finding eigenvectors given eigenvalues

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.

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.

• what do you mean by underdetermined system? can this be interpreted to mean something about the properties of the linear transform represented by $A$? – lgd Sep 7 '15 at 19:37
• An underdetermined system is a system of equations that has more equations than unknowns. So any solutions will depend on one or more parameters. PS: Sorry, I forgot to answer the second question. In this case, the rank of $A$ is 2, so we know that, as it has two eigenvalues, the dimension of each eigenspace has to be 1 (as the sum has to be 2). – Miguel Mars Sep 7 '15 at 19:39
• so $[-10\ 5]$ is an eigenvector too but we generally go with smallest? – lgd Sep 7 '15 at 19:43
• Yes, $(-10, 5)$ is another eigenvector in the same eigenspace as $(-2,1)$. – Miguel Mars Sep 7 '15 at 19:45
• mkspk: "An underdetermined system is a system of equations that has more equations than unknowns." Was that a typo? An undetermined system has more unknowns than equations! – user247327 Apr 22 '18 at 1:24

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.