# How to show for a normal matrix $A$, $(A,\lambda,x) \Leftrightarrow (A^*,\bar{\lambda},x)$?

A matrix $$A$$ is normal if $$AA^*=A^*A$$. Suppose $$(\lambda,x)$$ is an eigenpair of $$A$$, i.e., $$Ax = \lambda x$$. Proof for a normal matrix $$A$$, $$(\lambda,x)$$ is an eigenpair of $$A$$ if and only if $$(\bar{\lambda},x)$$ is an eigenpair of $$A^*$$.

My try:

We have $$Ax = \lambda x$$ and want to show $$A^*x=\bar{\lambda}x$$. Taking the conjugate transpose we have $$x^*A^* = \bar{\lambda} x^*$$. Now we can write $$x^*A^* A= \bar{\lambda} x^*A$$, using the fact that $$A$$ is normal, we have $$x^*A A^*= \bar{\lambda} x^*A$$. Now I do not know how to get $$A^*x=\bar{\lambda}x$$.

Proving one direction is enough. Think in norm. Equality in normed space means zero in norm: $$\Vert (A-\lambda I) x \Vert = 0$$, then take transpose inside: $$\Vert x^* (A^*- \lambda I) \Vert = 0$$. Square both sides and appy the given condition that $$A$$ is normal to get $$x^* AA^* x +\Vert \lambda\Vert^2 x^* x -\bar{\lambda} x^* A x - \lambda x^* A^* x.$$ Factorize this into $$\Vert (A^*-I) x \Vert^2 = 0$$.