Take n-tuples from the field of 4 elements. Define a Hermitian inner product (u,v)H by setting (u,v)H equal to the sum of all u * v^2 in the same place (its basically the same as a regular inner product, but the second term has its entries squared). Prove that this is nondegenerate; that is, no nonzero n-tuple u has (u,v)H = 0 for all v.
I'm not really sure how to go about this. Can I just show it is nonzero for one v? For example, let u have a nonzero element in its ith place. Then take v such that the ith entry in v is 1 and the rest are 0's. Then (u,v)H would be equal to whatever the ith entry of u is. Does that prove it? Seems too simple to be that.