1
$\begingroup$

I am trying to find the eigenvectors of the following 3x3 matrix:

\begin{bmatrix} -9 & 4 & 4 \\ -8 & 3 & 4 \\ -16 & 8 & 7 \\ \end{bmatrix}

The eigenvalues for the matrix are $\lambda_1=3$, $\lambda_2=-1$, $\lambda_3=-1$.

$\lambda_1=3$ :

$$\lambda I_3-A=\begin{bmatrix} (3)+9 & -4 & -4 \\ 8 & (3)-3 & -4 \\ 16 & -8 & (3)-7 \\ \end{bmatrix}=\begin{bmatrix} 12 & -4 & -4 \\ 8 & 0 & -4 \\ 16 & -8 & -4 \\ \end{bmatrix}$$

$$8x-4z=0$$

Let $z=t$ $$8x=4t$$ $$x=\frac 12t$$ $$-4y=4t-12(\frac 12t)$$ $$y=\frac 12t$$ When $t=2$$$v_1=\begin{bmatrix} 1 \\ 1 \\ 2 \\ \end{bmatrix}$$

But now I'm struggling to find the eigenvectors when $\lambda_2=\lambda_3=-1$

$\lambda_2=\lambda_3=-1$ : $$\lambda I_3-A=\begin{bmatrix} (-1)+9 & -4 & -4 \\ 8 & (-1)-3 & -4 \\ 16 & -8 & (-1)-7 \\ \end{bmatrix}=\begin{bmatrix} 8 & -4 & -4 \\ 8 & -4 & -4 \\ 16 & -8 & -8 \\ \end{bmatrix}$$

This matrix gives me 3 linear equations that are all multiples of each other so I effectively only have one equation: $$2x-y-z=0$$

According to WolframAlpha there should be two more eigenvectors: $$v_2=\begin{bmatrix} 1 \\ 0 \\ 2 \\ \end{bmatrix}$$ and $$v_3=\begin{bmatrix} 1 \\ 2 \\ 0 \\ \end{bmatrix}$$ but I have no idea how to find these. I'm pretty new to matrices so this is probably a really simple problem but any help would be much appreciated.

$\endgroup$
  • $\begingroup$ "The eigenvectors" don't exist, every vecor with $2x-y-z=0$ is by your calculation an eigenvector. WolframAlpha gives you two simple examples, which are linear independent. $\endgroup$ – user302982 Nov 24 '16 at 22:30
0
$\begingroup$

From the equation $2x-y-z=0$, you can finish your problem. All eigenvectors are of the form $(x,y,z)$ that satisfy that equation. That is, all vectors of the form $$\begin{bmatrix}x\\y\\2x-y\end{bmatrix} =x\begin{bmatrix}1\\0\\2\end{bmatrix} + y \begin{bmatrix}0\\1\\-1\end{bmatrix}.$$ Essentially, anything in the span of $(1,0,2)$ and $(0,1,-1)$ is an eigenvector.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.