Matrices are close if and only if null spaces are close

Question

Suppose $A$ and $B$ are $n \times m$ matrices with $n < m$. Assume $AG=BG=I_n$ (the identity matrix of dimension $n \times n$), where $G$ is an $m \times n$ matrix (with $Rank(G)=n$), and assume that $A H =0$, where $H$ is an $m \times (m-n)$ matrix and $0$ is an $n\times(m-n) $ matrix of $0$'s, and $Rank(H)=m-n$. Show that for all $\varepsilon>0$ there exists a $\delta>0$ such that

$$ ||AH-BH||< \delta \quad \Rightarrow \quad ||A-B||<\varepsilon. $$
where $||\cdot||$ is the operator norm. In words, if the null space of $A$ and the null space of $B$ are close then $A$ and $B$ are close.

Edit: I made some updates to reflect what I wanted from the question (specifically, $||AH-BH||< \delta \quad \Leftarrow \quad ||A-B||<\varepsilon$ is immediate). I will post my solution now. Any comments /confirmation is appreciated.

Answer 1

Note that $(G,H)$ is nonsingular because of the rank nullity theorem. Then,

\begin{align*} ||A-B||&=||[A(G,H)-B(G,H)](G,H)^{-1}||\\ &=||[(I_n,0)-(I_n,BH)](G,H)^{-1}||\\ &=||(0,BH)(G,H)^{-1}||\\ &\leq ||BH||\cdot||(G,H)^{-1}||\\ &=||AH-BH||\cdot||(G,H)^{-1}|| \end{align*}