1. ## Vector Spaces II

Let V be a vector space over Z_p with dim V = n. How many elements are in V?

2. Originally Posted by Coda202
Let V be a vector space over Z_p with dim V = n. How many elements are in V?
$p^n$

3. Do you see why that's true. Every vector in the space can be written as a linear combination of vectors in the basis: $a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n$. There are p possible values for each $a_i$ so by the "fundamental counting principle", there are $(p)(p)\cdot\cdot\cdot(p)$ (n times) so there are $p^n$ possible vectors.

4. Originally Posted by HallsofIvy
Do you see why that's true. Every vector in the space can be written as a linear combination of vectors in the basis: $a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n$. There are p possible values for each $a_i$ so by the "fundamental counting principle", there are $(p)(p)\cdot\cdot\cdot(p)$ (n times) so there are $p^n$ possible vectors.
Though it is evident, but just to make the above post from HallsofIvy more exact -
Every vector in the space can be written as a linear combination of vectors in the basis in a unique way

5. Originally Posted by aman_cc
Though it is evident, but just to make the above post from HallsofIvy more exact -
Every vector in the space can be written as a linear combination of vectors in the basis in a unique way
I think that what he meant by saying " There are p possible values for each $a_{i}$"

6. Originally Posted by Raoh
I think that what he meant by saying " There are p possible values for each $a_{i}$"
Hi Raoh - Sorry I didn't follow your question completely. Let me try to explain what I was trying to emphasize in my post.

What I meant was that each of the linear combination (and there are p^n linear combination because of the reason you mentioned) is in one-one and onto maping with all the vectors in the vector space. And the reason that happens is only because v1,v2.....vn form a basis. The phrase 'in a unique way' is important to stress the one-one part of the mapping. 'Every' specifies the onto part.