# How do I compute these eigenvectors with repeated multiplicity?

Find the eigenvectors of $$A = \left[ {\begin{array}{ccc} 2 & 0 & 0 \\ 4 & 2 & 3 \\ 3 & 0 & -1 \end{array} } \right]$$

Step 1. Find the eigenvalues.

$$\left| {\begin{array}{ccc} 2-\lambda & 0 & 0 \\ 4 & 2-\lambda & 3 \\ 3 & 0 & -1-\lambda \end{array} } \right| = (2-\lambda)^2(-1-\lambda)$$

So the eigenvalues $$\lambda = 2, -1$$, with repeated multiplicity for $$2$$.

Step 2. The eigenvector for $$\lambda = 2$$ is

$$\left[ {\begin{array}{ccc} 0 & 0 & 0 \\ 4 & 0 & 3 \\ 3 & 0 & -3 \end{array} } \right] \left[ {\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} } \right] = \left[ {\begin{array}{c} 0 \\ 0 \\ 0 \end{array} } \right]$$

This gives the system of equations

\begin{align*} 4x_1 + 3x_3 &= 0 \\ 3x_1 - 3x_3 &= 0 \end{align*}

Which gives $$(x_1, x_2, x_2)$$ as an "eigenvector." However, this seems extremely defective and I don't know how to proceed.

The equations imply $$x_1=x_3$$ and $$x_1=-3/4x_3$$ So $$x_3 = 0, x_1=0$$ and an eigenvector is $$(0,1,0)$$.
To find another generalized eigenvector for $$\lambda =2$$ you can set $$(A-2I)^2x = 0$$ and do the same thing.