How should I calculate the intersection of subspaces?

Question

How should I calculate the intersection of subspaces? In Z5.

span{(1, 4, 4), (2, 3, 4)} ∩ span{(1, 1, 4), (2, 4, 0)}

And how many vectors it includes?

Answer 1

Suppose that $(x,y,z)$ is an element of the interseaction. So you can express it in two ways:

$$\alpha_1(1,4,4)+\beta_1(2,3,4)=(x,y,z)=\alpha_2(1,1,4)+\beta_2(2,4,0)$$

Thus: $$\alpha_1(1,4,4)+\beta_1(2,3,4)-\alpha_2(1,1,4)+\beta_2(2,4,0)=0$$

Solve this system of the equations (over $\mathbb{Z}_5)$, and you will get that: $$\alpha_1=0$$$$3\alpha_2=\beta_2$$$$\beta_1=\alpha_2$$ So vectors in the intersection are of the form:

$$(x,y,z)=\beta_1(2,3,4)$$ Check all the values of $\beta_1$ from $0$ to $4$:

$\beta_1=0:(x,y,z)=(0,0,0)$

$\beta_1=1:(x,y,z)=(2,3,4)$

$\beta_1=2:(x,y,z)=(4,1,3)$

$\beta_1=3:(x,y,z)=(1,4,2)$

$\beta_1=4:(x,y,z)=(3,2,1)$

So the intersection is:

$$\{(0,0,0),(2,3,4),(4,2,3),(1,4,2),(3,2,1)\}$$