# trouble with inner product space problem...

• Nov 18th 2009, 04:35 PM
dannyboycurtis
trouble with inner product space problem...
I am having trouble proving the following...
Let B be a basis for a finite dimensional inner product space V.
Prove that <x,z>=0 implies x=0 (the zero vector) for all z in B.
Also prove that <x,z>=<y,z> implies x=y for all z in B.
So far all I have is proof that at least one element of z must be nonzero, hence one element of x is zero, cant get any farther though.
Any help would be appreciated.
Thanks
• Nov 18th 2009, 07:09 PM
Jose27
I think you need to learn to use quantifiers (read and write them) since what you wrote makes little sense as it is.

Anyway from what I get you're trying to prove that if $\displaystyle <x,z>=0$ for all $\displaystyle z\in V$ then $\displaystyle x=0$. If it's so then $\displaystyle <z,z>= \Vert z \Vert ^2 =0$ but a norm is zero iff $\displaystyle z=0$, for the second one use the first one with $\displaystyle <x-y,z>=0$ for all $\displaystyle z$
• Nov 18th 2009, 07:21 PM
dannyboycurtis
I actually quantified z to be a vector in B, the basis for V
what I did forget (thanks for pointing it out) is that x and y are vectors in V
• Nov 18th 2009, 07:27 PM
Jose27
But the order is important, your quantifier should go before the implication (check it!), unless of course I understood something else in which case let me know.

Anyway my proof assumes z was in V but it still works since if <,z> is zero on a basis of V then <,w>=0 for all w in V by bilinearity