# If $Q_kR_k$ converges to $QR$, where this represents their respective $QR$ decompositions, then $Q_k \rightarrow Q$ and $R_k \rightarrow R$?

Suppose $$Q_kR_k \rightarrow QR$$ as $$k \rightarrow \infty$$, where $$Q_k, Q$$ are orthogonal matrices and $$R_k, R$$ are upper triangular with positive diagonal entries, then would the uniqueness of the $$QR$$ decomposition imply that $$Q_k \rightarrow Q$$ and $$R_k \rightarrow R?$$ I need this detail for a proof, but I wasn't able to prove it.

It would suffice to show that if $$Q_kR_k \rightarrow I$$, then $$Q_k \rightarrow I$$ and $$R_k \rightarrow I$$.

Edit: If the limit of $$Q_k$$ and $$R_k$$ exist, then they must be $$I$$, but how would one show that these limits exist, if they do?

Since the $$Q_k$$ are orthogonal, they must be bounded, so we can apply Bolzano-Weierstrass. As you mentioned in your edit, any convergent subsequence would have to converge to $$I$$.
If $$Q_k$$ does not converge to $$I$$, then we can take the subsequence of all $$Q_k$$ that are at least $$\epsilon$$ away from $$I$$. Bolzano-Weierstrass then tells us there would be a subsequence that converges to something other than $$I$$, which leads to a contradiction.
Convergence of $$Q_k$$ and $$Q_kR_k$$, along with invertibility of the $$Q_k$$, then implies convergence of the $$R_k$$.