Answer
See below.
Work Step by Step
The characteristic equation of the matrix is
$det(xI_2-A)=
det\left(\begin{bmatrix}
x-2& 1&1 \\
1& x-2&1\\
1&1&x-2\\
\end{bmatrix} \right)=0$
Hence $x^3-6x^2+9x-4=0\\(x-1)^2(x-4)=0$
Thus the eigenvalues are $x=1$ with multiplicity $2$ and $x=4$ with multiplicity 1. A is symmetric; thus by Theorem 7.7, the corresponding $x=1$'s eigenspace will have a dimension of $2$ and the corresponding $x=4$'s eigenspace will have a dimension of $1$.
