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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?

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

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  • $\begingroup$ row reduced form to calculate rank. n×n matrix has n eigen vectors and values. rank determines number of 0 eigen vectors $\endgroup$ – shai horowitz May 22 '16 at 20:58
  • $\begingroup$ @moo But they are first and third 0??? $\endgroup$ – Laja May 22 '16 at 21:26
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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 $(A-x_3 I)^2 \vec{v_3} = 0$.

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