Differential Equations - Systems: Eigenvalues & Eigenvectors - ii

Example 3 Find the eigenvalues and eigenvectors of the following matrix.$$A=\left(\begin{array}{cc}-4& -17\\ 2& 2\end{array}\right)$$

So, we’ll start with the eigenvalues.
$$\begin{array}{cc}det\left(A-\lambda I\right)& =\left|\begin{array}{cc}-4-\lambda & -17\\ 2& 2-\lambda \end{array}\right|\\ & =\left(-4-\lambda \right)\left(2-\lambda \right)+34\\ & ={\lambda }^2+2\lambda +26\end{array}$$This doesn’t factor, so upon using the quadratic formula we arrive at,
$${\lambda }_{1,2}=-1\pm 5i$$In this case we get complex eigenvalues which are definitely a fact of life with eigenvalue/eigenvector problems so get used to them.
Finding eigenvectors for complex eigenvalues is identical to the previous two examples, but it will be somewhat messier. So, let’s do that.
${\lambda }_1=-1+5i$ :
The system that we need to solve this time is
$$\begin{array}{cc}\left(\begin{array}{cc}-4-\left(-1+5i\right)& -17\\ 2& 2-\left(-1+5i\right)\end{array}\right)\left(\begin{array}{c}{\eta }_1\\ {\eta }_2\end{array}\right)& =\left(\begin{array}{c}0\\ 0\end{array}\right)\\ \left(\begin{array}{cc}-3-5i& -17\\ 2& 3-5i\end{array}\right)\left(\begin{array}{c}{\eta }_1\\ {\eta }_2\end{array}\right)& =\left(\begin{array}{c}0\\ 0\end{array}\right)\end{array}$$Now, it’s not super clear that the rows are multiples of each other, but they are. In this case we have,
$$R_1=-\frac{1}{2}\left(3+5i\right)R_2$$This is not something that you need to worry about, we just wanted to make the point. For the work that we’ll be doing later on with differential equations we will just assume that we’ve done everything correctly and we’ve got two rows that are multiples of each other. Therefore, all that we need to do here is pick one of the rows and work with it.
We’ll work with the second row this time.
$$2{\eta }_1+\left(3-5i\right){\eta }_2=0$$Now we can solve for either of the two variables. However, again looking forward to differential equations, we are going to need the “$i$” in the numerator so solve the equation in such a way as this will happen. Doing this gives,
$$\begin{array}{cc}2{\eta }_1& =-\left(3-5i\right){\eta }_2\\ {\eta }_1& =-\frac{1}{2}\left(3-5i\right){\eta }_2\end{array}$$So, the eigenvector in this case is
$$\overrightarrow{\eta }=\left(\begin{array}{c}{\eta }_1\\ {\eta }_2\end{array}\right)=\left(\begin{array}{c}-\frac{1}{2}\left(3-5i\right){\eta }_2\\ {\eta }_2\end{array}\right),{\eta }_2\ne 0$$$${\overrightarrow{\eta }}^{\left(1\right)}=\left(\begin{array}{c}-3+5i\\ 2\end{array}\right),{\eta }_2=2$$As with the previous example we choose the value of the variable to clear out the fraction.
Now, the work for the second eigenvector is almost identical and so we’ll not dwell on that too much.
${\lambda }_2=-1-5i$ :
The system that we need to solve here is
$$\begin{array}{cc}\left(\begin{array}{cc}-4-\left(-1-5i\right)& -17\\ 2& 2-\left(-1-5i\right)\end{array}\right)\left(\begin{array}{c}{\eta }_1\\ {\eta }_2\end{array}\right)& =\left(\begin{array}{c}0\\ 0\end{array}\right)\\ \left(\begin{array}{cc}-3+5i& -17\\ 2& 3+5i\end{array}\right)\left(\begin{array}{c}{\eta }_1\\ {\eta }_2\end{array}\right)& =\left(\begin{array}{c}0\\ 0\end{array}\right)\end{array}$$Working with the second row again gives,
$$2{\eta }_1+\left(3+5i\right){\eta }_2=0\Rightarrow {\eta }_1=-\frac{1}{2}\left(3+5i\right){\eta }_2$$The eigenvector in this case is
$$\overrightarrow{\eta }=\left(\begin{array}{c}{\eta }_1\\ {\eta }_2\end{array}\right)=\left(\begin{array}{c}-\frac{1}{2}\left(3+5i\right){\eta }_2\\ {\eta }_2\end{array}\right),{\eta }_2\ne 0$$$${\overrightarrow{\eta }}^{\left(2\right)}=\left(\begin{array}{c}-3-5i\\ 2\end{array}\right),{\eta }_2=2$$Summarizing,
$$\begin{array}{cccc}{\lambda }_1& =-1+5i& {\overrightarrow{\eta }}^{\left(1\right)}& =\left(\begin{array}{c}-3+5i\\ 2\end{array}\right)\\ {\lambda }_2& =-1-5i& {\overrightarrow{\eta }}^{\left(2\right)}& =\left(\begin{array}{c}-3-5i\\ 2\end{array}\right)\end{array}$$

