Let $V$ be a vector space over a finite field $F_q$ and let $\{v_1, v_2, . . . , v_k\}$ be a basis of $V$. Show that the following two statements are equivalent:
(i) $v\cdot v^\prime = 0$ for all $v, v^\prime\in V$,
(ii) $v_i\cdot v_j = 0$ for all $i, j \in \{1, 2, . . . , k\}$. (Note: this shows that it suffices to check (ii) when we need to determine whether a given linear code is self-orthogonal.) (u.v is inner product) (i)$\Rightarrow$(ii) is obvious as product of basis are $0$ then product of vector is.
{{ Let $v = (v_1, v_2, . . . , v_n),w = (w_1,w_2, . . . , w_n) \in F_q^n$. The scalar product (also known as the dot product or the Euclidean inner product) of $v$ and $w$ is defined as $$v \cdot w = v_1w_1 + ·· ·+v_nw_n \in F_q .$$ The two vectors $v$ and $w$ are said to be orthogonal if $v \cdot w = 0.$}}
but the other side is confusing.
Thanks for help
Every vector's a linear combination of basis vectors, and the standard properties of the dot product (ditributivity and the like) show that if the basis vectors are orthogonal (including each being orthogonal to itself) then all pairs of vectors are orthogonal.
Just to give a small example, if $v\cdot v=v\cdot w=w\cdot w=0$, then $$(av+bw)\cdot(cv+dw)=(ac)v\cdot v+(ad)v\cdot w+(bc)w\cdot v+(bd)w\cdot w=0+0+0+0=0$$