The correct answer is $\boxed{4}$.
To find the eigenvalues of a matrix, we can use the following formula:
$$\lambda = \frac{tr(A) – \det(A)}{2}$$
where $tr(A)$ is the trace of $A$ and $\det(A)$ is the determinant of $A$.
In this case, we have:
$$tr(A) = 4 + 4 = 8$$
$$\det(A) = 4^2 – 2 \cdot 4 \cdot 2 = 0$$
Therefore, the eigenvalues of $A$ are $\frac{8 – 0}{2} = 4$.
To find the eigenvector corresponding to the eigenvalue $\lambda = 4$, we can use the following formula:
$$v = \frac{1}{\det(A – \lambda I)}(A – \lambda I)u$$
where $u$ is any non-zero vector.
In this case, we can choose $u = \left[ {\begin{array}{*{20}{c}} {101} \ {101} \end{array}} \right]$.
Therefore, the eigenvector corresponding to the eigenvalue $\lambda = 4$ is $\boxed{\left[ {\begin{array}{*{20}{c}} {101} \ {101} \end{array}} \right]}$.
Here is a brief explanation of each option:
- Option A: $2$ is not an eigenvalue of $A$.
- Option B: $4$ is an eigenvalue of $A$.
- Option C: $6$ is not an eigenvalue of $A$.
- Option D: $8$ is not an eigenvalue of $A$.