# Finding multiple linearly independent eigenvalues for a single eigenvalues

I have a question where I do not understand the answer, nor how it was derived. The problem asks to find the eigenvalues and corresponding eigenvector of the matrix:

$$\begin{pmatrix} 3 & 2 & 4 \\ 2 & 0 & 2 \\ 4 & 2 & 3 \\ \end{pmatrix}$$

Solving the characteristic equation allowed me to find the eignevalues λ = -1 and λ = 8.

Finding the associated eigenvector for λ = -1 is where I have problems, as I end up (with gaussian elimination) with a matrix:

$$\begin{pmatrix} 2 & 1 & 2 & | & 0\\ 0 & 0 & 0 & | & 0\\ 0 & 0 & 0 & | &0\\ \end{pmatrix}$$

Leaving (using variables of $x_1, x_2, x_3$): $2x_1+x_2+2x_3 = 0$, giving $x_1 = 1/2 x_2 - x_3$, which is as far as I get. *

The answers I have to this question provided details that it gives $$v_1 = -\frac{1}{2} v_2 - v_3$$ for the eigenvector components, and that because there are two degrees of freedom we can construct two eigenvectors.

It says that we can write $\begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix}$ as $-\frac{1}{2} v_2 \begin{bmatrix} -1 & 2 & 0 \end{bmatrix} + v_3 \begin{bmatrix} -1 & 0 & 1 \end{bmatrix}$ which allows us to identify the eigenvectors $\begin{bmatrix} -1 & 2 & 0 \end{bmatrix}^T$ and $\begin{bmatrix} -1 & 0 & 1 \end{bmatrix}^T$, as this is one of the cases where one eigenvalue has two linearly independent associated eigenvectors.

I'm looking to understand how they arrived at those numbers as I can't see any obvious explanation

Where you wrote $2x_1+x_2+x_3=0$, you meant to write $2x_1+x_2+2x_3=0$. Solving for $x_1$ gives $x_1=-(1/2)x_2-x_3$, which is exactly what's in the given answers, except for using $x$ instead of $v$.
Now you can take $x_2=2$, $x_3=0$, giving $x_1=-1$, so $(x_1,x_2,x_3)=(-1,2,0)$ and that's one eigenvector. You can take $x_2=0$, $x_3=1$, giving $x_1=-1$, so $(x_1,x_2,x_3)=(-1,0,1)$, your other eigenvector.
You could take other values of $x_2,x_3$ --- there's nothing special about the values in the answers --- all you need is two linearly independent results.
• Ok, I think I get it. So you just use any two values for $x_2, x_3$ that produces something easy to work with, and make sure they are linearly independent. Would $x_2 = 2, x_3 = 1$ therefore $e_1$ = (-2, 2, 1) and $x_2 = 0, x_3 = 1$ therefore $e_2$ = (-1, 0, 1) be a valid answer? I've had the same problem in another question, so using this would (1,1,0) and (0,1,0) be ok for {{1,1/2,-1/2},{1/2,1,1/2},{-1/2,1/2,1}}? – Adam M-W Oct 13 '12 at 12:18