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