Proof of Elementary Identities in a Vector Space

The axioms of a vector space are used in the form of Vector Space over a Field (i), Vector Space over a Field (ii), Vector Space over a Field (v), Vector Space over a Field (vi), Vector Space over a Field (vii) and Vector Space over a Field (viii). By The Zero Vector, Negatives and Differences in a Vector Space (i), $x+0_{V}=x$ for every $x\in V$. By The Zero Vector, Negatives and Differences in a Vector Space (ii), for $x,y\in V$,

$$x+(-x)=0_{V},\qquad\text{and}\qquad x+y=0_{V}\ \text{ implies }\ y=-x. \tag{N}$$

Cancellation. Adding $-w$ to both sides of $u+w=v+w$ and using associativity and (N),

$$u=u+0_{V}=u+\bigl(w+(-w)\bigr)=(u+w)+(-w)=(v+w)+(-w)=v+\bigl(w+(-w)\bigr)=v+0_{V}=v .$$

The zero scalar. Since $0+0=0$ in $K$, distributivity over the addition of scalars gives

$$0v+0v=(0+0)v=0v=0v+0_{V}=0_{V}+0v,$$

and cancellation of $0v$ gives $0v=0_{V}$.

Multiples of the zero vector. Since $0_{V}+0_{V}=0_{V}$, distributivity over the addition of vectors gives

$$\lambda0_{V}+\lambda0_{V}=\lambda(0_{V}+0_{V})=\lambda0_{V}=0_{V}+\lambda0_{V},$$

and cancellation gives $\lambda0_{V}=0_{V}$.

Negatives as multiples. By distributivity, (ii) and (iii),

$$\lambda v+(-\lambda)v=\bigl(\lambda+(-\lambda)\bigr)v=0v=0_{V},\qquad\lambda v+\lambda(-v)=\lambda\bigl(v+(-v)\bigr)=\lambda0_{V}=0_{V},$$

so $(-\lambda)v=-(\lambda v)$ and $\lambda(-v)=-(\lambda v)$ by (N). For $\lambda=1$ this gives $(-1)v=-(1v)=-v$, by the unit axiom.

Negative of a negative. By commutativity and (N), $(-v)+v=v+(-v)=0_{V}$, so $v=-(-v)$ by (N).

Negative of a sum. Rearranging by associativity and commutativity,

$$(u+v)+\bigl((-u)+(-v)\bigr)=\bigl(u+(-u)\bigr)+\bigl(v+(-v)\bigr)=0_{V}+0_{V}=0_{V},$$

so $(-u)+(-v)=-(u+v)$ by (N).

Difference of a vector with itself. $v-v=v+(-v)=0_{V}$ by (N).

Adding back. Since $(-v)+v=0_{V}$, as shown for (v),

$$(u-v)+v=u+\bigl((-v)+v\bigr)=u+0_{V}=u .$$

Multiples of a difference. By distributivity and (iv), $\lambda(u-v)=\lambda u+\lambda(-v)=\lambda u+\bigl(-(\lambda v)\bigr)=\lambda u-\lambda v$.

Chaining differences. By associativity and $(-v)+v=0_{V}$,

$$(u-v)+(v-w)=u+\bigl((-v)+v\bigr)+(-w)=u+0_{V}+(-w)=u-w .$$

Negative of a difference. By (x) with $w=u$ and (vii), $(u-v)+(v-u)=u-u=0_{V}$, so $v-u=-(u-v)$ by (N).

No zero divisors. Suppose $\lambda v=0_{V}$ and $\lambda\neq0$, and let $\lambda^{-1}$ be the reciprocal of $\lambda$, so that $\lambda^{-1}\cdot\lambda=1$. By the unit axiom, associativity of scalar multiplication and (iii),

$$v=1v=(\lambda^{-1}\cdot\lambda)v=\lambda^{-1}(\lambda v)=\lambda^{-1}0_{V}=0_{V}.$$

Hence $\lambda=0$ or $v=0_{V}$.