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This question already has an answer here:

let $A = \left( \begin{array}{} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array} \right)$

$A$ has one eigenvalue $\lambda=1$ with algebraic multiplicity $3$ but it has only two linearly independent eigenvectors $\left( \begin{array}{} 1 \\ 0 \\ 0 \end{array} \right),\left( \begin{array}{} 0 \\ 1 \\ 0 \end{array} \right)$(this can be easily checked since $A-\lambda I=A-I=\left( \begin{array}{} 0 & 0 & 3 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{array} \right)$ and both vectors are in $ker(A-I)$), so the geometric multiplicity is $2$ which is less than 3, the algebraic multiplicity.

and from here, I was told that I need to compute something like $(A-I)w=v$ to find $w$ where $v$ is the eigenvector. but I have two $v$'s, so I don't know which $v$ I should take.

and as I put $A$ in the Wolfram Alpha, I found its Jordan decomposition and there was no $\left( \begin{array}{} 1 \\ 0 \\ 0 \end{array} \right)$ in the column of $S$, such that $A=SJS^{-1}$, where $J$ is the Jordan form.

exactly how to find $S$? I'm so confused.

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marked as duplicate by Dietrich Burde, user91500, Claude Leibovici, Daniel W. Farlow, Vladhagen Jan 21 '17 at 23:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @DietrichBurde, in your link that 3 by 3 matrix A has only one true eigenvector v. I know how to proceed in that case, like finding w in (A-I)w=v and so on. but my example has 2 eigenvactors, that's why i'm confused. and neither of them give nonzero w. $\endgroup$ – user159234 Jan 20 '17 at 10:23
  • $\begingroup$ You can also find the $S$ just from the matrix equation $AS=SJ$, since you know $J$. $\endgroup$ – Dietrich Burde Jan 20 '17 at 10:28
  • $\begingroup$ @DietrichBurde No, I don't. I used to find all generalized eigenvectors first, and then find J using S. that's how I learned it and sadly I don't know the way to get J first. $\endgroup$ – user159234 Jan 20 '17 at 10:32
  • $\begingroup$ Ah, sorry. Your title says "Finding $S$". Finding $J$ is another duplicate, I admit. See for example here, or here, or here, etc. $\endgroup$ – Dietrich Burde Jan 20 '17 at 10:34
  • $\begingroup$ @DietrichBurde I read all examples you linked, but none of them treated multiple eigenvectors case. $\endgroup$ – user159234 Jan 20 '17 at 10:49
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Hint: the Jordan form of $A$ is given by $$ J= \begin{pmatrix} 1 & 0 & 0 \cr 0 & 1 & 1 \cr 0 & 0 & 1 \end{pmatrix}. $$ Then the matrix equation $AS=SJ$ can be easily solved by linear equations in the coefficients of $S$, e.g., $$ S= \begin{pmatrix} 3 & 3 & 0 \cr 3 & 2 & 0 \cr 0 & 0 & 1 \end{pmatrix}. $$

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  • $\begingroup$ I think the first column of S was (0, 1, 0) in the Wolfram Alpha. is that S not unique? $\endgroup$ – user159234 Jan 20 '17 at 10:35
  • $\begingroup$ No, take say $A=I$. Then any invertible $S$ can be taken. Also $S^{-1}$ can be taken as $S$. $\endgroup$ – Dietrich Burde Jan 20 '17 at 10:37
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After you've found the eigenvectors, the next step is to find vectors $v$ for which $(A-I)v\ne0$ and $(A-I)^2v=0$ (this basically means that $(A-I)v$ is an eigenvector. For the given matrix $A$ we have $$ (A-I)^2=\begin{pmatrix} 0&0&3\\0&0&2\\0&0&0 \end{pmatrix}^2=\begin{pmatrix} 0&0&0\\0&0&0\\0&0&0 \end{pmatrix}, $$ so any vector $v$ such that $(A-I)v\ne0$ would work. Take $v=(0,0,1)^T$, it'll be our first basis vector. Then take $u=(A-I)v=(3,2,0)^T$, it'll be the second basis vector. And for the third one we can take any other eigenvector which isn't collinear with $u$. Take $w=(1,0,0)^T$. So, the matrix $$ S=\begin{pmatrix} 0&3&1\\0&2&0\\1&0&0 \end{pmatrix} $$ is such that $$ A=S\begin{pmatrix} 1&0&0\\1&1&0\\0&0&1 \end{pmatrix}S^{-1}. $$

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