The correct answer is $\boxed{\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \ {\frac{{ – 1}}{{\sqrt 2 }}} \end{array}} \right)}$.
To find the eigenvalues and eigenvectors of a matrix, we can use the following formula:
$$\lambda v = A v$$
where $\lambda$ is the eigenvalue, $v$ is the eigenvector, and $A$ is the matrix.
In this case, we have the following matrix:
$$A = \left[ {\begin{array}{*{20}{c}} 5&3 \ 1&3 \end{array}} \right]$$
To find the eigenvalues, we can solve the following equation:
$$| A – \lambda I | = 0$$
where $I$ is the identity matrix.
Solving this equation, we find that the eigenvalues are $\lambda = 5$ and $\lambda = 3$.
To find the eigenvectors corresponding to each eigenvalue, we can substitute each eigenvalue into the following equation:
$$(A – \lambda I) v = 0$$
In this case, we have the following equations:
$$\left[ {\begin{array}{*{20}{c}} 5 – \lambda &3 \ 1&3 – \lambda \end{array}} \right] v = 0$$
For the eigenvalue $\lambda = 5$, we have the following equation:
$$\left[ {\begin{array}{*{20}{c}} 5 – 5 &3 \ 1&3 – 5 \end{array}} \right] v = 0$$
Solving this equation, we find that the eigenvector corresponding to $\lambda = 5$ is $v = \left( {\begin{array}{*{20}{c}} 1 \ 0 \end{array}} \right)$.
For the eigenvalue $\lambda = 3$, we have the following equation:
$$\left[ {\begin{array}{*{20}{c}} 3 – 3 &3 \ 1&3 – 3 \end{array}} \right] v = 0$$
Solving this equation, we find that the eigenvector corresponding to $\lambda = 3$ is $v = \left( {\begin{array}{*{20}{c}} 0 \ 1 \end{array}} \right)$.
To normalize the eigenvectors, we divide them by their norm. The norm of a vector is given by the following formula:
$$\| v \| = \sqrt{v^T v}$$
In this case, we have the following eigenvectors:
$$v_1 = \left( {\begin{array}{*{20}{c}} 1 \ 0 \end{array}} \right)$$
$$v_2 = \left( {\begin{array}{*{20}{c}} 0 \ 1 \end{array}} \right)$$
The norms of these vectors are $\| v_1 \| = 1$ and $\| v_2 \| = 1$. Therefore, the normalized eigenvectors are given by:
$$v_1 = \left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \ {\frac{{ – 1}}{{\sqrt 2 }}} \end{array}} \right)$$
$$v_2 = \left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \ {\frac{{ 1}}{{\sqrt 2 }}} \end{array}} \right)$$
Therefore, one of the normalized eigenvectors of the matrix $A$ is $\boxed{\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \ {\frac{{ – 1}}{{\sqrt 2 }}} \end{array}} \right)}$.