Effect of orthonormal projection on rank

  • Thread starter Thread starter kalleC
  • Start date Start date
  • Tags Tags
    Projection rank
kalleC
Messages
9
Reaction score
0

Homework Statement


Given rank(R) and a QR factorization A = QR, what is the rank(A)


Homework Equations





The Attempt at a Solution


I want to know if multiplication by a full rank orthonormal matrix Q and an upper trapezoidal matrix R yields rank(R)=rank(Q*R)=rank(A)

This is mostly guesswork by me but I'd like to use it for a question I need to answer.
 
Physics news on Phys.org
Well, the rank of a matrix is the dimension of the image, right? If the image of R is a subspace S of dimension rank(R), then what's the dimension of Q(S) if Q is full rank?
 
They are equal? As the only way Q(S) would be dissimilar would be if rank(Q)<rank(R).

But does not the reason for this have anything to do with Q being orthnormal? Otherwise couldn't Q act on R and cause some of the image to overlap effectively reducing the rank?
 
Q is full rank, so it's one to one. So yes, rank(QR)=dim(Q(S))=dim(S)=rank(R). So rank(QR)=rank(R).
 
Thank you very much =)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top