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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.

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  • $\begingroup$ What are the dimensions of $B$? $\endgroup$ – artificial_moonlet Aug 14 '18 at 14:42
  • $\begingroup$ Sorry about that, I've edited it...$B \in \mathbb{R}^{m_2 \times n}$ and also there was a mistake...$Q_2 \in \mathbb{R}^{(m_1+m_2) \times n}$. $\endgroup$ – TylerMasthay Aug 14 '18 at 16:08
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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 $Q_0 Q_3 = Q_2$.

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  • $\begingroup$ Great answer. Is there more going in the construction of $Q_0$ other than "it's just a trick with the block matrices" or is that essentially all? Also, the dimensions of $Q_0$ are $\mathbb{R}^{(m_1+m_2) \times (n+m_2)}$, correct? $\endgroup$ – TylerMasthay Aug 14 '18 at 16:12
  • $\begingroup$ Fixed the dimensions on $Q_0$. Note that this is not an "if and only if" statement-- you asked for some conditions that imply that $R_2 = R_3$. But I think these conditions are quite natural. You're concatenating $B$ with $A$ and again with a piece of $A$ and then asking how they are related. They must of course be related by the other piece of $A$-- $Q_1$-- and the trick with the block matrices gives you precisely that. It may also help to consider why $Q_0$ is orthonormal in the first place. ;) $\endgroup$ – artificial_moonlet Aug 15 '18 at 8:14

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