I guess v will have n-q element, but I am not sure.Suppose that K is a finite field with q elements. Let V be an n-dimensional vector space over k. How munch element does v have?
and if I am right how can I show that v have n-q element???
Then look at simple cases. Way back in the first response TheEmptySet asked you "if V is 1 dim vector space how many distinct elements are there?" You did not answer that- instead you responded about n- dimensions.
If V is a one dimensional vector space, then it has a single basis vector, v, say. All vectors are of the form av where a is from K. If, for example, K is a finite field with 3 elements, 0, 1, 2, say, then the only vectors in the space are 0v, 1v, and 2v- exactly 3 vectors.
If V is a two dimensional vector space, then a basis contains two vectors, u and v, say. All vectors are of the form au+ bv where a and b are from K. Again, if K has only 0, 1, and 2 then the only vectors are 0u+ 0v, 0u+ 1v, 0u+ 2v, 1u+ 0v, 1u+ 1v, 1u+ 2v, 2u+ 0v, 2u+ 1v, and 2u+ 2v- exactly 9 vectors.
Do you get the point? How many vectors would there be if V were three dimensional?