# Inner Product Space of two orthogonal Vectors is 0 , Is this defined as it is ?

1. Dec 27, 2012

### vish_maths

This may be a very silly question, but still apologies, I read in Sheldon Axler, that the inner product of two orthogonal vectors is DEFINED to be 0.
Let u,v belong to C^n. I am unable to find a direction of proof which proves that for an nth dimension vector space, if u perp. to v, then <u,v> = 0
Is it really just defined ? Or it can be proved to be 0 ?

Last edited: Dec 27, 2012
2. Dec 27, 2012

### micromass

Staff Emeritus
This is not correct. Rather, we define two vectors to be orthogonal if their inner product is 0.
So given $u,v\in \mathbb{C}^n$, we say that u and v are orthogonal iff <u,v>=0. This is a definition.

3. Dec 27, 2012

### HallsofIvy

Staff Emeritus
It can be shown, in R2 or R3, where we have a geometric definition of "orthogonal", that two vectors are orthogonal if and only if their dot product is 0. For higher dimension Euclidean spaces or more general vector spaces, it is simplest to take "inner product is 0" as the definition of "orthogonal".

4. Dec 27, 2012

### vish_maths

Thanks a lot :). There has been just one more question which has been lingering in my mind.

If V is a complex inner product space and T is an operator on V such that <Tv,v> = 0 for all v belongs to V. Then T =0.
Though, it's proof is somewhat convincing , it has left me confused about
a) the existence of orthogonality in n dimensional vectors belong to C^n
b) if the answer above is yes, then why can't Tv be orthogonal to v. ( even if it's not visual, I mean in the mathematical sense)