Answer
See below.
Work Step by Step
The characteristic equation of the matrix is
$det(xI_2-A)=
det\left(\begin{bmatrix}
x& -2&-2 \\
-2& x&-2\\
-2&-2&x\\
\end{bmatrix} \right)=0$
Hence $x^3-12x-16=0\\(x+2)^2(x-4)=0$
Thus the eigenvalues are $x=-2$ with multiplicity $2$ and $x=4$ with multiplicity 1. A is symmetric; thus by Theorem 7.7, the corresponding $x=-2$'s eigenspace will have a dimension of $2$ and the corresponding $x=4$'s eigenspace will have a dimension of $1$.
