DLA Theory

Section 15.6 Theory

In this section.

Subsection 15.6.1 Uniqueness of the zero vector and of negatives

Notice that Axiom A 4 and Axiom A 5 only say that there is a zero vector, and that every vector has some negative — they don’t say that there is only one zero vector, or that every vector has only one negative. There is no need to make these axioms that strong — we can instead just logically deduce these properties from the weaker axioms we already have.

Proposition 15.6.1.

In a vector space, there is one unique zero vector.
A vector in a vector space has one unique negative vector.

Proof of Statement 1.

Suppose there were two vectors, \(\zerovec_1\) and \(\zerovec_2\text{,}\) that could each fulfill the requirement of Axiom A 4. But then we would have both

\begin{align*} \zerovec_1 + \zerovec_2 \amp= \zerovec_1 \amp \amp\text{(i)},\\ \end{align*}

and

\begin{align*} \zerovec_1 + \zerovec_2 \amp= \zerovec_2 + \zerovec_1 \amp \amp\text{(ii)}\\ \amp= \zerovec_2 \amp \amp\text{(iii)}, \end{align*}

with justifications

Axiom A 4 with \(\uvec{v} = \zerovec_1\) and \(\zerovec = \zerovec_2\text{;}\)
Axiom A 2; and
Axiom A 4 with \(\uvec{v} = \zerovec_2\) and \(\zerovec = \zerovec_1\text{.}\)

Since \(\zerovec_1+\zerovec_2\) equals both \(\zerovec_1\) and \(\zerovec_2\text{,}\) we must have \(\zerovec_1=\zerovec_2\text{.}\) So there can’t really be more than one zero vector, since multiple zero vectors would end up having to be equal to each other.

Proof of Statement 2.

Suppose a vector \(\uvec{v}\) could have two negatives, \((-\uvec{v})_1\) and \((-\uvec{v})_2\text{.}\) But then,

\begin{align*} (-\uvec{v})_2 \amp= (-\uvec{v})_2 + \zerovec \amp \amp\text{(i)}\\ \amp= (-\uvec{v})_2 + \bbrac{\uvec{v} + (-\uvec{v})_1} \amp \amp\text{(ii)}\\ \amp= \bbrac{(-\uvec{v})_2 + \uvec{v}} + (-\uvec{v})_1 \amp \amp\text{(iii)}\\ \amp= \bbrac{\uvec{v} + (-\uvec{v})_2} + (-\uvec{v})_1 \amp \amp\text{(iv)}\\ \amp= \zerovec + (-\uvec{v})_1 \amp \amp\text{(v)}\\ \amp= (-\uvec{v})_1 + \zerovec \amp \amp\text{(vi)}\\ \amp= (-\uvec{v})_1 \amp \amp\text{(vii)}, \end{align*}

with justifications

Axiom A 4;
Axiom A 5 with \(-\uvec{v} = (-\uvec{v})_1\text{;}\)
Axiom A 3;
Axiom A 2;
Axiom A 5 with \(-\uvec{v} = (-\uvec{v})_2\text{;}\)
Axiom A 2; and
Axiom A 4.

So \(\uvec{v}\) can’t really have more than one negative vector, since multiple negative vectors would end up having to be equal to each other.

Subsection 15.6.2 Basic vector algebra rules

There was also no need to include the condition \(\zerovec + \uvec{v} = \uvec{v}\) in Axiom A 4 or the condition \(-\uvec{v} + \uvec{v} = \zerovec\) in Axiom A 5, as these can be deduced from the axioms we have, as we did in Discovery 15.4.a and Discovery 15.4.b. Let’s record these properties, and some others that can be deduced from the axioms.

Proposition 15.6.2.

Suppose that \(\uvec{u},\uvec{v},\uvec{w}\) are vectors in a vector space, and that \(k\) is a scalar. Then the following are always true.

Additional rules of the zero vector.
1. \(\zerovec + \uvec{v} = \uvec{v}\text{.}\)
2. \(0\uvec{v} = \zerovec\text{.}\)
3. \(-\zerovec = \zerovec\text{.}\)
4. \(k\zerovec = \zerovec\text{.}\)
5. If \(k\uvec{v} = \zerovec\text{,}\) then either \(k=0\) or \(\uvec{v} = \zerovec\) (or both).
Additional rules of vector negatives.
1. \(-\uvec{v} + \uvec{v} = \zerovec\text{.}\)
2. \(-(\uvec{u} + \uvec{v}) = (-\uvec{u}) + (-\uvec{v})\text{.}\)
3. \(-(-\uvec{v}) = \uvec{v}\text{.}\)
4. \(-(k \uvec{v}) = k (-\uvec{v})\text{.}\)
5. \((-1)\uvec{v} = -\uvec{v}\text{.}\)
6. (Cancellation) If \(\uvec{u} + \uvec{w} = \uvec{v} + \uvec{w}\text{,}\) then \(\uvec{u} = \uvec{v}\text{.}\)

Proofs of Rules 1.a–1.b, Rule 2.c, and Rule 2.e.

We have already considered these rules in Discovery 15.4.

Proof of Rule 1.c.

