# Is a linear vector space a vector space?

On the first page of the classical book "Ordinary Differential Equations" by Jack Hale (Revised Edition, 1980) there is the following definition:

An abstract linear vector space (or linear space) $\mathcal{X}$ over $\mathbb{R}$ is a collection of elements $\{x,y,\ldots\}$ such that for each $x,y \in \mathcal{X}$, the sum $x+y$ is defined, $x+y \in \mathcal{X}$, $x+y=y+x$ and there is an element $0 \in \mathcal{X}$ such that $x+0=x$ for all $x \in \mathcal{X}$. Also, for any number $a,b \in\mathbb{R}$, scalar multiplication $ax$ is defined, $ax \in \mathcal{X}$ and $1 \cdot x = x$, $(ab)x=a(bx)=b(ax)$, $(a+b)x=ax+bx$ for all $x,y \in \mathcal{X}$.

The terminology linear vector space is the same as vector space (i.e., without the adjective linear)? I am asking this because a classical axiom of vector spaces is missing here: given an $x \in \mathcal{X}$ there is an element $z \in \mathcal{X}$ such that $x+z=0$, where the element $0$ was defined above.

Question improvement: with respect to the definition of vector space, more axioms seem to be missing too, namely the associativity under $+$ and the scalar distributivity as $a(x+y) = ax + ay$. This was mentioned by more than one comment/post of contributors.

Why is that?

• Is the object Hale defines associative under $+$? Dec 5, 2017 at 17:22
• Actually it seems to me that these axioms might not be good enough after all. Specifically, while it's clear how to show that additive inverses exist if we know $0\cdot x=0$ for all vectors $x$, I don't see how to prove that from the axioms given. (More specifically: it's easy to show that $0\cdot x+x=x$, but I don't see how to show $0\cdot x+y=y$ for arbitrary $y$ from the axioms given.) Additionally, those axioms don't seem to include associativity for vector addition, which is pretty important. I don't immediately see how to build a counterexample, though, so I might be missing something. Dec 5, 2017 at 17:30
• Sorry I just meant the sort of object Hale defines in the OP's post. Dec 5, 2017 at 17:43
• Also missing from the usual axioms is $a(x+y)=ax+ay$. Dec 5, 2017 at 17:44
• @EricWofsey Oh wow, yeah, I missed that. This is terrible (unless "abstract linear vector space" means something wildly different from what it should). Dec 5, 2017 at 17:45

Here is a counterexample. Consider $\mathcal{X}=\{0,1\}$, with addition defined by $x+y=\max(x,y)$ and scalar multiplication defined by $ax=x$ for all $a\in\mathbb{R}$ and $x\in\mathcal{X}$. This satisfies all of Hale's axioms, but $1$ has no additive inverse.

Here's a slightly less trivial example. Consider the set $\mathcal{X}=\mathbb{R}\cup\{z\}$, with addition and scalar multiplication defined as usual for elements of $\mathbb{R}$, $x+z=z+x=x$ for all $x\in\mathcal{X}$, and $az=z$ for all $a\in\mathbb{R}$. This satisfies the given axioms, with $z$ as the zero element. However, no element other than $z$ has an additive inverse.

(In fact, any example without additive inverses contains a copy of the first counterexample. If $\mathcal{X}$ satisfies Hale's axioms and $x\in\mathcal{X}$ has no additive inverse, then $\{0,0\cdot x\}\subseteq \mathcal{X}$ will be closed under addition and scalar multiplication and isomorphic to the first example, sending $0\cdot x$ to $1$. We must have $0\cdot x\neq 0$ since $x+(-1)\cdot x=0\cdot x$ so $x$ would have an additive inverse if $0\cdot x=0$.)

Note, though, that additive inverses aren't the only axiom that is missing. Associativity of $+$ and $a(x+y)=ax+ay$ are missing too! Here's an example that has additive inverses but which fails associativity. Let $\mathcal{X}=\mathbb{R}\times\{0,1\}\setminus\{(0,1)\}$. We define addition by $(x,i)+(y,j)=(x+y,\max(i,j))$ and scalar multiplication by $a(x,i)=(ax,i)$, except that if either operation gives an output of $(0,1)$, we change it to $(0,0)$ instead (so for instance, $0(x,1)=(0,0)$ for any $x$). This satisfies Hale's axioms, and has additive inverses ($(-x,i)$ is the inverse of $(x,i)$). However, it fails associativity, since $$((x,0)+(-x,0))+(x,1)=(0,0)+(x,1)=(x,1)$$ whereas $$(x,0)+((-x,0)+(x,1))=(x,0)+(0,0)=(x,0)$$ for any $x\neq 0$.

The author almost certainly does not intend to give a different definition from the usual one, though--this is just an error in the book. It is definitely not standard to use the term "linear vector space" to refer to this weaker definition.

• Nice minimal counterexample and inclusion proof. Too bad I can't give you an second +1. Dec 5, 2017 at 18:07

The definition is indeed missing something for a vector space, but I suspect that is not intentional. “Linear space” is a common synonym of “vector space”, probably because it is linear functions that respect the structure of a vector space.

To see that the conditions are not sufficient, consider $\mathcal X = \mathbb R\times \mathbb N$ with the addition $(a,m)+(b,n) = (a+b,\max\{m,n\})$ and the multiplication $a(b,n)=(ab,n)$.

• That's a nice example. It contains my example as a substructure (namely, $\{(0,0)\}\cup\{n\}\times\mathbb{R}$ for any $n>0$). Dec 5, 2017 at 17:55
• @EricWofsey: Thanks. Actually, I suspect your example is the minimal one. Dec 5, 2017 at 18:00
• I was just thinking about that question, and it's not! I'm about to add that to my answer. :) Dec 5, 2017 at 18:01