\\ 0.6 & - 0.6 j \end{pmatrix}.$

Suppose $[x_1, x_2]^T$ is the corresponding eigenvector for the eigenvalue $\lambda = 0.8 + 0.6 j$ .

Thus I can say

$\begin{pmatrix} - 0.6 j&-0.6\\ 0.6 & - 0.6 j \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}.$

This means

$- 0.6j x_1 - 0.6 x_2 = 0$

and

$0.6 x_1 - 0.6j x_2 = 0.$

So I get two different equations as

(1) $\begin{equation*}j x_1 + x_2 = 0\end{equation*}$

and

(2) $\begin{equation*} x_1 -j x_2 = 0.\end{equation*}$

From equation (2), I can say

$x_1 = j x_2.$

Thus the eigenvector will be

$\begin{pmatrix} x_1\\ x_2 \end{pmatrix} = \begin{pmatrix} j x_2\\ x_2 \end{pmatrix} = x_2 \begin{pmatrix} j\\ 1 \end{pmatrix} .$

Therefore I can say for $\lambda = 0.8 + 0.6 j$ , the corresponding eigenvector is $[j, 1]^T.$

Now I’ll find out the eigenvector corresponding to $\lambda_2 = 0.8 - 0.6 j$ .

Step 3

For $\lambda_2 = 0.8 - 0.6 j$ , the matrix $(A - \lambda I)$ is

$A - \lambda I = \begin{pmatrix} 0.8 - 0.8 + 0.6 j&-0.6\\ 0.6 &0.8- 0.8 + 0.6 j \end{pmatrix}.$

This means

$A - \lambda I = \begin{pmatrix} 0.6 j&-0.6\\ 0.6 & 0.6 j \end{pmatrix}.$

Suppose $[x_1, x_2]^T$ is the corresponding eigenvector for the eigenvalue $\lambda = 0.8 - 0.6 j$ .

Thus I can say

$\begin{pmatrix} 0.6 j&-0.6\\ 0.6 & 0.6 j \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}.$

This means

$0.6j x_1 - 0.6 x_2 = 0$

and

$0.6 x_1 + 0.6j x_2 = 0.$

So I get two different equations as

(3) $\begin{equation*} j x_1 - x_2 = 0\end{equation*}$

and

(4) $\begin{equation*} x_1 + j x_2 = 0.\end{equation*}$

From equation (4), I can say

$x_1 = - j x_2.$

Thus the eigenvector will be

$\begin{pmatrix} x_1\\ x_2 \end{pmatrix} = \begin{pmatrix} - j x_2\\ x_2 \end{pmatrix} = x_2 \begin{pmatrix} - j\\ 1 \end{pmatrix} .$

Therefore I can say for $\lambda = 0.8 - 0.6 j$ , the corresponding eigenvector is $[-j, 1]^T.$

Hence I can conclude that the eigenvalues of the matrix are $0.8 \pm 0.6 j$ and the corresponding eigenvectors are $[j, 1]^T, [-j, 1]^T$ .

This is the answer to this example.

Now I’ll solve another example on complex eigenvalues and eigenvectors of a matrix.

Example 2

According to Kreyszig (2005) “Find the eigenvalues and eigenvectors of the following matrices.

$\begin{pmatrix} \cos \theta&-\sin \theta\\ \sin \theta &\cos \theta \end{pmatrix}.$

”

Solution

Here the given matrix is

$\begin{pmatrix} \cos \theta&-\sin \theta\\ \sin \theta &\cos \theta \end{pmatrix}.$

First of all, I’ll give it a name, say, $A$ .

Therefore it will be

$\begin{pmatrix} \cos \theta&-\sin \theta\\ \sin \theta &\cos \theta \end{pmatrix}.$

Now I’ll start with the eigenvalues of the matrix $A$ .

Step 1

In the beginning, comes the characteristic equation of the matrix $A$ .

So it will be

$|A - \lambda I| = 0.$

Here $\lambda$ is the eigenvalue of the matrix $A$ .

Thus $|A - \lambda I|$ will be

$|A - \lambda I| = \left|\begin{array}{cc} \cos \theta - \lambda&-\sin \theta\\ \sin \theta &\cos \theta - \lambda \end{array}\right|.$

Next, I’ll evaluate the determinant.

So it will be

$|A - \lambda I| = (\cos \theta - \lambda)(\cos \theta - \lambda) - (-\sin \theta)(\sin \theta).$

Now this gives

$\begin{eqnarray*} |A - \lambda I| &=& (\cos \theta - \lambda)^2 + \sin^2 \theta\\&=& \cos^2 \theta + \lambda^2 - 2 \cos \theta \lambda + \sin^2 \theta\\&=& \lambda^2 - 2 \cos \theta \lambda + (\cos^2 \theta + \sin^2 \theta)\\&=& \lambda^2 - 2 \cos \theta \lambda + 1.\end{eqnarray*}$

Since $|A - \lambda I| = 0$ , this means $\lambda^2 - 2 \cos \theta \lambda +1 = 0$ .

Now I’ll solve this equation to get the values of $\lambda$ .

