# How do I find generalized eigenvector? [duplicate]

I have matrix: $A=\begin{pmatrix} 5 & 1 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}$I found 2 eigenvectors: $\vec{v_{1} } =\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \vec{v_{2} } =\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$Formula for generalized eigenvector is: $\left(A-\lambda I\right)\vec{x}=\vec{v}$How do I find generalized eigenvector and how do I know how much of them there are?If I put in this formula, I get $x_2=1$(for 1. eigenvector) and $x_2=0$ (for 2. eigenvector)? What will I do with $x_3$? Put the numbers I want or?

## marked as duplicate by JMoravitz, colormegone, Leucippus, Daniel W. Farlow, ShaileshMay 23 '16 at 1:45

Since A is 3x3 and linearly independent, you will have 3 (generalized) eigenvectors. Assuming that $x_1,\vec{v_1}$ and $x_2, \vec{v_2}$ are correct, then you'll have one more pair of a generalized eigenvalue and generalized eigenvector, $x_3,\vec{v_3}$. You can find this by pair solving $(A-x_3 I)^2 \vec{v_3} = 0$.