True or False: P-Dimensional Subspace and Basis for R^n

  • Thread starter Thread starter Dwolfson
  • Start date Start date
  • Tags Tags
    Subspace
Dwolfson
Messages
9
Reaction score
0

Homework Statement



If H is a p-dimensional subsapce for R^n and {v1,...vp}
is a spanning set of H, then {v1,...vp} is automatically a basis for H.


True or False


Homework Equations



I am unsure of my answer.

The Attempt at a Solution



I am under the impression that this is true due to the fact that since the subspace is p-dimensional that {v1,...vp} is a basis because it must be linearly independent because this set spans p-dimensions thus needs p vectors.
 
Physics news on Phys.org
That's pretty much it. If you have p vectors that are linearly independent, and that span a subspace of dimension p, then these vectors are a basis for that subspace.
 
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