Eigenvectors of $\left( \begin{array}{ccc} a & 0 \\ 0 & -b \end{array} \right)$ I calculated the eigenvalues of the following matrix to be $a$ and $-b$.
$J = \left( \begin{array}{ccc}
a & 0 \\
0 & -b \end{array} \right)$
But when I use the formula $(J - \lambda I)v = 0$ with either $a$ or $-b$ as eigenvalue to calculate the eigenvectors I get an indeterminate system of equations as they cancel with the $a$ and $-b$ in the matrix.
Yet Wolfram Alpha says the eigenvectors are $(1, 0)$ and $(0, 1)$?

if $a=b=0$
Then you have 
$$\begin{pmatrix}a&0\\0&-b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$$
This means that every basis $(v_1,v_2)$ of $\Bbb{R}^2$ is a basis of eigenvectors. One can verify this by seeing that the dimension of the kernel of $A$ is $2$ and the fact that multiplying the zero matrix by $v_1$ and $v_2$ we get the vector $(0,0)^T$.
If $a=-b$ Then we have 
$$(J-aI)=\begin{pmatrix}0&0\\0&-b-a\end{pmatrix}=\begin{pmatrix}0&0\\0&-b+b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$$
You can do the same reasoning as above to conclude that every basis of $\Bbb{R}^2$ is a basis of eigenvectors.
If $a\neq -b$ we have:
$$(J-aI)v_1=0 \iff\begin{pmatrix}0&0\\0&-b-a\end{pmatrix}v_1=0\implies v_1 = k\cdot \begin{pmatrix}1\\0\end{pmatrix},\quad k\in \Bbb{R},$$
because $$\begin{pmatrix}0&0\\0&-b-a\end{pmatrix} \begin{pmatrix}k\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}.$$
Do you get the second eigenvector?
