Solution: We recall that the eigenspace of the eigenvalue $\lambda $ is the null space of the matrix $A - \lambda I$. Here, \[ A - 3I = \begin{bmatrix} 11 & -4 & -8 \\ 4 & 1 & -4 \\ 8 & -4 & -5 \\ \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 8 & -4 & -8 \\ 4 & -2 & -4 \\ 8 & -4 & -8 \\ \end{bmatrix}. \]

Now we need to find the null space of the matrix $A - 3I$. We will use the row reduction to find the null space of $A - 3I$. \begin{align*} A- 3I = \begin{bmatrix} 8 & -4 & -8 \\ 4 & -2 & -4 \\ 8 & -4 & -8 \\ \end{bmatrix} \xrightarrow[R_3 \to R_3 - R_1 ]{R_2 \to R_2 - \frac{1}{2}R_1} \begin{bmatrix} 8 & -4 & -8 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} = 4\begin{bmatrix} 2 & -1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}. \end{align*} Therefore, any vector $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix}$ which is in the null space must satisfy \[ 2x_1 - x_2 - 2x_3 = 0 \implies x_2 = 2x_1 - 2x_3. \] Thus, \[ \mathbf{x} = x_1 \begin{bmatrix} 1 \\ 2 \\ 0 \\ \end{bmatrix} + x_3 \begin{bmatrix} 0 \\ -2 \\ 1 \\ \end{bmatrix}. \] Therefore, the eigenspace of $A$ corresponding to the eigenvalue $\lambda =3$ will be \[ \operatorname{span}\left\{ % \begin{bmatrix} 1 \\ 2 \\ 0 \\ \end{bmatrix}, \begin{bmatrix} 0 \\ -2 \\ 1 \\ \end{bmatrix} \right\} . \]