What matrices have property $\mathcal{N}(A)=\mathcal{N}(A^*)$? In one paper, to prove some linear algebra results, authors assumed that some matrix $A\in\mathbb{C}^{n\times n}$ have the following property: $$\mathcal{N}(A)=\mathcal{N}(A^*),\tag{1}$$ where $\mathcal{N}(A)$ is a null space of $A$, and $A^*$ is a conjugate transpose of $A$.
I want to understand this assumption, i.e., what matrices have this property?
Using SVD, we obtain $A=USV^*$, where $S$ is a diagonal matrix with singular values; $U,V$ are unitary. Then, if all singular values are strictly larger than $0$, $(1)$ always holds. So we need to consider only the case, when $S$ contains zero diagonal elements.
For any $x$ that belong to $\mathcal{N}(A)$ and $\mathcal{N}(A^*)$, we have $SV^*x=SU^*x=0.$ Let $y=U^*x$, then $SV^*Uy=Sy=0.$ I am not sure how to proceed, does it mean that we must have $V=U^*$?
 A: If $M$ is the matrix of a normal linear operator (on a finite-dimensional inner product space) with respect to some basis, then $\text{null}(M) = \text{null}(M^\ast)$.
Note: If $V$ is a finite-dimensional inner product space, we say it's normal if it commutes with its adjoint.
Context: this is a more abstract way of thinking about it than dealing directly with matrices... it takes some time to wrap your head around the idea of the adjoint of a linear map if you've never seen it before.  If $T:V\to W$, where $V,W$ are finite-dimensional inner product spaces, then the adjoint of $T$, denoted $T^\ast$, is the unique linear map $T^\ast: W \to V$ such that $\left\langle Tv, w\right\rangle = \left\langle v, T^\ast w\right\rangle$ for all $v\in V$ and $w\in W$.  The existence and uniqueness of $T^\ast$ is guaranteed by the Riesz representation theorem.  Once you commit to a basis in $V$ and $W$ and represent $T$ by a matrix $M$, then the matrix of $T^\ast$ is the conjugate transpose of $M$ (the definition you're probably used to).  There's actually quite a bit going on in the background!
