Prove ||u-v|| = 2^(1/2)

• Oct 2nd 2011, 08:52 AM
page929
Prove ||u-v|| = 2^(1/2)
I need to prove that ||u-v|| = 2^(1/2) if {u,v} is an orthonormal set in the vector space V.

Here is what I have so far, but do not know if I am even on the right page:

||u-v|| = <u-v, u-v>^(1/2), which is the definition of distance between vectors

But where do I go from here?

Thanks for any help.
• Oct 2nd 2011, 09:01 AM
Plato
Re: Prove ||u-v|| = 2^(1/2)
Quote:

Originally Posted by page929
I need to prove that ||u-v|| = 2^(1/2) if {u,v} is an orthonormal set in the vector space V.

$\displaystyle \|u-v\|^2=(u-v)\cdot (u-v)=u\cdot u-2u\cdot v+v\cdot v=~?$
• Oct 2nd 2011, 09:28 AM
page929
Re: Prove ||u-v|| = 2^(1/2)
Quote:

Originally Posted by Plato
$\displaystyle \|u-v\|^2=(u-v)\cdot (u-v)=u\cdot u-2u\cdot v+v\cdot v=~?$

u^2 - 2uv + v^2
||u||^2 - 2uv + ||v||^2
• Oct 2nd 2011, 09:29 AM
Deveno
Re: Prove ||u-v|| = 2^(1/2)
the fact that {u,v} is an orthonormal set tells you you have "special values" for <u,u>, <u,v> and <v,v>. what are they?
• Oct 2nd 2011, 09:48 AM
page929
Re: Prove ||u-v|| = 2^(1/2)
Quote:

Originally Posted by Deveno
the fact that {u,v} is an orthonormal set tells you you have "special values" for <u,u>, <u,v> and <v,v>. what are they?

<u,u> = 1, <u,v> = 0, <v,v> = 1
So, 1+0+1 = 2
but I need ||u-v|| not ||u-v||^2, so I take the square root and my answer is 2^(1/2)