Skip to main content

[Community Question] Linear-algebra: Calculating only the needed part of Q of thin QR decomposition

One of our user asked:

A rectangular, $A \in \mathbb{R}^{m \times n}$ matrix, where $m \ge n$, can be decomposed (QR factorization): $$A = \begin{bmatrix}Q_1 | Q_2 \end{bmatrix}\begin{bmatrix}R\\0\end{bmatrix}$$ where $Q_1$ and $Q_2$ has orthonormal columns, and $R$ is upper triangular.

I'm implementing a routine (based on Householder reflections) which calculates $Q_1$ and $R$ (so called thin/reduced QR decomposition).

My question is: is it possible to calculate $Q_1$ without calculating $Q_2$? The problem is that a Householder matrix is $\mathbb{R}^{m \times m}$, and $Q_1 \in \mathbb{R}^{m \times n}$, so I cannot multiply them. My routine currently calculates $Q=[Q_1|Q_2]$, and then throws away the $Q_2$ part.


Comments

Popular posts from this blog

[Community Question] Linear-algebra: non-negative matrix satisfying two conditions

One of our user asked: A real matrix $B$ is called non-negative if every entry is non-negative. We will denote this by $B\ge 0$ . I want to find a non-negative matrix $B$ satisfying the following two conditions: (1) $(I-B)^{-1}$ exists but not non-negative. Here $I$ is the identity matrix. (2) There is a non-zero and non-negative vector $\vec{d}$ such that $(I-B)^{-1}\vec{d}\ge 0$ . I tried all the $2\times 2$ matrices, but it did not work. I conjecture that such a $B$ does not exist, but don't know how to prove it.

[Community Question] Linear-algebra: Are linear transformations between infinite dimensional vector spaces always differentiable?

One of our user asked: In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).