TITLE: Show that $\lambda$ is eigenvalue of a normal $A$ if and only if $\bar \lambda$ is eigenvalue of $A^*$
QUESTION [6 upvotes]: I want to show that $\lambda$ is an eigenvalue of a normal matrix $A$ if and only if $\overline{\lambda}$ is an eigenvalue of $A^{*}$. I am trying to show it for a while and I guess there are some simple ideas to deal with it but I am just stuck. I know that a matrix is normal if and only if it is unitarily diagonalizable but I don't want to use this property cause I believe there must be other easier solution. Any advice?

REPLY [11 votes]: Even though this question is very old, I thought I would give an answer to show a different, more algebraic approach.
Since $A$ is a normal matrix it can be shown that:
1) $||A\overrightarrow{z}|| = ||A^*\overrightarrow{z}||$
2) $ A - \lambda I $ is normal
Given that $\lambda$ is an eigenvalue with some corresponding eigenvector $\overrightarrow{z}$, then observe the following:
$$A\overrightarrow{z} = \lambda \overrightarrow{z} $$
$$A\overrightarrow{z} - \lambda \overrightarrow{z} = \overrightarrow{0} $$
$$(A - \lambda I)\overrightarrow{z} = \overrightarrow{0} $$
$$||(A - \lambda I )\overrightarrow{z}|| = 0 $$
$$||(A - \lambda I)^*\overrightarrow{z}|| = 0 $$
$$||(A^* - \overline{\lambda} I)\overrightarrow{z}|| = 0 $$
$$||A^*\overrightarrow{z} - \overline{\lambda}\overrightarrow{z}|| = 0 $$
From which we can conclude that $A^*\overrightarrow{z} = \overline{\lambda}\overrightarrow{z}$ as required. The if and only if can be shown the exact same way essentially.
It is also interesting to observe that the eigenvector remains the same even when the eigenvalue changes.