TITLE: some subspace identity
QUESTION [0 upvotes]: Suppose $A\in\mathbb{R}^{n\times n}$ and $\mathcal{V}$ be any subspace of $\mathbb{R}^n$. Then could anyone help me to prove.
\begin{align*}
&(i)\hspace{0.1cm} (A^{-1}\mathcal{V})^{\perp}=A^T (\mathcal{V}^{\perp}).\\
&(ii) (A\mathcal{V})^{\perp}=A^{-T} (\mathcal{V}^{\perp}).\\
\end{align*}
$A\mathcal{M}$  $:=\{Ax: x\in \mathcal{M}\}$ the image of a set $\mathcal{M}\subseteq\mathbb{R}^n$ under $A\in \mathbb{R}^{m\times n}$. 
$A^{-1}\mathcal{M}:=\{x: Ax\in \mathcal{M}\}$ the pre-image of a set $\mathcal{M}\subseteq\mathbb{K}^n$ under $A\in \mathbb{R}^{m\times n}.$

REPLY [1 votes]: As I mentioned in a comment, (i) follows directly from (ii). Therefore, I will first prove (ii). Note that $x\in (A\mathcal{V})^{\perp}$ iff $x\cdot Ay = 0$ for all $y\in \mathcal{V}$. Also, $x\in (A^{\mathrm{T}})^{-1}\mathcal{V}^{\perp}$ iff $A^{\mathrm{T}}x\cdot y = 0$ for all $y\in \mathcal{V}$. But, $x\cdot Ay\equiv A^{\mathrm{T}}x\cdot y$, so $x\in (A\mathcal{V})^{\perp}$ iff $x\in (A^{\mathrm{T}})^{-1}\mathcal{V}^{\perp}$. This implies that $$(A\mathcal{V})^{\perp} = (A^{\mathrm{T}})^{-1}\mathcal{V}^{\perp}$$
Now, take $A^{\mathrm{T}}$ of both sides of that equation and substitute $A^{-1}\mathcal{V}$ for $\mathcal{V}$ to get $$(A^{-1}\mathcal{V})^{\perp} = A^{\mathrm{T}}\mathcal{V}^{\perp}$$
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