Eigenvalues and Eigenvectors

This section is relevant as prerequisite material for systems of ODEs discussed in Chapter 14.

Definition: Eigenvalues and Eigenvectors

For a square matrix $A$ we have a nonzero vector $\boldsymbol{x}$ and a scalar $\lambda$ such that

then $\lambda$ is an eigenvalue of $A$ and $\boldsymbol{x}$ is the corresponding eigenvector.

We find eigenvalues and eigenvectors firstly by rewriting (10.88) as,

This is a homogeneous equation and if $A-\lambda I$ is non-singular then the only solution is the trivial one given by $\boldsymbol{x}=\mathbf{0}$. We therefore require that

The Characteristic Equation

The characteristic equation of a matrix $A$ is given by,

Consider the $2 \times 2$ matrix,

To find the eigenvalues of $A$, we solve the characteristic equation (10.90). This is given by

which yields a quadratic equation for $\lambda$,

The solutions to (10.93) are $\lambda_{1}=2$ and $\lambda_{2}=3$. For each eigenvalue, we look for a corresponding eigenvector by solving for $\boldsymbol{x}$ in Eq. (10.89). We start with $\lambda_{1}=2$ and define $\boldsymbol{x}=\left(x_{1}, x_{2}\right)^{\top}$ in Eq. (10.89) which gives us the matrix equation,

Note that since we demanded that $A-\lambda I$ is singular, then we expect Eq. (10.94) to result in an under-determined system, i.e. 2 unknowns but only 1 independent equation. Equations (10.94) admit the solution $x_{2}=0$ and we are free to set $x_{1}$ since the system is satisfied with any value for $x_{1}$. We choose to set $x_{1}=c$ where $c$ is some arbitrary constant. This gives the eigenvector as

which represents an infinite set of solutions as $c \in \mathbb{R}$. Repeating the steps for the second eigenvalue, $\lambda_{2}=3$, we have

this gives us the solution $x_{1}=x_{2}$ and again we are free to set $x_{1}=d$ and hence $x_{2}=d$, where $d \in \mathbb{R}$. The eigenvector corresponding to $\lambda_{2}=3$ therefore is

Note that we can also write the eigenvectors simply as

and the understanding is that any constant multiple of the above eigenvectors is also an eigenvector.

Characteristic equation of a $2 \times 2$ matrix

Recall that the trace of a square matrix is given by the sum of its diagonal entries [see Eq. (10.11)]. We may calculate the characteristic equation of a $2 \times 2$ matrix using the trace and the determinant of the coefficient matrix $A$,

where $\tau$ and $\Delta$ are often used to denote the trace and determinant of a square matrix, respectively. The eigenvalues are then obtained from,

if $\tau^{2}-4 \Delta<0$ we have complex eigenvalues, if $\tau^{2}-4 \Delta>0$ we have real eigenvalues, and if $\tau^{2}-4 \Delta=0$ we have repeated eigenvalues. Note that in this course we focus on real coefficient matrices; it is still possible to find complex eigenvalues for real coefficient matrices but we need to take care when finding corresponding eigenvectors. Since the system we start with is real then the solutions to the real system must also be real. This case is discussed in more detail in Chapter 14 in the context of differential equations. Example 10.10 Find the eigenvalues and corresponding eigenvectors of the matrix,

Solution We start with computing the characteristic equation from

The eigenvalues are therefore 1,3 , and 4 . We now find the corresponding eigenvectors. For $\lambda=4$,

This system gives the following equations

Again we note that the equations are not linearly independent; we can add the first and the third one to obtain the second one. We have

by setting $x_{1}=c$, where $c$ is an arbitrary constant, we have the following eigenvector,

Following the same steps, the eigenvectors corresponding to the eigenvalues 1 and 3 are

where $d$ and $e$ are constants.

Exercises

Find the eigenvalues and eigenvectors of the following matrices:

1. $\left(\begin{array}{cc}-1 / 3 & 2 / 3 \\ 4 / 3 & 1 / 3\end{array}\right)$
2. $\left(\begin{array}{cc}-3 & -71 / 2 \\ -10 & 7\end{array}\right)$
3. $\left(\begin{array}{ccc}3 / 2 & 1 / 2 & 3 / 2 \\ 1 / 2 & 3 / 2 & -1 / 2 \\ 0 & 0 & 3\end{array}\right)$
4. $\left(\begin{array}{ccc}1 & -4 & -2 \\ 2 & -5 & -7 / 2 \\ 0 & 0 & 2\end{array}\right)$
5. $\left(\begin{array}{ccc}8 / 3 & 0 & -2 / 3 \\ 14 / 3 & -2 & -2 / 3 \\ -7 / 3 & 0 & 7 / 3\end{array}\right)$