Prove that for a square matrix A: $Im\left(A^{\top}\right)=Ker(A)^{\perp}$

Question

Question:

Let $A$ be a square matrix. Show that: $$ Im\left(A^{\top}\right)=Ker(A)^{\perp} $$ My take:

I was going for bi-directional inclusion.

The first direction is easy.

Let $v\in Im\left(A^{T}\right)$, so there exists $x \in \mathbb{R}^{m}$ such that $A^{T}x=v$.

Let $w\in Ker\left(A\right)$, then: $$ \left\langle v,w\right\rangle =\left\langle A^{T}x,w\right\rangle =\left\langle x,Aw\right\rangle =\left\langle x,0\right\rangle =0 $$ and we get that $Im\left(A^{T}\right)\subseteq Ker\left(A\right)^{\perp}$.

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I need help with showing that $Ker\left(A\right)^{\perp}\subseteq Im\left(A^{T}\right)$, I couldn't figure it out.

Answer 1

We have \begin{align} v \in \text{Ker }A &\Leftrightarrow A v = 0\\ &\Leftrightarrow \langle A v, w\rangle = 0 \quad \forall w \in \mathbb R^m\\ &\Leftrightarrow \langle v, A^T w\rangle = 0 \quad \forall w \in \mathbb R^m\\ &\Leftrightarrow v \in \text{Im }(A^T)^\perp \end{align} so that $\text{Ker }A = \text{Im }(A^T)^\perp$ and $\text{Ker }(A)^\perp = \text{Im }(A^T)$.