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Consider this matrix: $\begin{bmatrix}2-\lambda & -1 & 2\\-1&2-\lambda&2\\2&2&-1-\lambda\end{bmatrix}$ and the eigenvalues $-3$,$3$ and $3$.

The corresponding eigenvectors are $\begin{pmatrix}-1\\-1\\2\end{pmatrix}$, $\begin{pmatrix}1\\-1\\0\end{pmatrix}$ and $\begin{pmatrix}2\\0\\1\end{pmatrix}$.

The solution to the linear system can be $\begin{pmatrix}-a+2b\\a\\b\end{pmatrix}$, $\begin{pmatrix}a\\2b-a\\b\end{pmatrix}$ and $\begin{pmatrix}a\\b\\\frac{a+b}{2}\end{pmatrix}$.

My question is, why are there two eigenvectors for the eigenvalue $3$? Since there are two variables $a$ and $b$, aren't there infinite linearly independent vectors? So why pick these two in particular? And why two and not one or three or more?

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Whenever there is a double root in the characteristic equation giving a repeated eigenvalue as you have here, then there are infinitely many eigenvectors, correspondong to this eigenvalue, all of which are coplanar. You can therefore choose any two independent (non-parallel) vectors in this plane to be the eigenvectors.

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    $\begingroup$ @Arthur If you think about it, you even do this in the one-dimensional case. Every non-zero vector of the form $c(-1,-1,2)$ is an eigenvector for $-3$, but you only report one representative. If you don't see this as analogous then you might be misunderstanding "linearly independent". $\endgroup$ – Erick Wong May 4 '15 at 18:00
  • $\begingroup$ @Erick Wong: This, i do understand. But I think you both misunderstood me. For example: $(1,1,1), (3,3,2), (5,3,4)$ would be some of the infinite set of linearly independent vectors which are possible solutions for the linear system above, hence there is an infinite amount of linearly independent eigenvectors for the eigenvalue $3$. Correct me if I'm wrong. Hence my question: Does it matter, how many vectors I pick for a basis and which vectors in particular? $\endgroup$ – Arthur May 6 '15 at 0:54
  • $\begingroup$ @Arthur If you truly believe there is any infinite set of linearly independent vectors in $\mathbb R^3$, then you have a fundamental misunderstanding of linear independence... Make sure you aren't confusing "linearly independent" with "mutually non-parallel": the first conditon is much, much stronger than the second. $\endgroup$ – Erick Wong May 6 '15 at 7:07
  • $\begingroup$ @Erick Wong Ok, I see my mistake now. I somehow confused infinite set of linearly independent vectors with infinite set of bases. $\endgroup$ – Arthur May 11 '15 at 10:09

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