Range space of $A$ and Null space of $A^*$

Question

Let $A$ be $p \times q$ matrix with complex entries. Then we define $R(A) = \{Ax : x \in \mathbb{C}^q\, \}$ and $Ker(A) = \{ x \in \mathbb{C}^q : Ax = 0\}$. Let $A^*$ denote conjugate transpose of $A$ i.e. Hermitian operator.

I have to prove that $R(A) \cap Ker(A^*) = \{0\}$

To do this $y$ be in the intersection. Then there is a $x$ in $\mathbb{C}^q$ such that $y = Ax$ and $A^*y=0$. Then $A^*(Ax)=0$. How can I proceed further?

Answer 1

Hint: consider $\langle y, y \rangle$ making use of the facts:

1. $y = Ax$
2. $A^*y = 0$
3. $\langle Au, v \rangle = \langle u, A^*v \rangle$ for any $u, v$.