Cardinality vs. Dimension, Solution of homogeneous equations

[Locoism]

Homework Statement

Show that the number of distinct solutions of a system of linear equations (in any number of equations, and unknowns) over the field Zp is either 0, or a power of p.

The Attempt at a Solution

First off, I was wondering whether there is any difference between "cardinality" of a vector space and "dimension". Aren't both just the size of the basis? (the cardinality of V = dim(V)??)
This is just because my prof keeps switching between both and confusing the rest of us.

for the question, suppose the system is m equations in n unknowns.
The case of 0 is trivial, so if we take the subset W of Z_pⁿ of solutions over Z_p, W is a vector space, and dim(W) = n.
I'm not sure how to put this technically, but for each unknown more than the number of equations (say we parametrize them) we have p possible choices, and thus p^t solutions.

There's something missing, but am I on the right track?

 

[morphism]

No. The cardinality of V isn't equal to the dimension of V. Easy counterexample: an n-dimensional vector space over Z/pZ has p^n elements (which is essentially what you've noticed with your W). What is true is that the cardinality of a basis for V is equal to the dimension of V (and this is the definition of dimension).

Anyway, you've basically solved the problem, although you've said some questionable things. You've noticed W is a finite-dimensional vector space over Z/pZ, which is correct. If the dimension of W is t (it's not necessarily n), then you can choose a basis w_1,...,w_t for W. Any element of W will thus look like a_1w_1+...+a_tw_t with a_i in Z/pZ. Now count these suckers.

 

[Locoism]

No. The cardinality of V isn't equal to the dimension of V. Easy counterexample: an n-dimensional vector space over Z/pZ has p^n elements (which is essentially what you've noticed with your W). What is true is that the cardinality of a basis for V is equal to the dimension of V (and this is the definition of dimension).

Oh ok thank you. So what is cardinality then (talking about a vector space)?

If the dimension of W is t (it's not necessarily n), then you can choose a basis w_1,...,w_t for W. Any element of W will thus look like a_1w_1+...+a_tw_t with a_i in Z/pZ. Now count these suckers.

So then the dimension is p^t! It makes sense to say dim(W) = t, but I don't understand why it isn't n:
The system is in n unknowns, and as you said, you can form a basis {w1, w2,...,wt}. But in the standard basis, you would have {e1,...,en}, thus dim(W) = n since all bases of W have the same cardinality. Why not n?

 

[morphism]

Oh ok thank you. So what is cardinality then (talking about a vector space)?

The cardinality of a vector space is just its cardinality as a set.

So then the dimension is p^t! It makes sense to say dim(W) = t, but I don't understand why it isn't n:
The system is in n unknowns, and as you said, you can form a basis {w1, w2,...,wt}. But in the standard basis, you would have {e1,...,en}, thus dim(W) = n since all bases of W have the same cardinality. Why not n?

Try solving
x_1 + x_2 = 0
x_1 + x_2 + x_3 = 0
in Z/2Z, for example. Is dimW=3?

 

[Locoism]

The cardinality of a vector space is just its cardinality as a set.

Thank you, makes perfect sense.

Try solving
x_1 + x_2 = 0
x_1 + x_2 + x_3 = 0
in Z/2Z, for example. Is dimW=3?

Oh alright. Essentially saying dimW = t is just saying it is finite dimensional.