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.
4 Answers
This is entirely expected. Your system for $\lambda = 9$ will have a one dimensional solution space, since if $x$ is an eigenvector of $A$, then any non-zero multiple of $x$ will also be an eigenvector of $A$.
You see that your two equations basically say the same thing, one of them is just a multiple of the other. Just pick a value for $v_1$ and calculate what $v_2$ should be. Take $v_1 = 2$ and you get $v_2 = -1$, so you can take $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$ as your eigenvector.
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 $[b,2b]^T$.
You get an underdetermined system. In fact, you can see both equations are essentially the same (the one below is the upper multiplied by two). So we have $$-v_1-2v_2=0$$ That leads to $$v_1=-2v_2$$ And the vectors in the eigenspace for $\lambda=9$ will be of the form $$\left( \begin{array}{c} -2v_2\\ v_2\\ \end{array} \right)$$ For example, for $v_2=1$, you have that one eigenvector for the eigenvalue $\lambda=9$ is $$\left( \begin{array}{c} -2\\ 1\\ \end{array} \right)$$ It is easy to do this analogously for the other eigenvalue.
Your last system is equivalent to $v_1=-2v_2$(the second line is a multiple of the first), so $(-2,1)$ is an eigenvector.
