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.


Labels:

Comments

Post a Comment

Popular posts from this blog

[Community Question] Calculus: Manifold with boundary - finding the boundary

One of our user asked: I have the manifold with boundary $M:= \lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1\geq 0, x_1^2+x_2^2+x_3^2=1\rbrace \cup\lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1= 0, x_1^2+x_2^2+x_3^2\leq1\rbrace$ and I need to find the boundary of this manifold. I think it is $\lbrace (x_1,x_2,x_3) \in \mathbb R^n : x_1= 0, x_2^2+x_3^2=1\rbrace$ , the other option is that the boundary is the empty set? I think the first is right? Am I wrong?
Post a Comment
Read more

[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).
Post a Comment
Read more

Order of elements of the Prüfer groups $\mathbb{Z}(p^{\infty})$

Let $\mathbb{Z}(p^{\infty})$ be defined by $\mathbb{Z}(p^{\infty}) = \{ \overline{a/b} \in \mathbb{Q}/ \mathbb{Z} / a,b \in \mathbb{Z}, b=p^i$ $ with$ $ i \in \mathbb{N} \}$ , I wish show that any element in $\mathbb{Z}(p^{\infty})$ has order $p^n$ with $n \in \mathbb{N}$ . i try several ways but I have not been successful, some help ?? thank you
Post a Comment
Read more