Thus it will be

$\lambda = \frac{2 \cos \theta \pm \sqrt{(2 \cos \theta )^2 - 4(1)(1)}}{2}.$

So I’ll simplify it to get the values of $\lambda$ as

$\begin{eqnarray*} \lambda &=& \frac{2 \cos \theta \pm \sqrt{4\cos^2 \theta - 4}}{2}\\&=& \frac{2 \cos \theta \pm 2\sqrt{-\sin^2 \theta}}{2}\\&=& \cos \theta \pm \sin \theta j.\end{eqnarray*}$

Now I have two values of $\lambda$ – one is $\lambda_1 = \cos \theta + \sin \theta j$ and the other one is $\lambda_2 = \cos \theta - \sin \theta j$ .

So both are the complex conjugate numbers.

Hence these are the complex eigenvalues of a matrix with real numbers.

Now I’ll find out the eigenvectors corresponding to each eigenvalue.

Step 2

First of all, I’ll get the eigenvector corresponding to $\lambda_1 = \cos \theta + \sin \theta j$ .

For $\lambda_1 = \cos \theta + \sin \theta j$ , the matrix $(A - \lambda I)$ is

$A - \lambda I = \begin{pmatrix} \cos \theta - \cos \theta - \sin \theta j &-\sin \theta\\ \sin \theta &\cos \theta - \cos \theta - \sin \theta j\\ \end{pmatrix}.$

This means

$A - \lambda I = \begin{pmatrix} - \sin \theta j &-\sin \theta\\ \sin \theta & - \sin \theta j \end{pmatrix}.$

Suppose $[x_1, x_2]^T$ is the corresponding eigenvector for the eigenvalue $\lambda = \cos \theta + \sin \theta j$ .

Thus I can say

$\begin{pmatrix} - \sin \theta j &-\sin \theta\\ \sin \theta & - \sin \theta j \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}.$

This means

$- \sin \theta j x_1 - \sin \theta x_2 = 0$

and

$\sin \theta x_1 - \sin \theta j x_2 = 0.$

So I get two different equations as

(5) $\begin{equation*}j x_1 + x_2 = 0\end{equation*}$

and

(6) $\begin{equation*} x_1 -j x_2 = 0.\end{equation*}$

From equation (6), I can say

$x_1 = j x_2.$

Thus the eigenvector will be

$\begin{pmatrix} x_1\\ x_2 \end{pmatrix} = \begin{pmatrix} j x_2\\ x_2 \end{pmatrix} = x_2 \begin{pmatrix} j\\ 1 \end{pmatrix} .$

Therefore I can say for $\lambda = \cos \theta + \sin \theta j$ , the corresponding eigenvector is $[j, 1]^T.$

Now I’ll find out the eigenvector corresponding to $\lambda_2 = \cos \theta - \sin \theta j$ .

Step 3

For $\lambda_2 = \cos \theta - \sin \theta j$ , the matrix $(A - \lambda I)$ is

$A - \lambda I = \begin{pmatrix} \cos \theta - \cos \theta + \sin \theta j &-\sin \theta\\ \sin \theta &\cos \theta - \cos \theta + \sin \theta j\\ \end{pmatrix}.$

This means

$A - \lambda I = \begin{pmatrix} \sin \theta j &-\sin \theta\\ \sin \theta & \sin \theta j \end{pmatrix}.$

Suppose $[x_1, x_2]^T$ is the corresponding eigenvector for the eigenvalue $\lambda = \cos \theta - \sin \theta j$ .

Thus I can say

$\begin{pmatrix} \sin \theta j &-\sin \theta\\ \sin \theta & \sin \theta j \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}.$

This means

$\sin \theta j x_1 - \sin \theta x_2 = 0$

and

$\sin \theta x_1 + \sin \theta j x_2 = 0.$

So I get two different equations as

(7) $\begin{equation*}j x_1 - x_2 = 0\end{equation*}$

and

(8) $\begin{equation*} x_1 + j x_2 = 0.\end{equation*}$

From equation (8), I can say

$x_1 = - j x_2.$

Thus the eigenvector will be

$\begin{pmatrix} x_1\\ x_2 \end{pmatrix} = \begin{pmatrix} - j x_2\\ x_2 \end{pmatrix} = x_2 \begin{pmatrix} - j\\ 1 \end{pmatrix} .$

Therefore I can say for $\lambda = \cos \theta + \sin \theta j$ , the corresponding eigenvector is $[- j, 1]^T.$

Hence I can conclude that the eigenvalues of the matrix are $\cos \theta \pm \sin \theta j$ and the corresponding eigenvectors are $[j, 1]^T, [-j, 1]^T$ .

This is the answer to this example.

Dear friends, this is the end of my today’s post on complex eigenvalues and eigenvectors of a matrix. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

Kreyszig, E. (2005): Advanced Engineering Mathematics: International Edition, John Wiley & Sons, 9th Edition, 29th December 2005, Chapter 2, Second-order linear ODEs, p. 338, Problem set 2.2, Q. 1(Example 7), Q. 2 (Example 10).

Complex eigenvalues and eigenvectors of a matrix