I am learning linear algebra, and have a question about the eight laws in the definition of a vector space.
A non-empty set $V$ is a vector space on field $F$, if addition $+:V\times V\to V$ and scaler multiplication $\cdot:F\times V\to V$ are defined such that
$\forall \alpha,\beta,\gamma\in V,k,l\in F$,
- $\alpha+\beta=\beta+\alpha$
- $(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)$
- $\alpha+0=\alpha$
- $\alpha+(-\alpha)=0$
- $1\cdot\alpha=\alpha$
- $k\cdot(\alpha+\beta)=k\cdot\alpha+k\cdot\beta$
- $(k+l)\cdot\alpha=k\cdot\alpha+l\cdot\alpha$
- $k\cdot(l\cdot\alpha)=(kl)\cdot\alpha$
However,these eight laws are not independent. For example, Law 1 can be proved by the rest seven.
Here is the proof from a textbook.
Let $V$ be a vector space on field $F$. $\forall\alpha,\beta\in V$.
- Step 1. Let $\delta$ be an additive inverse of $\alpha$. $$\begin{align}\alpha+\delta &=0 &&[4] \tag{1.1}\\ \delta+(\alpha+\delta) &=\delta+0 \tag{1.2}\\ (\delta+\alpha)+\delta &=\delta &&[2,3] \tag{1.3} \end{align}$$ Let $\eta$ is a additive inverse of $\delta$. $$\begin{align} (\delta+\alpha)+\delta+\eta &=\delta+\eta \tag{1.4} \\ (\delta+\alpha)+0 &=0 &&[2,4] \tag{1.5}\\ \delta+\alpha &=0 &&[3] \tag{1.6} \end{align}$$
- Step 2. $$\begin{align} 0+\alpha &=(\alpha+\delta)+\alpha &&[4] \tag{2.1}\\ &=\alpha+(\delta+\alpha) &&[2] \tag{2.2}\\ &=\alpha+0 &&[\text{Step 1}] \tag{2.3}\\ &=\alpha &&[3] \tag{2.4} \end{align}$$
- Step 3. Let $0'$ also be an additive identity. $$\begin{align} 0 &=0+0' &&[3] \tag{3.1}\\ &=0' &&[\text{Step 2}] \tag{3.2} \end{align}$$ Thus the additive identity is unique.
- Step 4. Let $\delta'$ also be an additive inverse of $\alpha$. $$\begin{align} \delta' &=0+\delta' &&[\text{Step 2}] \tag{4.1}\\ &=(\delta+\alpha)+\delta' &&[\text{Step 1}] \tag{4.2}\\ &=\delta+(\alpha+\delta') &&[2] \tag{4.3}\\ &=\delta+0 &&[4] \tag{4.4}\\ &=\delta \tag{4.5} \end{align}$$ So the additive inverse of $\alpha$ is unique. Thus the additive inverse of any element is unique.
- Step 5. $$\begin{align} 0\alpha+0\alpha &=(0+0)\alpha &&[7] \tag{5.1}\\ &=0\alpha \tag{5.2}\\ (0\alpha+0\alpha)+(-0\alpha) & =0\alpha+(-0\alpha) \tag{5.3}\\ 0\alpha &=0 &&[2,3,4] \tag{5.4} \end{align}$$
- Step 6. $$\begin{align} \alpha+(-1)\alpha &=1\alpha+(-1)\alpha &&[5] \tag{6.1}\\ &=(1+(-1))\alpha &&[7] \tag{6.2}\\ &=0\alpha \tag{6.3}\\ &=0 &&[\text{Step 5}] \tag{6.4}\\ (-1)\alpha &=-\alpha &&[4, \text{Step 4}] \tag{6.5} \end{align}$$
- Step 7. $$\begin{align} (\alpha+\beta)+(-(\beta+\alpha)) &=(\alpha+\beta)+(-1(\beta+\alpha)) &&[\text{Step 6}] \tag{7.1}\\ &=(\alpha+\beta)+(-1\beta+(-1\alpha)) &&[6] \tag{7.2}\\ &=(\alpha+\beta)+(-\beta+(-\alpha)) &&[\text{Step 6}] \tag{7.3}\\ &=\alpha+(\beta+(-\beta))+(-\alpha) &&[2] \tag{7.4}\\ &=(\alpha+0)+(-\alpha) &&[4] \tag{7.5}\\ &=\alpha+(-\alpha) &&[3] \tag{7.6}\\ &=0 &&[4] \tag{7.7} \end{align}$$
- Step 8. $$\begin{align} \beta+\alpha &=0+(\beta+\alpha) &&[\text{Step 2}] \tag{8.1}\\ &=((\alpha+\beta)+(-(\beta+\alpha)))+(\beta+\alpha) &&[\text{Step 7}] \tag{8.2}\\ &=(\alpha+\beta)+((-(\beta+\alpha))+(\beta+\alpha)) &&[2] \tag{8.3}\\ &=(\alpha+\beta)+0 &&[\text{Step 1}] \tag{8.4}\\ &=\alpha+\beta &&[3] \tag{8.5} \end{align}$$
So my questions are:
Can you give some examples that breaks exactly one of the necessary laws?
What is the relationship among these eight laws? Which of the laws can be removed so that the definition remains unchanged while others are indeed necessary? In other words, which of the laws can be proved by the other seven?
The first question has been answered in HERE. That answer showed that Law 1,2,4',5,6,7,8(in their notations) are independent because they give examples that breaks exactly one law of seven the while Law 3 is proved by the others.
However, I still do not think they give a satisfying solution to the my question. My question is different from theirs, for example, we proved Law 1 from Law 2,3,4,5,6,7,8, but Law 1 is independent of Law 2,3,4',5,6,7,8. To see this, note that their "All but 1" example satisfies Law 4' but not Law 4.
Thank you very much.
