Okay, so my eigenvalue is 1. It has multiplicity 3.

$\displaystyle det(xI-A)=\left( \begin{array}{ccc} x+3 & 0 & 8 \\ -4 & x-1 & -8 \\ -2 & 0 & x-5 \end{array} \right)$

After plugging in my eigenvalue, I find all three rows are equal, so I cancel out two of them. I'm left with:

$\displaystyle -2x+0y-4z=0$

I find the eigenvector to be:

$\displaystyle t\left( \begin{array}{ccc} -2 \\ 0 \\ 1 \end{array} \right)$

The question asks for the eigenvalue (which is 1) and the eigenspace. I still don't understand what the eigenspace is? It also has room for two different answers (of 3 rows and 1 column.)

I tried entering $\displaystyle \left( \begin{array}{ccc} -2 \\ 0 \\ 1 \end{array} \right)$ but it rejected it.

Help!