Exercise 4.2.2

Question

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ewcommand{\T}{^\mathsf{T}}$
Let $A$ be a square matrix with real entries, and let $\lambda$ be its complex eigenvalue. Suppose $\mathbf{v} = (v_1, v_2, ..., v_n)\T$ is a corresponding eigenvector, $A\mathbf{v} = \lambda \mathbf{v}$. Prove that the $\bar{\lambda}$ is an eigenvalue of $A$ and $A \bar{\mathbf{v}} = \bar{\lambda} \bar{\mathbf{v}}$. Here $\bar{\mathbf{v}}$ is the complex conjugate of the vector $\mathbf{v}$, $\bar{\mathbf{v}} := (\bar{v_1}, \bar{v_2}, ..., \bar{v_n})\T$.

Jing Huang · Accepted Answer

\begin{proof}
$A$ is real matrix. Then $\bar{A} \bar{\mathbf{v}} = A \bar{\mathbf{v}}$. In the same time $\bar{A} \bar{\mathbf{v}} = \overline{A \mathbf{v}} = \overline{\lambda \mathbf{v}} = \bar{\lambda} \bar{\mathbf{v}}$. Thus $A \bar{\mathbf{v}} = \bar{\lambda} \bar{\mathbf{v}}$.
\end{proof}

Exercise 4.2.2

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Exercise 4.2.2

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