# Inner product space proof question

• March 16th 2010, 07:27 PM
firebio
Inner product space proof question
Prove that {u,v}=0 for all v belongs to V iff u=0.

{u,v}= $\Sigma$ u*v, where u* is conjugate of u

If u=0 then {u,v} is obviously 0.

now im not sure how to prove it the other way

If {u,v}=0 then u=0

$\Sigma$ u*v=0...

• March 16th 2010, 07:31 PM
tonio
Quote:

Originally Posted by firebio
Prove that {u,v}=0 for all v belongs to V iff u=0.

{u,v}= $\Sigma$ u*v, where u* is conjugate of u

If u=0 then {u,v} is obviously 0.

now im not sure how to prove it the other way

If {u,v}=0 then u=0

$\Sigma$ u*v=0...

Apparently you're using {u,v} to denote inner product...(Punch) Anyway, if it is true that $=0\,\,\forall v\in V$ then this is true for $v=u$ as well, so apply now positiveness of inner product to get that it must be $u=0$ .

Tonio
• March 16th 2010, 07:33 PM
Drexel28
Quote:

Originally Posted by firebio
Prove that {u,v}=0 for all v belongs to V iff u=0.

{u,v}= $\Sigma$ u*v, where u* is conjugate of u

If u=0 then {u,v} is obviously 0.

now im not sure how to prove it the other way

If {u,v}=0 then u=0

$\Sigma$ u*v=0...