There is a nice fact that we can use to simplify the work when we get complex eigenvalues. We need a bit of terminology first however.

If we start with a complex number,

$$z=a+bi$$

then the complex conjugate of $z$ is

$$\overline{z}=a-bi$$

To compute the complex conjugate of a complex number we simply change the sign on the term that contains the “$i$”. The complex conjugate of a vector is just the conjugate of each of the vector’s components.

We now have the following fact about complex eigenvalues and eigenvectors.

Fact

If $A$ is an $n\times n$ matrix with only real numbers and if ${\lambda }_1=a+bi$ is an eigenvalue with eigenvector ${\overrightarrow{\eta }}^{\left(1\right)}$. Then ${\lambda }_2=\overline{{\lambda }_1}=a-bi$ is also an eigenvalue and its eigenvector is the conjugate of ${\overrightarrow{\eta }}^{\left(1\right)}$.

This fact is something that you should feel free to use as you need to in our work.

Now, we need to work one final eigenvalue/eigenvector problem. To this point we’ve only worked with $2\times 2$ matrices and we should work at least one that isn’t $2\times 2$. Also, we need to work one in which we get an eigenvalue of multiplicity greater than one that has more than one linearly independent eigenvector.

Example 4 Find the eigenvalues and eigenvectors of the following matrix.$$A=\left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)$$

