Show that if $A \in \mathbb{C}^{m \times n}$ is of full rank, then null$(A^*)$ is the orthogonal complement of range$(A)$.

Question

Show that if $A \in \mathbb{C}^{m \times n}$ is of full rank, then null$(A^*)$ is the orthogonal complement of range$(A)$.

I saw the above fact listed in this linked MSE proof.

My attempt to prove it:

proof

Assume $m \geq n$.

First, we can show that for any vector in null$(A)$ that that vector is orthogonal to any vector in range$(A)$. So let $y \in \text{null}(A)$. Then

$$y^*Ax = (A^*y)^*x = 0^*x = 0$$

Now here's where I'm a little lost. So the following is not proof but a description of what I think needs to happen.

Since $A \in \mathbb{C}^{m \times n}$ and $A$ has full rank then the range of $A$, $Ax$ can be written

$$x_1A_1 + \cdots+ x_nA_n$$

where $A_i$ denotes a column of $A$. That is, the span$(A) = \mathbb{C}^n$ but range$(A) \subset \mathbb{C}^m$ so I think we need to show that span$(A^*) = m-n$.

Answer 1

You don't need the hypothesis.

Assume $z \perp \mathrm{range}(A)$. Consider any $x \in \mathbb{C}^n$. Since $\langle z,Ax \rangle_{\mathbb{C}^m} = 0$ you have $\langle A^*z,x \rangle_{\mathbb{C}^n}=0$. Setting $x=e_i,i=1,\dots,n$, you find $A^*z=0$, i.e. $z$ is in the null space of $A^*$. Thus the orthogonal complement of the range of $A$ is contained in the null space of $A^*$.

To finish we do the argument in reverse: assume $A^*z=0$, then for any $x \in \mathbb{C}^n$ you have $\langle A^*z,x \rangle_{\mathbb{C}^n}=0$. Equivalently $\langle z,Ax \rangle_{\mathbb{C}^m}=0$. Since $x$ was arbitrary you conclude the reverse containment you want.