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 $\bar{A^T}v=0\iff v\in\ker(\bar{A^T})$.

  • $\begingroup$ What is the difference between $v*r_i = [0]$ and $v^T \bar{r}_i = 0$? $\endgroup$ – financial_physician Jan 13 at 21:29
  • 1
    $\begingroup$ I used $*$ to denote the conjugate transpose, $\bar{ }$ to denote the conjugate and $^T$ to denote the transpose. $\endgroup$ – Shubham Johri Jan 13 at 21:30

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