Despite the fact that this is a $3\times 3$ matrix, it still works the same as the $2\times 2$ matrices that we’ve been working with. So, start with the eigenvalues
$$\begin{array}{cc}det\left(A-\lambda I\right)& =\left|\begin{array}{ccc}-\lambda & 1& 1\\ 1& -\lambda & 1\\ 1& 1& -\lambda \end{array}\right|\\ & =-{\lambda }^3+3\lambda +2\\ & =\left(\lambda -2\right){\left(\lambda +1\right)}^2{\lambda }_1=2,{\lambda }_{2,3}=-1\end{array}$$So, we’ve got a simple eigenvalue and an eigenvalue of multiplicity 2. Note that we used the same method of computing the determinant of a $3\times 3$ matrix that we used in the previous section. We just didn’t show the work.
Let’s now get the eigenvectors. We’ll start with the simple eigenvector.
${\lambda }_1=2$ :
Here we’ll need to solve,
$$\left(\begin{array}{ccc}-2& 1& 1\\ 1& -2& 1\\ 1& 1& -2\end{array}\right)\left(\begin{array}{c}{\eta }_1\\ {\eta }_2\\ {\eta }_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$$This time, unlike the $2\times 2$ cases we worked earlier, we actually need to solve the system. So let’s do that.
$$\left(\begin{array}{cccc}-2& 1& 1& 0\\ 1& -2& 1& 0\\ 1& 1& -2& 0\end{array}\right)\begin{array}{c}{R}_1\leftrightarrow R_2\\ \Rightarrow \end{array}\left(\begin{array}{cccc}1& -2& 1& 0\\ -2& 1& 1& 0\\ 1& 1& -2& 0\end{array}\right)\begin{array}{c}{R}_2+2R_1\\ R_3-R_1\\ \Rightarrow \end{array}\left(\begin{array}{cccc}1& -2& 1& 0\\ 0& -3& 3& 0\\ 0& 3& -3& 0\end{array}\right)$$$$\begin{array}{c}-\frac{1}{3}{R}_2\\ \Rightarrow \end{array}\left(\begin{array}{cccc}1& -2& 1& 0\\ 0& 1& -1& 0\\ 0& 3& -3& 0\end{array}\right)\begin{array}{c}{R}_3-3R_2\\ R_1+2R_2\\ \Rightarrow \end{array}\left(\begin{array}{cccc}1& 0& -1& 0\\ 0& 1& -1& 0\\ 0& 0& 0& 0\end{array}\right)$$Going back to equations gives,
$$\begin{array}{cccc}{\eta }_1-{\eta }_3& =0& \Rightarrow {\eta }_1& ={\eta }_3\\ {\eta }_2-{\eta }_3& =0& \Rightarrow {\eta }_2& ={\eta }_3\end{array}$$So, again we get infinitely many solutions as we should for eigenvectors. The eigenvector is then,
$$\overrightarrow{\eta }=\left(\begin{array}{c}{\eta }_1\\ {\eta }_2\\ {\eta }_3\end{array}\right)=\left(\begin{array}{c}{\eta }_3\\ {\eta }_3\\ {\eta }_3\end{array}\right),{\eta }_3\ne 0$$$${\overrightarrow{\eta }}^{\left(1\right)}=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right),{\eta }_3=1$$Now, let’s do the other eigenvalue.
${\lambda }_2=-1$ :
Here we’ll need to solve,
$$\left(\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right)\left(\begin{array}{c}{\eta }_1\\ {\eta }_2\\ {\eta }_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$$Okay, in this case is clear that all three rows are the same and so there isn’t any reason to actually solve the system since we can clear out the bottom two rows to all zeroes in one step. The equation that we get then is,
$${\eta }_1+{\eta }_2+{\eta }_3=0\Rightarrow {\eta }_1=-{\eta }_2-{\eta }_3$$So, in this case we get to pick two of the values for free and will still get infinitely many solutions. Here is the general eigenvector for this case,
$$\overrightarrow{\eta }=\left(\begin{array}{c}{\eta }_1\\ {\eta }_2\\ {\eta }_3\end{array}\right)=\left(\begin{array}{c}-{\eta }_2-{\eta }_3\\ {\eta }_2\\ {\eta }_3\end{array}\right),{\eta }_2\ne 0\text{and}{\eta }_3\ne 0\text{at the same time}$$Notice the restriction this time. Recall that we only require that the eigenvector not be the zero vector. This means that we can allow one or the other of the two variables to be zero, we just can’t allow both of them to be zero at the same time!
What this means for us is that we are going to get two linearly independent eigenvectors this time. Here they are.
$${\overrightarrow{\eta }}^{\left(2\right)}=\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right){\eta }_2=0\text{and}{\eta }_3=1$$$${\overrightarrow{\eta }}^{\left(3\right)}=\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right){\eta }_2=1\text{and}{\eta }_3=0$$Now when we talked about linear independent vectors in the last section we only looked at $n$ vectors each with $n$ components. We can still talk about linear independence in this case however. Recall back with we did linear independence for functions we saw at the time that if two functions were linearly dependent then they were multiples of each other. Well the same thing holds true for vectors. Two vectors will be linearly dependent if they are multiples of each other. In this case there is no way to get ${\overrightarrow{\eta }}^{\left(2\right)}$ by multiplying ${\overrightarrow{\eta }}^{\left(3\right)}$ by a constant. Therefore, these two vectors must be linearly independent.
So, summarizing up, here are the eigenvalues and eigenvectors for this matrix
$$\begin{array}{cccc}{\lambda }_1& =2& {\overrightarrow{\eta }}^{\left(1\right)}& =\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)\\ {\lambda }_2& =-1& {\overrightarrow{\eta }}^{\left(2\right)}& =\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)\\ {\lambda }_3& =-1& {\overrightarrow{\eta }}^{\left(3\right)}& =\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right)\end{array}$$