The correct answer is: A. S1 implies S2.
A real 4 Ã 4 matrix $M$ has 4 linearly independent eigenvectors if and only if it has 4 distinct eigenvalues. This is because if $v_1, v_2, v_3, v_4$ are linearly independent eigenvectors of $M$ corresponding to the eigenvalues $\lambda_1, \lambda_2, \lambda_3, \lambda_4$, then we can write $M v_i = \lambda_i v_i$ for $i = 1, 2, 3, 4$. If $\lambda_1 = \lambda_2$, then $v_1 – v_2$ is also an eigenvector of $M$ corresponding to the eigenvalue $\lambda_1$. This means that $v_1$ and $v_2$ are not linearly independent. Similarly, if $\lambda_i = \lambda_j$ for some $i \neq j$, then $v_i$ and $v_j$ are not linearly independent. Therefore, if $M$ has 4 linearly independent eigenvectors, then the eigenvalues $\lambda_1, \lambda_2, \lambda_3, \lambda_4$ must be distinct.
Conversely, if $M$ has 4 distinct eigenvalues, then we can choose a basis for $\mathbb{R}^4$ consisting of eigenvectors of $M$. These eigenvectors will be linearly independent, since the eigenvalues are distinct. Therefore, if $M$ has 4 distinct eigenvalues, then it has 4 linearly independent eigenvectors.
Therefore, S1 implies S2.