how many different outcomes for the inner product over finite fields

Question

defining the inner product $<u ,v >=\sum_{i=1}^ku_iv_i (mod \ p)$ when $u,v \in GF(p)^k$ ($p$ is a prime). for any non-zero, fixed $u$ and for every $v$ how can I show that the outcome will be every element in $GF(p)$ exactly $p^{k-1}$ times.

Answer 1

Fix a vector $u\in GF(p)^n$. Then taking the inner product with $u$ is a linear map $\phi_u:GF(p)^n\to GF(p)$. That is, for vectors $v,w\in GF(p)^n$, $$\langle u, av+w\rangle=\sum u_i(av_i+w_i)=a\sum u_iv_i+\sum_iu_iw_i=a\langle u,v \rangle +\langle u,w \rangle.$$

Now let's show three things for non-zero $u$:

1. The linear map $\phi_u$ is surjective.
2. The kernel $K\subset GF(p)^n$ of this map has $p^{n-1}$ elements. (Hint: Rank-Nullity theorem)
3. Every coset of $K$ in $GF(p)^n$ has exactly $p^{n-1}$ elements.

Conclude your result. Let me know if you need help.