Proof that multiplying by the scalar 1 does not change the vector in a normed vector space.

Question

I'm beginning a self-study of functional analysis, and I seem to have come to a halt trying to solve the first problem in the first problem set, and was wondering if someone could give me a pointer.

The problem asks to show that, while the property $1\mathbf{a}=\mathbf{a}$ must be included in the axioms of a vector space, it is a property that can be derived from the axioms of a normed vector space.

The axioms for a vector space $V$ over the field $\mathbf{R}$ given in the book are as follow:

1. $V$ is a group.
2. For each $\alpha\in\mathbf{R}$ and $\mathbf{a}\in V$, $\alpha\mathbf{a}\in V$
3. $\alpha(\mathbf{a}+\mathbf{b})=\alpha\mathbf{a}+\alpha\mathbf{b}$
4. $(\alpha+\beta)\mathbf{a}=\alpha\mathbf{a}+\beta\mathbf{a}$
5. $\alpha(\beta\mathbf{a})=(\alpha\beta)\mathbf{a}$
6. $1\mathbf{a}=\mathbf{a}$

The axioms for a normed vector space are 1--5 plus the additional axiom that there is a map $\Vert\cdot\Vert:V\to\mathbf{R}$ that satisfies all the properties of a norm.

The problem formally stated now is to show that 1--5 do not imply 6, but 1--5 together with the norm do imply 6.

This problem seems similar to the question posed here, and from the discussion of that question it seems an easy example of a vector space satisfying 1--5 but not 6 would be one where scalar multiplication always yields the zero vector. This cannot be the case in a normed vector space, because for $\mathbf{a}\neq0$ and $\alpha\neq0$ we know $\Vert\mathbf{a}\Vert\neq0$, and a norm satisfies $\Vert\alpha\mathbf{a}\Vert=\vert\alpha\vert\Vert\mathbf{a}\Vert\neq0$, but we also have $\Vert\alpha\mathbf{a}\Vert=\Vert\mathbf{0}\Vert=0$ from our definition of scalar multiplication, which is a contradiction.

It is trivial that $1\mathbf{a}=\mathbf{a}$ if every vector in $V$ can be written as a scalar multiple of a vector in $V$, since then $\mathbf{a}=\alpha\mathbf{b}=(1\alpha)\mathbf{b}=1(\alpha\mathbf{b})=1\mathbf{a}$, but while the norm has ruled out the strange case of scalar multiplication that means only $\mathbf{0}$ can be written as a scalar multiple of a vector, I haven't been able to convince myself that other cases where there exist vectors that are not scalar multiples of a vector have been ruled out.

Can anyone offer a pointer for how to show every vector in a normed vector space can be written as a scalar multiple of a vector in the space, or perhaps point me to a more natural way of approaching this problem?

Thanks.

Answer 1

Nice question, this took me more thought than I expected.

Consider $a\in V$. Note that $0=\|1a-1a\|=\|1a-(1)(1a)\|=\|(1)(a-1a)\|=\|a-1a\|$.

Thus $a=1a$.