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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.

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$y \in C(A)$ 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., $x \in N(A^T)$.

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