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