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Hi i'm confused about this homework question:

"Let $V$ be a vector space over the field $F$ and let $a ∈ F$ and $x, y ∈ V$ .

Show that $a(x − y) = ax − ay$ in $V$."

What I did is below but i'm confused because I feel as though it was too simple.

So from the vector space axioms there is a distributivity property that states that $a(\alpha+\beta)=a\alpha+a\beta$ where $a ∈ F$ and $\alpha, \beta$ $ ∈ V$ so I thought if I set $ \alpha=x$ and $\beta=-y$ then the equality above just becomes: $a(x+(-y))=ax+a(-y))$$=ax+(-a)y)$$=ax-ay$.

Is that ok what I've done?

Many thanks

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    Have you already proved that $-y = (-1)y$? You might need to prove that first (and either way you should include it as an explicit step). Otherwise it looks good. – 2017-02-21
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    If you have shown that $(-a) = (-1)a$ then that is fine. – 2017-02-21

2 Answers 2

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Assuming you've already proven that $-y = (-1)y$ (if not, prove it first), I'd recommend adding just a couple of extra steps (in red) so that you can justify each step by a single axiom/ previous lemma:

$$\begin{align}a(x − y) &= a(x+(-y)) & \text{Def of vector subtraction} \\ &= ax + a(-y) &\text{Distributivity over vector addition} \\ &\color{red}{= ax + a[(-1)y]} &\text{Previous lemma: $-y = (-1)y$} \\ &= ax + (a\cdot -1)y &\text{Compatibility of field and scalar multiplication$^\dagger$} \\ &\color{red}{= ax + (-1\cdot a)y} &\text{Commutativity of field multiplication} \\ &\color{red}{= ax + (-1)(ay)} &\text{Compatibility of field and scalar multiplication$^\dagger$} \\ &= ax - ay &\text{Def of vector subtraction}\end{align}$$

$\dagger:$ Meaning this axiom $a(by) = (ab)y$.

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(Responding to the suggestions in the comments.)

Recall that anything multiplied by $0$ is $0$.

If unfamiliar, you may observe that $0x + 0x = (0 + 0)x = 0x$, and use a right cancellation law on the equation $0x + 0x = 0x$ to conclude $0x = 0$.

Next, observe that $(-1)y + y = (-1)y + 1y = (-1 + 1)y = 0y = 0$.

In particular, $(-1)y + y = 0$, so adding $-y$ on the right for both expressions, and again using a right cancellation law, we have $(-1)y = -y$ as desired.