Show dim U <= n in Vector Subspace of C2n

[brru25]

Homework Statement

Let U be a vector subspace of C²ⁿ such that
sum(x_i*y_i) = 0 for 1 <= i <= 2n for any x, y ∈ U. Show that dim U <= n. Give an example of such a subspace U with dim U = n

2. The attempt at a solution

I tried just writing out the summation and was thinking along the lines of linear independence but I don't think that applies here (maybe it does, I'm not sure). Could I think of a linear map contained in U that maps two vectors x and y to be the sum = 0? I think I'm confusing myself here.

 

[lanedance]

i'm not sure i understand the question correctly... so is that effectively the compex innner product of 2 vectors in the subsapce is always zero?

$$<\texbf{x},\texbf{y}> = \sum_i x_i^* y_i$$

but if that were the case, as U is a vector space, if x is in U, then so is c.x, but
$$<\texbf{x},\texbf{cx}> = c||x||^2$$
contradiction?

 

[Hurkyl]

Maybe by * he meant multiplication rather than complex conjugation? We'll have to wait for him to clarify, I guess.

 

[Mark44]

Glad you guys (lanedance and Hurkyl) jumped in on this one. I was thinking along the lines that lanedance described, except I was thinking of this product of a vector with itself.
$$\sum_{i = 1}^{2n} x_i*x_i~=~0$$
which suggests that all the x_i's are 0.

 

[brru25]

it's multiplication not conjugate (sorry about the mix-up everybody!)

 

[Mark44]

brru25, You're sure you have given us the exact problem description, right?

 

[brru25]

positive, word-for-word...see why I'm confused? :-)

 

[lanedance]

ok, think I'm getting it now, sounds like what Hurkyl was thinking...

I haven't worked it, but would start with an example in the 2D case in $\mathbb{C}^2$, so n = 1

so say you have a vector (a,b) which is in U, it satisfies the rule with itself
$$\sum_i x_i^2 = a^2 + b^2 = 0$$

so, first can you find a vector that satisfies above... and 2nd can you show given a vector in U, there can be no other linearly independent vectors in U?