For the matrix \[\left[ {\begin{array}{*{20}{c}} 4&2 \\ 2&4 \end{array}} \right]\] the eigen value corresponding to the eigen vector \[\left[ {\begin{array}{*{20}{c}} {101} \\ {101} \end{array}} \right]\] is A. 2 B. 4 C. 6 D. 8

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 $A$.