Conditions for upper-triangular factor in QR factorization

Question

Let $A,Q_1 \in \mathbb{R}^{m_1 \times n}$, $B \in \mathbb{R}^{m_2 \times n}$ $Q_2 \in \mathbb{R}^{(m_1+m_2) \times n}$, and $Q_3 \in \mathbb{R}^{ (n+m_2) \times n }$ with $Q_i$ having orthonormal columns for $i=1,2,3$. Let $R_1,R_2,R_3 \in \mathbb{R}^{n \times n}$ be upper triangular. Assume

\begin{equation*} A = Q_1R_1 \\ \begin{bmatrix} B \\ A \end{bmatrix} = Q_2R_2 \\ \begin{bmatrix} B \\ R_1 \end{bmatrix} = Q_3R_3 \end{equation*}

Under certain conditions, $R_2 = R_3$. What are these conditions? As a side note, the problem is sort of saying that the "upper-triangular factor obtained in $QR$ factorization when augmenting the matrix B is invariant under the unitary transformation $Q_1^{T}$."

Note that it is implicit that $m_1,m_2 \geq n$ given that $Q_i$ have orthonormal columns. The only thing I can think to try is to multiply the equations by $Q_2^{T},Q_3^{T}$ respectively and see under what constraints that makes $R_2 = R_3$. I tried expanding $Q_i$ as block matrix...this didn't lead me anywhere (but might work). I'm unsure how to exploit that $R_i$ are upper triangular.

Answer 1

Let $I$ be the $m_2 \times m_2$ identity. Then the matrix $$ Q_0:= \begin{bmatrix} I & 0 \\ 0 & Q_1 \end{bmatrix} \in \mathbb{R}^{(m_1 + m_2) \times (m_2 \times n)} $$ has orthonormal columns. (For why, think about $Q_0$ in terms of its indiviual columns and how the 2-norm is calculated.)

Furthermore, $$ Q_0 Q_3 R_3 = Q_0 \begin{bmatrix} B \\ R_1 \end{bmatrix} = \begin{bmatrix} B \\ Q_1 R_1 \end{bmatrix} = \begin{bmatrix} B \\ A \end{bmatrix} = Q_2 R_2 $$ Since each $Q_i$ has full rank ($Q_0$ by construction; the others by assumption), then $R_3 = R_2$ follows from requring that $Q_0 Q_3 = Q_2$.