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For a matrix of coefficients that is:

$$ \begin{bmatrix} 1 & 0 \\ 1 & -1 \\ \end{bmatrix} $$

I find the eigenvalues to be -1 and 1.

Eigenvalue 1 gives an eigenvector of

$$ \begin{bmatrix} 2 \\ 1 \\ \end{bmatrix} $$

but I'm having trouble finding the eigenvector for the corresponding eigenvalue -1, where I seem to get

$$ \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$

Wolfram Alpha is saying it should be

$$ \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} $$

Any ideas on how it found that?

Thank you.

EDIT - the eigenvector corresponding to an eigenvalue of 1 was the wrong way round.

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    $\begingroup$ If you show your work leading up to $(0,0)$ we might be able to help you. $\endgroup$ – B. Goddard Apr 17 '17 at 13:52
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    $\begingroup$ Just use the definition of eigenvalue and write down the equation system you are supposed to solve if $\lambda =-1$. $\endgroup$ – mathreadler Apr 17 '17 at 13:54
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    $\begingroup$ And, by the way, $[{}^1_2]$ is not an eigenvector; multiplied by the matrix it gives $[{}^{\;1}_{-1}]$. So perhaps you should also show how you got that. $\endgroup$ – Henning Makholm Apr 17 '17 at 13:58
  • $\begingroup$ The second column of the matrix is $-1\cdot(0,1)^T$, so... $\endgroup$ – amd Apr 17 '17 at 17:57
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You need to solve the equation:

$Av = -v$

Where $A$ is your matrix and $v$ is the eigenvector you're looking for. So (using $x$ for the first coordinate and $y$ for the second):

$1x + 0y = -x$

$1x - 1y = -y$

From the first expression we have that $x = 0$. So:

$-y = -y$

That's a tautology, so any $y$ will satisfy the equation. So we may as well pick the normalized one, which has $y = 1$.

As an aside, the $0$ vector will always satisfy the eigenvalue equation, but it's excluded from the definition of eigenvectors.

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  • $\begingroup$ Thank you, the idea that any y will satisfy the equation was what I found online, but I never came across that before so was unsure if it was valid to write. $\endgroup$ – freja Apr 18 '17 at 19:05
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You are on the right track. The matrix $$ A= \left( \begin{array}{cc} 1 & 0 \\ 1 & -1 \\ \end{array} \right) $$ has characteristic function $$ p(\lambda) = \lambda^{2} - 1 $$ whose roots are the eigenvalues $\lambda_{\pm} = \pm 1$.

For the eigenvalue $\lambda_{-} =-1$, $$ A v_{-} = \left( \begin{array}{rr} 1 & 0 \\ 1 & -1 \\ \end{array} \right) \left( \begin{array}{cc} 0 \\ 1 \\ \end{array} \right) = \left( \begin{array}{r} 0 \\ -1 \\ \end{array} \right) = - \left( \begin{array}{r} 0 \\ 1 \\ \end{array} \right) = \lambda_{-} v_{-} $$

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