# 33. Eigenexamples

## 33. Eigenexamples

Example:

Let $A=\begin{pmatrix}2&1\\ 1&1\end{pmatrix}$ . The characteristic polynomial is $\det(A-tI)=\det\begin{pmatrix}2-t&1\\ 1&1-t\end{pmatrix}=(2-t)(1-t)-1=t^{2}-3t+1.$ The roots of this polynomial are $\frac{3\pm\sqrt{9-4}}{2}=\frac{3\pm\sqrt{5}}{2}$ .

Rather than doing the $\pm$ cases separately, I'll just leave $\pm$ in the notation; at the end, the answer will end up with a $\pm$ sign in and you pick the sign corresponding to the eigenvalue you want. The eigenvectors for the eigenvalue $\frac{3\pm\sqrt{5}}{2}$ are the solutions to $Av=\lambda v$ , i.e. $\begin{pmatrix}2&1\\ 1&1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\frac{3\pm\sqrt{5}}{2}\begin{pmatrix}x\\ y\end{pmatrix}.$ Multiplying out, this gives the equations $2x+y=\frac{3\pm\sqrt{5}}{2}x,\qquad x+y=\frac{3\pm\sqrt{5}}{2}y.$ Rearranging, we get $y=\frac{-1\pm\sqrt{5}}{2}x,\qquad x=\frac{1\pm\sqrt{5}}{2}y.$ As usual, the second equation follows from the first (check it!), so our eigenvectors are those of the form $\begin{pmatrix}x\\ \frac{-1\pm\sqrt{5}}{2}x\end{pmatrix}$ .

What this notation means is that the eigenvectors for $\frac{3+\sqrt{5}}{2}$ are $\begin{pmatrix}x\\ \frac{-1+\sqrt{5}}{2}x\end{pmatrix}$ and the eigenvectors for $\frac{3-\sqrt{5}}{2}$ are $\begin{pmatrix}x\\ \frac{-1-\sqrt{5}}{2}x\end{pmatrix}$ . Another useful piece of notation is $\mp$ (which means $-\pm$ ).

Example:

Let $A=\begin{pmatrix}\frac{3}{2}&\frac{5}{2}&3\\ -\frac{1}{2}&-\frac{3}{2}&-3\\ 1&1&2\end{pmatrix}$ . The characteristic polynomial is: $\det(A-tI)=\det\begin{pmatrix}\frac{3}{2}-t&\frac{5}{2}&3\\ -\frac{1}{2}&-\frac{3}{2}-t&-3\\ 1&1&2-t\end{pmatrix}$ $=\left(\frac{3}{2}-t\right)\det\begin{pmatrix}-\frac{3}{2}-t&-3\\ 1&2-t\end{pmatrix}-\frac{5}{2}\det\begin{pmatrix}-\frac{1}{2}&-3\\ 1&2-t\end{pmatrix}$ $\qquad\qquad+3\det\begin{pmatrix}-\frac{1}{2}&-\frac{3}{2}-t\\ 1&1\end{pmatrix}$ $=\left(\frac{3}{2}-t\right)\left(-\left(\frac{3}{2}+t\right)\left(2-t\right)+3% \right)-\frac{5}{2}\left(-\frac{1}{2}(2-t)+3\right)$ $\qquad\qquad+3\left(-\frac{1}{2}+\frac{3}{2}+t\right)$ $=\left(\frac{3}{2}-t\right)(t^{2}-t/2)-\frac{5}{2}(t/2+2)+3t+3$ $=-t^{3}+2t^{2}+t-2.$ To find the roots of this cubic, we use a standard trick: we guess one of them! Let's guess $t=1$ . Amazingly this is a root (funny how often that happens in carefully designed examples) because $-1^{3}+2(1^{2})+1-2=0$ . Once we have this, we know that our cubic factors as $(t-1)\times\mbox{ a quadratic}$ . Now we get the quadratic by doing polynomial long division. I don't know a good way of typesetting this, so you'll have to look at the video for the working! Note the similarities between long division and row reduction! These are both special cases of a general technique called "Groebner bases".

The result is: $-t^{2}+t+2.$ This has roots $\frac{-1\pm\sqrt{1+8}}{-2}$ , i.e. $-1$ or $2$ .

This means that the eigenvalues are $-1,1,2$ . What are the eigenvectors? I'll just work out the eigenvectors for $\lambda=1$ and leave the others as a fun exercise (answers at the end of the video).

We want to solve $Av=v$ , which means $3x/2+5y/2+3z=x,$ $-x/2-3y/2-3z=y,$ $x+y+2z=z.$ The first two rows are both equivalent to $x+5y+6z=0$ , the third means $z=-x-y$ , so overall we get $y=-5x,\qquad z=4x.$ In other words, the eigenvectors have the form $\begin{pmatrix}x\\ -5x\\ 4x\end{pmatrix}.$ Answers for $\lambda=2,-1$ from .