determining linear dependence and space spanned for a case of infinite fields

Question

For vectors $(1,7)$ and $(2,-1)$, one can look at the rationals and the prime fields:

$\mathbb{Z/pZ}$ (in which $\mathbb{p}$ is prime)

For every case, determine linear dependence and the space spanned.

Hint: While at first glance there it looks as if one must check an infinite number of cases, there are only a few distinctive cases that actually need to be checked.

Answer 1

Let's consider an aribtrary linear combination of these vectors that results in zero,

$$ a\cdot(1,7) + b\cdot(2,-1) = 0 $$

This means that

$$ a + 2b = 0 \ (mod \ p) \\ 7a - b = 0 \ (mod \ p) $$

and therefore

$$ b = 7a \ (mod \ p) \\ 0 = a + 2b = a + 14a = 15a \ (mod \ p) \\ $$

Now if $15$ is invertible modulo $p$, this inmediatly implies $a = b = 0$ and therefore the vectors are linearly independent, so if $p \notin \{3,5\}$ the vectors are linearly independent. If $p = 3$, you can choose $a = b = 1$, so they are not independent. As for $p = 5$, we can pick $a = 2$ and $b = -1$, again they are not independent. I'm confident you can answer which space they span yourself.

Answer 2

The vector $(2,-1)$ is non-zero (because $-1 \neq 0$) no matter what field you are working with. Assume that $(2,-1),(1,7)$ are linearly dependent. Then we can find $c$ in the field we are working with such that $(2,-1) = c(1,7)$. Comparing the first coordinates, we get $c = 2$ and then by looking at the second coordinates we get $7c = 14 = -1$ which can happen only if $15 = 5 \cdot 3 = 0$. Because we are in a field, this implies that either $5 = 0$ or $3 = 0$. Hence:

1. If we are working over $\mathbb{Z} / 3 \mathbb{Z}$ then indeed $(2,-1) = 2(1,7)$ because $2 \cdot 7 = 2 \cdot (3 + 3 + 1) = 2 = -1$. In this case the vectors are linearly dependent.
2. If we are working over $\mathbb{Z} / 5 \mathbb{Z}$ then indeed $(2,-1) = 2(1,7)$ because $2 \cdot (1,7)$ because $2 \cdot 7 = 2 \cdot (5 + 2) = 2 \cdot 2 = 4 = -1$. Again, in this case the vectors are linearly dependent.
3. Over any other $\mathbb{Z} / p \mathbb{Z}$ or over $\mathbb{Q}$ we have $5 \neq 0$ and $3 \neq 0$, a contradiction. In this case, the vectors are linearly independent.