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The matrix A=$\begin{pmatrix}0 & 1 &1 \\1& 0&1 \\ 1 & 1 & 0\end{pmatrix}$

has eigenvalues $\lambda=2$ with algebriac multiplicity $1$ and $\lambda=-1$ with multiplicity $2$

For $\lambda=-1$

$A+I=0 \implies$$\begin{pmatrix}1 & 1 &1 \\1& 1&1 \\ 1 & 1 & 1\end{pmatrix}\begin{pmatrix}x_1 \\x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix}0 \\0 \\ 0 \end{pmatrix}$

so $x_1+x_2+x_3=0$

$3$ possible cases of eigenvectors are:

$\begin{pmatrix}-2 \\1 \\ 1 \end{pmatrix}$, $\begin{pmatrix}1 \\-2 \\ 1 \end{pmatrix}$, $\begin{pmatrix}1 \\-1 \\ 0 \end{pmatrix}$

which are all linearly independent.

So we can have more than $2$ distinct (non-linearly dependent) eigenvectors, even if the algebraic multiplicity is only $2$?

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    $\begingroup$ Please clarify how you got your three eigenvectors. $\endgroup$ – vadim123 Apr 17 '15 at 14:36
  • $\begingroup$ Also what you mean to claim (which is false) is not "non-linearly dependent" but "linearly independent". $\endgroup$ – mathreadler Apr 17 '15 at 14:54
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All of the vectors you give belong to $\lambda = -1$. This set is not linearly independent; the rank of $$ \begin{pmatrix} -2 & 1 & 1 \\ 1 & -2 & -1 \\ 1 & -1 & 0 \end{pmatrix} $$ is 2, which means each two of them are sufficient to span the 2-dimensional eigenspace for this eigenvalue.

The eigenvector for $\lambda = 2$ is $$ \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}. $$


Extension due to the question's update:

The equation $x_1+x_2+x_3=0$ has infinitely many solutions, even if you disregard scaling (i.e. impose a constraint like $x_1 = 1$). Because $\lambda = -1$ has algebraic multiplicity 2, there is no unique (up to a scalar multiple) corresponding eigenvector, but an eigenspace of dimension 2, and each vector in that eigenspace is an eigenvector. You can pick any two linearly independent vectors out of that space to construct a basis of eigenvectors (plus the one for $\lambda = 2$).

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  • $\begingroup$ But eigenvectors are never unique, as they span other eigenvectors. $\endgroup$ – GFauxPas Apr 17 '15 at 16:18
  • $\begingroup$ @GFauxPas, an eigenvector for an eigenvalue with algebraic multiplicity 1 is unique except for scaling and flipping. That's why in my answer I referred to a constraint. If you have a word that means "unique except for scaling and flipping" I'd be happy to use it. $\endgroup$ – A. Donda Apr 17 '15 at 16:20
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    $\begingroup$ Oh, that's what you meant. I always say "unique up to a scalar multiple". $\endgroup$ – GFauxPas Apr 17 '15 at 16:24
  • $\begingroup$ @GFauxPas, great, I edited the answer. $\endgroup$ – A. Donda Apr 17 '15 at 16:26
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Compute

$$\begin{pmatrix}-2 \\1 \\ 1 \end{pmatrix} - \begin{pmatrix}1 \\-2 \\ 1 \end{pmatrix} + 3\begin{pmatrix}1 \\-1 \\ 0 \end{pmatrix}$$

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