The zero vector is a special vector, but it’s still a vector so it must have a negative because of Axiom A 5. Now, Statement 2 of Proposition 15.6.1 with \(\uvec{v} = \zerovec\) tells us that the only way to fill the blank in

\begin{equation*} \zerovec \: + \:\boxed{\phantom{X}}\: = \:\zerovec \end{equation*}

is with the negative \(-\zerovec\text{.}\) But Axiom A 4 with \(\uvec{v}=\zerovec\) says that we may also fill this blank with plain \(\zerovec\text{.}\) Therefore, we must have \(\zerovec = -\zerovec\text{,}\) as desired.

Proof of Rule 1.d.

We need to verify the equality \(k\zerovec = \zerovec\text{:}\)

\begin{align*} \text{LHS} \amp= k\zerovec \\ \amp= k\bbrac{\zerovec + (-\zerovec)} \amp \amp\text{(i)}\\ \amp= k\bbrac{\zerovec + (-1)\zerovec} \amp \amp\text{(ii)}\\ \amp= k\zerovec + k\bbrac{(-1)\zerovec} \amp \amp\text{(iii)}\\ \amp= k\zerovec + (-k)\zerovec \amp \amp\text{(iv)}\\ \amp= \bbrac{k+(-k)}\zerovec \amp \amp\text{(v)}\\ \amp= 0\zerovec \amp \amp\text{(vi)}\\ \amp= \zerovec \amp \amp\text{(vii)}\\ \amp= \text{RHS}, \end{align*}

as desired, with justifications

Axiom A 5 with \(\uvec{v}=\zerovec\text{;}\)
Rule 2.e;
Axiom S 2;
Axiom S 4;
Axiom S 3;
algebra of numbers; and
Rule 1.b.

Proof of Rule 1.e.

Suppose \(k\uvec{v} = \zerovec\text{.}\) Regardless of this starting assumption, either \(k\) is equal to \(0\) or it is not. If it is, then the desired conclusion “either \(k=0\) or \(\uvec{v}=\zerovec\)” is true, regardless of whether \(\uvec{v}\) is zero of not. On the other hand, if \(k\) is not equal to \(0\text{,}\) then the reciprocal \(\inv{k}\) exists, and so we can use it to compute

\begin{align*} \uvec{v} \amp= 1\uvec{v} \amp \amp\text{(i)}\\ \amp= (\inv{k}k)\uvec{v} \amp \amp\text{(ii)}\\ \amp= \inv{k} (k\uvec{v}) \amp \amp\text{(iii)}\\ \amp= \inv{k} \zerovec \amp \amp\text{(iv)}\\ \amp= \zerovec, \amp \amp\text{(v)} \end{align*}

with justifications

Axiom S 5;
algebra of numbers;
Axiom S 4;
assumption \(k\uvec{v}=\zerovec\text{;}\) and
Rule 1.d.

In this case, the desired conclusion “either \(k=0\) or \(\uvec{v}=\zerovec\)” is true again.

Proof of Rule 2.f.

Suppose \(\uvec{u} + \uvec{w} = \uvec{v} + \uvec{w}\text{.}\) Starting with \(\uvec{u}\text{,}\) we can use the axioms to work in a \(\uvec{w}\) and then convert to \(\uvec{v}\text{:}\)

\begin{align*} \uvec{u} \amp= \uvec{u} + \zerovec \amp \amp\text{(i)}\\ \amp= \uvec{u} + \bbrac{\uvec{w}+(-\uvec{w})} \amp \amp\text{(ii)}\\ \amp= (\uvec{u} + \uvec{w}) + (-\uvec{w}) \amp \amp\text{(iii)}\\ \amp= (\uvec{v} + \uvec{w}) + (-\uvec{w}) \amp \amp\text{(iv)}\\ \amp= \uvec{v} + \bbrac{\uvec{w}) + (-\uvec{w})} \amp \amp\text{(v)}\\ \amp= \uvec{v} + \zerovec \amp \amp\text{(vi)}\\ \amp= \uvec{v}, \amp \amp\text{(vii)} \end{align*}

as desired, with justifications

Axiom A 4;
Axiom A 5;
Axiom A 3;
assumption \(\uvec{u}+\uvec{w} = \uvec{v}+\uvec{w}\text{;}\)
Axiom A 3;
Axiom A 5; and
Axiom A 4.

Remark 15.6.3.

Again, keep in mind the difference between the left- and right-hand sides in Rule 2.e in the proposition above. The left-hand side is the scalar multiple of \(\uvec{v}\) by the scalar \(-1\text{,}\) while the right-hand side is the special negative vector that adds with \(\uvec{v}\) to the zero vector. These are two different processes of obtaining a new vector from the old vector \(\uvec{v}\text{,}\) and the point of the rule is to verify our intuition that these two processes should always return the same result. One of the advantages of this rule is that it eliminates any ambiguity in our definition of vector subtraction, since now it doesn’t matter if we interpret \(\uvec{v}-\uvec{w}\) to mean \(\uvec{v}+(-\uvec{w})\) or \(\uvec{v}+(-1)\uvec{w}\text{.}\)

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