Let $A$ be a real $m\times n$ matrix with $n>m$. Let $Q_1$ be a $n\times m$ with orthonormal columns such that $$ AQ_1 = L $$ where $L$ is of dimension $m\times m$ and lower triangular.
Question: Is it always possible to find a $n\times (n-m)$ matrix $Q_2$ such that $Q:=[Q_1\, |\, Q_2]$ is orthogonal and $$ AQ = [L\, |\, \mathbf{0}_{m\times (n-m)}], $$ where $\mathbf{0}_{m\times (n-m)}$ denotes a $m\times (n-m)$ zero matrix?
The matrix $A$ has maximum rank $m$. So rank of its null space $\ge n-m$. Hence we can get $n-m$ independent vectors from $Null(A)$ which constitutes $Q_2$ such that $AQ_2=0$. The columns of $Q_1$ and $Q_2$ will be independent. We can carry out a Gram Schmidt orthogonalization to make the matrix $Q$ orthogonal.