Show that $wt(y)+ \sum_{\lambda \in F_q}wt(x+ \lambda y)=q(n-q)$

Question

I need to help with this excercise :)

Get $x,y$ two linearly independent words in $F_q^n$ and get $k$ is the number of coordinates in which x and y are both zero.

a) Show that $wt(y)+ \sum_{\lambda \in F_q}wt(x+ \lambda y)=q(n-k)$

I think that,, $wt(x)\leq n-k$ , $wt(y)\leq n-k$

But $wt(x+ \lambda y) \leq n-k $???

Then

$\sum_{\lambda \in F_q}wt(x+ \lambda y)=wt(x+ \lambda_1 y)+wt(x+ \lambda_2 y)+...+wt(x+ \lambda_q y) \leq q(n-k)$

But i dont know how putting weight of y also believe there is wrong in $wt(x+ \lambda y) \leq n-k $,

stressed :(

Answer 1

If $x_i=y_i=0,$ so is $x_i+\lambda y_i$ for any $\lambda$ so $$wt(x+\lambda y)\leq n-k$$ holds as needed to complete the proof.