TITLE: finding eigenvectors given eigenvalues
QUESTION [7 upvotes]: i have the matrix:
$A = \begin{bmatrix}8 & -2\\-2 & 5\end{bmatrix}$
i want to find its eigenvectors and eigenvalues. by the characteristic equation:
$\textrm{det}(A - \lambda I) = 0$
expanding the determinant:
$\begin{bmatrix}8 - \lambda & -2\\-2 & 5-\lambda\end{bmatrix} = \lambda^2 - 13\lambda + 36 = 0$
using the quadratic formula, $\lambda = 9$ or $\lambda = 4$, so the two eigenvalues are $\{9,4\}$.
when i try to get the eigenvectors, i run into problems. i plugin $\lambda = 9$ into the characteristic polynomial equation:
$\begin{bmatrix}-1 & -2\\-2 & -4\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix} = 0$
resulting in:   
$-v_1-2v_2 = 0$
$-2v_1 - 4v_2 = 0$
How can this be solved to get the eigenvector: $[v_1 v_2]^T$? I just get $v_1 = v_2 = 0$ which does not help.

REPLY [2 votes]: The zero vector is always a solution of $(A-\lambda I)v=0$, which is one reason why it’s not considered an eigenvector, but you’re on the right track. Remember that for any eigenvector $v$ of $A$, a scalar multiple of of it is also an eigenvector of $A$: $A(kv) = k(Av) = k(\lambda v)=\lambda (kv)$.
The equations you’ve derived so far tell you that $v_1=-2v_2$, so any vector of the form $[-2a,a]^T$ is an eigenvector corresponding to the eigenvalue 9. Similarly, for the eigenvalue 5, you get $v_2=2v_1$, so its eigenvectors will be of the form $[b,2b]^T$.