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I have the matrix \begin{bmatrix}1&0&0\\2&2&-1\\0&1&0\end{bmatrix} I know that the only eigenvalue is 1 with multiplicity 3

I solved for the first eigenvalue and got \begin{bmatrix}0\\1\\1\end{bmatrix}

How do I find the other two? I know they are \begin{bmatrix}0\\1\\0\end{bmatrix} and \begin{bmatrix}1/2\\0\\0\end{bmatrix} but when I do $(A-\lambda I)v_2 = v_1$, I get the system of equations $2x + y -z = 1$, $y -z =1$. I don't see how that gives the second eigenvector.

Thanks

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    $\begingroup$ Have you heard of generalized eigenvectors and methods of finding those? Your particular example requires two generalized eigenvectors and there are many such examples on MSE. For example, cfm.brown.edu/people/sg/classnotes1.pdf $\endgroup$ – Moo Apr 21 '17 at 12:01
  • $\begingroup$ What makes you think there are $3$ independent eigenvectors ? If there were, then $A$ would be diagonalizable, with only eigenvalue $1$, which would mean $A$ would be unity matrix, which it is definitely not :-) Your system for kernel of $A-\lambda I$ has two equations, so the eigenspace should have dimension $1$. $\endgroup$ – Nicolas FRANCOIS Apr 21 '17 at 12:18
  • $\begingroup$ Here is an example where algebraic multiplicity and geometric multiplicity are different. Those are important to know at least probably later course. $\endgroup$ – mathreadler Apr 21 '17 at 12:30
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All you need is just to solve the system. The system of equations $$ \begin{cases} 2x+y-z=1,\\ \qquad y-z=1 \end{cases} $$ has the solution $$ \begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 0\\1+t\\t \end{bmatrix},\quad t\in\Bbb R. $$ Now take $t=0$.

The second vector is obtained similarly: solve $$ \begin{cases} 2x+y-z=1,\\ \qquad y-z=0 \end{cases} $$ to get $$ \begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 1/2\\-t\\t \end{bmatrix},\quad t\in\Bbb R $$ and set $t=0$.

P.S. Those other two vectors are not eigenvectors. They are called generalized eigenvectors.

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If you use a Linear Algebra tool like in MATLAB, or a libraries for programming languages, like NumPy with Python or MathNet Numerics with C#, you can find out that the eigenvectors calculated by these tools are:

[0, 0, 0] [(√2)/2, (√2)/2, (√2)/2] [(√2)/2, (√2)/2, (√2)/2]

The eigenvalues are as stated:

(1,0) (1,0) (1,0)

1 of multiplicity 3.

We should bear in mind that eigenvalues could be complex numbers.

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  • $\begingroup$ Please use MathJax for mathematical equations and expressions: math.meta.stackexchange.com/questions/5020 $\endgroup$ – jvdhooft Jun 27 '17 at 21:23
  • $\begingroup$ @jvdhooft: Thanks a lot for the advice on MathJax, will do! $\endgroup$ – GEN Jun 27 '17 at 21:42

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