Let $A:\mathbb R^m \to \mathbb R^n$ be a linear transformation. If $W$ be a subspace of $\mathbb R^n.$ define $$W^\perp=\{y\in \mathbb R^n|\langle x,y\rangle=0 \text{ for all } x\in W\}$$ Then which one of the following is true?
(a) $R(A)\subset (Ker(A^T))^\perp$
(b) $R(A)=(Ker(A^T))^\perp$
(c) None of the above
Attempt:- Claim:- $R(A)\subset (Ker(A^T))^\perp$
$z\in R(A)\implies \exists x\in \mathbb R^m: z=A(x)$
We know that $A^T:(\mathbb R^n)^* \to (\mathbb R^m)^*$. $W^*$ denotes the dual space of $W.$ Let So, $Ker(A^T)=\{g \in (\mathbb R^n)^*: A^T(g)=0\}.$ So, $(Ker(A^T))^\perp=\{g \in (\mathbb R^n)^*| \langle g, h\rangle =0,\forall h\in Ker(A^T)\}.$ Let $g\in Ker(A^T). $ We keed to prove that $\langle g, z\rangle=0 \iff \langle g, A(x)\rangle=0 $. I am not able to proceed. Please help me. It was a question appeared in NBHM Examination, India. How do I find this complicated answer quickly?
You show that $Ker(A^{T})=R(A)^{\perp}$ and then you use that in $\mathbb{R}^n$ you have for any subspace $W$ that $W^{\perp\perp} = W$.
$$y \in Ker(A^{T}) \Leftrightarrow \langle \underbrace{A^Ty}_{=o},x \rangle = 0\: (\forall x\in \mathbb{R}^n)$$ $$\Leftrightarrow \langle y,Ax \rangle = 0\: (\forall x\in \mathbb{R}^n)$$ $$\Leftrightarrow \langle y,z \rangle = 0\: (\forall z\in R(A))$$ $$\Leftrightarrow y \in R(A)^{\perp}$$
Now, you take the orthogonal complement of $Ker(A^{T})=R(A)^{\perp} \Leftrightarrow Ker(A^{T})^{\perp}=R(A)^{\perp\perp}= R(A)$.