Show that $\lambda$ is eigenvalue of a normal $A$ if and only if $\bar \lambda$ is eigenvalue of $A^*$ 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?
 A: 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.
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.
A: Since $A$ is normal, $A^{*}A=AA^{*}$.   Now, suppose $\lambda$ is an eigenvalue of $A$. Then $Ax=\lambda x$, where $x$ is the eigenvector associated with $\lambda$. Premultiplying with $A^{*}$, we get $A^{*}Ax=\lambda A^{*}x$ or equivalently  $AA^{*}x=\lambda A^{*}x$. This implies that $A^{*}x$ is also an eigenvector associated with $\lambda$. Now we know that $ A^{*}x$ and $x$ are both eigenvectors associated with $\lambda$. Thus $A^{*}x=kx$ for some $k\in \mathbb{C}$. We are left to show that $k=\bar{\lambda}$. Now we pre-multiply $x^{*}$. So we get $x^{*}A^{*}x=kx^{*}x$. or $\frac{x^{*}A^{*}x}{x^{*}x}=k$ Now, taking the conjugate transpose of both sides we get $\frac{x^{*}Ax}{x^{*}x}=\bar{k}$ but from the 1st equation we know that $\frac{x^{*}Ax}{x^{*}x}=\lambda$. This then implies that $\lambda=\bar{k}$ and thus $k=\bar{\lambda}$. The converse can be shown in a similar manner.
