Show that $\lambda$ is eigenvalue of a normal $A$ if and only if $\bar \lambda$ is eigenvalue of $A^*$

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.

You can simply use that a matrix $A$ and its transpose have the same eigenvalues (they have the same characteristic and minimal polynomials), and that conjugating a matrix gives the complex conjugate characteristic polynomial, whose roots are complex conjugates of those of the original. So the stated property holds for any complex matrix$~A$; being normal is a red herring.