Linear Subspace of R^n: Arithmetic Progressions Verification

  • Thread starter Thread starter KaiserBrandon
  • Start date Start date
  • Tags Tags
    Linear Subspace
KaiserBrandon
Messages
51
Reaction score
0

Homework Statement



Is the set of all vectors in R^n whose components form an arithmetic progression a linear subspace of R^n?

Homework Equations



none

The Attempt at a Solution



I basically need one thing verified: would (0,0,0,...,0) be considered an arithmetic progression. The definition says that an arithmetic progression is one where the difference between any two consecutive members of the sequence is constant. Since 0-0=0, it would seem like it is an arithmetic sequence, however, is there a condition that the difference must be non-zero? If not, then (1,1,...,1), (2,2,...,2), etc. would all be arithmetic progressions, and that doesn't seem right to me.
 
Physics news on Phys.org
No, I don't think there's any condition on an arithmetic sequence saying the difference can't be zero.
 
alright, so in that case it is a linear subspace since it meets the three requirements to be a linear subspace. Thanks.
 
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...

Similar threads

Back
Top