Answer
See below.
Work Step by Step
The characteristic equation of the matrix is
$det(xI_2-A)=
det\left(\begin{bmatrix}
x& -4&-4 \\
-4& x-2&0\\
-4&0&x+2\\
\end{bmatrix} \right)=0$
Hence $x^3-36x=0\\x(x+6)(x-6)=0$
Thus the eigenvalues are $x=-6$ with multiplicity $1$, $x=6$ with multiplicity $1$, and $x=0$ with multiplicity 1. A is symmetric; thus by Theorem 7.7, the corresponding $x=-6$'s eigenspace will have a dimension of $1$, the corresponding $x=6$'s eigenspace will have a dimension of $1$ and the corresponding $x=4$'s eigenspace will have a dimension of $1$.
