Let $A(t)$ be a smooth family of invertible $n \times n$ matrices with $A(0) = I$, and let $A(t) = Q(t) R(t)$ be the $QR$ decomposition. Given $\dot{A}(0)$, what is an algorithm to compute $\dot{Q}(0), \dot{R}(0)$ in $O(n^2)$ run-time?
I can find using the product rule that $$\dot{A}(0) = \dot{Q}(0) + \dot{R}(0)$$ and that $\dot{Q}(0)$ must be antisymmetric. But I don't know how to proceed from here.
Since $R(t)$ is upper triangular, $R'(0)$ must be upper triangular as well. Split $$ A'(0) = L + U $$ into $L$ strictly lower triangular and $U$ upper triangular matrices. Then it follows that the lower triangular part of $Q'(0)$ must be equal to $L$, hence $Q'(0)$ is given by $$ Q'(0) = L - L^H. $$ This implies $$ R'(0) = U +L^H. $$ Thats $O(n^2)$ run-time :)