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For the matrix

$$ A= \begin{bmatrix} 2&-1\\ -1&2 \end{bmatrix} $$

the eigenvalues are $1$ and $3$. This gives the eigenvector $(1,1)$ when $\lambda=1$.

When $\lambda=3$

$$ (A-3I)x= \begin{bmatrix} -1&-1\\ -1&-1 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} 0\\0 \end{bmatrix} \mathrm{gives\;eigenvector} \begin{bmatrix} 1\\-1 \end{bmatrix} $$

However can we say there is a second eigenvector $(-1,1)$ as this also solves $x+y=0$ when $\lambda=3$?

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    $\begingroup$ $\begin{bmatrix}1\\-1\end{bmatrix}$ and $\begin{bmatrix}-1\\1\end{bmatrix}$ are linearly dependent, so it does not count as another eigenvector. $\endgroup$ – DHMO Apr 17 '17 at 8:59
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    $\begingroup$ If you have $n$ linearly independent linear equations and $k$ variables then you will have a $k-n$ dimensional solution space. $k=2$ and $n=1$ in this case so $k-n=2-1=1$ $\endgroup$ – mathreadler Apr 17 '17 at 9:00
  • $\begingroup$ So if only one counts as an eigenvector when $\lambda=3$, does it matter which is chosen? $\endgroup$ – clicky Apr 17 '17 at 9:00
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    $\begingroup$ @clicky No, it does not matter. $\endgroup$ – DHMO Apr 17 '17 at 9:01
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    $\begingroup$ Any scalar constant times $[1\,\,-1]^T$ will do. $\endgroup$ – mathreadler Apr 17 '17 at 9:01
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No, you cannot say that there is a second eigenvector $\begin{bmatrix}-1\\1\end{bmatrix}$ , because $\begin{bmatrix}1\\-1\end{bmatrix}$ and $\begin{bmatrix}-1\\1\end{bmatrix}$ are linearly dependent and thus form the same eigenspace, which means it will not span the space of your matrix (for which two linearly independent eigenvectors, such the ones you have found, are needed).

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Actually, to each eigenvalue there belongs an eigenspace as you can easily show that if $v$ and $w$ are eigenvectors to the same eigenvalue of $A$, then $\alpha v+\beta w$ is eigenvector as well.

Therefore what we usually mean when we speak of the eigenvectors to a given eigenvalue is actually a basis for the corresponding eigenspace.

In your specific example, the eigenspace to eigenvalue $3$ is one-dimensional, therefore it has a single non-zero vector as basis. Any non-zero vector in that eigenspace will do.

Note that the eigenspace is not always one-dimensional. For example, take the identity matrix. It maps all vectors to themselves, therefore all vectors are solutions of the eigenvalue equation for eigenvalue $1$, that is, the eigenspace to eigenvalue $1$ is the full vector space. Therefore any basis of the eigenspace is a basis of the full vector space. So for the $n\times n$ identity matrix, you've got $n$ eigenvectors to the eigenvalue $1$.

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All vectors solving the equation

$$x+y=0$$

are eigenvectors with eigenvalues $3$. Basically eigenvectors for an eigenvalue $\lambda$ is not unique in general, but you always find an entire vector subspace of eigenvector.

Read again the definition of eigenvector with eigenvalue $\lambda$. They are all vector $v$ such that $Av=\lambda v$ that is $Av-\lambda v=0$. In this way you find an omogeneous system and the solution space is the eigenspace relative to $\lambda$. Any element inside this space is an eigenvector.

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