# Showing the orthogonal subspace to the Range of A is the null space of the complex conjugate of A

I have matrices $$A \in \mathbb{C}^{m \times n}$$

My goal is to show that $$R^{\perp}(A) = N(A')$$.

Conceptually I get that the orthogonal space to the range of the matrix is going to be the null space because otherwise it'd be in the row space but I'm having a hard time expressing it in math.

I haven't done too much complex matrix work so I'm not sure what pitfalls I need to be wary of

The column space (range) of $$A$$ is the row space of $$A^T$$. Thus $$v\in R^\perp(A)\iff\forall i,v^*r_i=[0]\iff v^T\bar{r_i}=0$$ where $$r_i$$ is a row vector of $$A^T$$. This gives $$\bar{A^T}v=0\iff v\in\ker(\bar{A^T})$$.
• What is the difference between $v*r_i = [0]$ and $v^T \bar{r}_i = 0$? – financial_physician Jan 13 at 21:29
• I used $*$ to denote the conjugate transpose, $\bar{ }$ to denote the conjugate and $^T$ to denote the transpose. – Shubham Johri Jan 13 at 21:30