# Proof Verification: the orthogonal complement of the column space is the left nullspace

Can someone please check my proof and my definitions.

Let $$A \in \mathbb{R}^{n \times m}$$ be my matrix.

The left null space of $$A$$ is written as,

$$\mathcal{N}(A^\top) = \{x \in \mathbb{R}^n| A^\top x = 0\}$$

The orthogonal complement of the column space $$\mathcal{C}(A)$$ is written as,

$$\mathcal{C}(A)^\perp = \{x \in \mathbb{R}^n | x^\top y = 0, \forall y \in \mathcal{C}(A)\}$$

We want to show that $$\mathcal{N}(A^\top) = \mathcal{C}(A)^\perp$$

First, we show, $$\mathcal{N}(A^\top) \subseteq \mathcal{C}(A)^\perp$$

Let $$x \in \mathcal{N}(A^\top)$$, then $$A^\top x = 0 \implies x^\top A = 0^\top \implies x^\top Av= 0^\top v, \forall v \in \mathcal{C}(A) \implies x^\top y = 0 , y = Av$$, $$\implies x \in C(A)^\perp$$.

Next, we show, $$\mathcal{N}(A^\top) \supseteq \mathcal{C}(A)^\perp$$

Let $$x \in C(A)^\perp$$, then $$x^\top y = 0$$, forall $$y \in C(A)$$. But $$y = Av, \forall v \in \mathbb{R}^n$$. Hence, $$x^\top y = x^\top Av = v^\top A^\top x.$$ For all $$v \neq 0, A^\top x = 0$$, hence $$x \in \mathcal{N}(A^\top)$$.

I'm pretty confident about the first proof. But the second proof is a bit more rough. Can someone please check for me.

$$y \in C(A)$$ means that there exists (at least one) $$v$$ of appropriate dimension such that $$y = Av$$.

So we can say: For $$x \in C(A)^{\perp}$$, then $$x^T y = 0$$ for every $$y \in C(A)$$. For every $$y \in C(A)$$, we can express $$y = Av$$ for some (nonzero) $$v$$. So we can always express $$x^T y$$ as $$x^T Av$$. So $$x^T y = x^T (A v) = (x^T A) v = (A^T x)^T v = 0^T v = 0$$ for $$v \neq 0$$, so we must have $$A^T x = 0$$, i.e., $$x \in N(A^T)$$.