The correct assertions are:
a) \(a \lambda\) is an eigenvalue of \(a L\) for any scalar \(a\).
d) If \(\lambda=a+i b\), with \(a, b \equiv 0\) some real numbers, is an eigenvalue of \(L\), then \(\bar{\lambda}=a-i b\) is also an eigenvalue of \(L\).
Explanation:
a) If \(v \) is an eigenvector of \( L \) corresponding to \(\lambda\), then \(a v \) is an eigenvector of \( a L \) corresponding to \( a\) \(\lambda \).
d) If \(\lambda \) is an eigenvalue of \( L \) with eigenvector \( v \), then $\bar{\lambda}$ is an eigenvalue of \( L \) with eigenvector $\bar{v}$.