Finding eigenvectors of a 3x3 matrix 2
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.
$\endgroup$ 11 Answer
$\begingroup$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.
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