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Let \(L\) be an \(n \times n\) matrix with real entries and let \(\lambda\) be an eigenvalue of \(L\). In the following list, identify all the assertions that are correct.
a) \(a \lambda\) is an eigenvalue of \(a L\) for any scalar \(a\).
b) \(\lambda^{2}\) is an eigenvalue of \(L^{2}\).
c) \(\lambda^{2}+a \lambda+b\) is an eigenvalue of \(L^{2}+a L+b I_{n}\) for all real scalars \(a\) and \(b\).
d) If \(\lambda=a+i b\), with \(a, b \neq 0\) some real numbers, is an eigenvalue of \(L\), then \(\bar{\lambda}=a-i b\) is also an eigenvalue of \(L\).
in Mathematics by Platinum (164,920 points) | 146 views

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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 $\bar{v}$.

by Diamond (78,343 points)

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