A LinAlg Proof Involving Orthogonal Complement

  • Thread starter Thread starter jpcjr
  • Start date Start date
  • Tags Tags
    Orthogonal Proof
jpcjr
Messages
16
Reaction score
0

Homework Statement



Here is the problem and my complete answer.

Am I OK?

Thanks!

http://www.d-series.org/forums/members/52170-albums1546-picture8143.jpg


Homework Equations





The Attempt at a Solution


 
Last edited by a moderator:
Physics news on Phys.org
No, it is not ok. You seem to prove that if u,v\in S^\bot, that cu+v is an element of S. But this is simply not true at all.
Likewise, you take a,b\in S^{\bot \bot} and you conclude that these are in S^\bot. But this is also not true.

In short, it is NOT true that

S^{\bot \bot}\subseteq S^\bot \subseteq S

How do you prove the theorem, well you need to prove two things:

1) S^{\bot \bot} is a subspace.
2) S\subseteq S^{\bot \bot}.
 
Thank you!

By the skin of my teeth, some help from you, and the grace of God, I received the best grade I could have expected in Linear Algebra.

Thanks, again!

Joe
 
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