If A is normal, then the nullspace of A is the nullspace of A* Suppose $A$ is a normal matrix. Prove that $x$ is in the nullspace of $A$ if and only if $x$ is in the nullspace of $A^{*}$.
This isn't a homework problem. It was on a test I took recently, and I'd like to know if I was on the right track in solving it.
 A: I suppose that $A \in \mathbb {C}^{n \times n}$. Then
$$\left\Vert Ax \right\Vert^2 =  \left\langle Ax , Ax \right\rangle = \left\langle A^* Ax , x \right\rangle  = \left\langle AA^* x , x \right\rangle = \left\langle A^*x , A^*x \right\rangle = \left\Vert A^*x \right\Vert^2 .$$
Hence, 
$$Ax =0 \Leftrightarrow \left\Vert Ax \right\Vert = 0 \Leftrightarrow \left\Vert A^*x \right\Vert = 0 \Leftrightarrow A^*x = 0 $$
A: $\DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\dim}{dim}$We have in fact the more general statement $\ker(A) = \ker(A^*A)$.
The direction $\subseteq$ is clear. We know that the dimension of the nullspace of a matrix is the number of the zero eigenvalues it has, if $A = U D U^*$, then we see that $A^*A = U D^2 U^*$, i.e. all non-zero eigenvalues of $A^*A$ correspond to the non-zero eigenvalues of $A$, in particular we must have
$$\dim \ker A = \dim \ker(A^*A)$$
which means, the inclusion $\ker(A) \to \ker(A^*A)$ must be a bijection, i.e. $\ker(A) = \ker(A^*A)$.
The desired result follows from $A^*A = AA^*$, i.e. $\ker A = \ker(A^*A) = \ker(AA^*) = \ker A^*$.
