Trouble finding eigenvector for corresponding eigenvalue -1 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.
 A: 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.  
A: 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 $\lambda_{-} =-1$,
$$
A v_{-} = 
\left(
\begin{array}{rr}
 1 & 0 \\
 1 & -1 \\
\end{array}
\right)
\left(
\begin{array}{cc}
 0 \\
 1 \\
\end{array}
\right)
=
\left(
\begin{array}{r}
 0 \\
 -1 \\
\end{array}
\right)
=
-
\left(
\begin{array}{r}
 0 \\
 1 \\
\end{array}
\right)
=
\lambda_{-} v_{-}
$$
