# 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?

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.

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.

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.

• You cannot conclude that $A^*x$ and $x$ are parallel, since the eigenspace associated with $\lambda$ might have dimension greater than 1. Jul 29, 2015 at 15:29
• You can fix that, however, by noticing that the eigenspace is invariant under $A^*$, as well as under $A$. By restricting the operators to this eigenspace, we have $A=\lambda I$ and therefore $A^* = \overline{\lambda} I$. Jul 29, 2015 at 15:31