# Do we need to check for closure of addition and multiplication when checking whether a set is a vector space

I am studying linear algebra using Axler's 3rd edition book.

When checking whether a set is a vector space, I refer to the definition on page 12. These are the definitions being used:

A vector space is a set V along with an addition on V and a scalar multiplication on V such that the following properties hold: commutativity, associativity, additive identity, additive inverse, multiplicative identity, distributive properties.

The book defines the operator "+" to be closed in set V and scalar multiplication to be closed in set V, therefore when I check whether a space is a "vector space" using these two operations, I only check whether the bolded properties hold.

Question 1: It appears that "+" and scalar multiplication are inhereited from the field which the vector space is over. Is this interpretation correct?

Question 2: Suppose now that there is an alternative definition of addition (call it +') and scalar multiplication (call it $$\cdot'$$) over a vector space candidate W, that does not correspond with our commonly known addition and multiplication in $$\mathbb{R}$$. To check whether W is a vector space, I know I have to go ahead and check the 7 properties, but I'm not sure whether checking that addition and scalar multiplication are closed is a) redundant or b) necessary. Put in a different way, does a set U following the 7 properties of a vector space imply "closed under addition and scalar multiplication?

The answer to Question 1 is no, addition and scalar multiplication are not inherited from the field. Let $$+_V$$ denote addition in the vector space $$V$$ and $$\cdot_V$$ scalar multiplication in $$V$$. Similarly, let $$+_F$$ and $$\cdot_F$$ denote addition and multiplication in the field $$F$$. Then the distributive properties allow you to relate $$+_V$$ and $$+_F$$ and $$\cdot_V$$ and $$\cdot _F$$. For instance, for all $$v \in V$$ and $$k_1,k_2 \in F$$, $$k_1 \cdot_V(k_2\cdot_V v) = (k_1 \cdot_F k_2)\cdot_V v$$ and $$(k_1 \cdot_V v) +_V (k_2 \cdot_V v )= (k_1+_Fk_2)\cdot_V v.$$ However, knowledge of the operations $$\cdot_F$$ and $$\cdot_F$$ do not tell you how to define $$+_V$$ and $$\cdot_V$$.

To illustrate this point, I could tell you that $$V=\mathbb{R}_+=\{x \in \mathbb{R}\mid x>0\}$$ is a vector space over $$F=\mathbb{R}$$. Do you know how to add vectors in $$V$$ now? It's not $$x+_Vy = x+_Fy$$. Rather, $$x+_Vy = xy$$. And scalar multiplication is not $$k \cdot_V x = kx$$ but rather $$k\cdot_V x = x^k$$.

So the answer to Question 2 is yes, you must check for closure of $$V$$ under the vector space operations.

• You said: "Do you know how to add vectors in V now? It's not x+Vy=x+Fy. Rather, x+Vy=xy." Just wanted to confirm that your point here is that the vector addition could be anything, which has no bearing as to how it's defined in $\mathbb{R}$? Oct 3 '19 at 4:38
• @jh1001 Right, my point is that just knowing $V$ is a vector field over $F$ and knowing how to add and multiply in $F$ does not tell you how to add and scalar multiply in $V$. I wouldn't say vector addition could be "anything" since it still has to follow certain axioms, but it is not determined by the field $F$.
– kccu
Oct 3 '19 at 21:55

Consider the union of the two axes in the $$xy$$-plane. (This is a cross shape, not the whole plane.)

$$W=\{(x,0)\mid x\in\mathbb R\}\cup\{(0,y)\mid y\in\mathbb R\}$$

With the usual definition of addition and scaling of vectors in the plane, this set satisfies all of your axioms, but is not closed under addition.

$$(3,0)+(0,2)=(3,2)\not\in W$$

So if we don't want to say that $$W$$ is a vector space, the axiom of closure is necessary.

When dealing with a real vector space (as in Axler) you do not get to change the definitions of addition and multiplication (of numbers) in $$\mathbb{R}$$. A vector space has addition of vectors and multiplication of a vector by a scalar as part of the definition. (Those definitions may sometimes look weird, or alternative, but they are fixed for that space.)

In order for a subset to be a subspace it must be closed under the operations in inherits from the containing space.

In general, the addition ($$+$$) and scalar multiplication ($$\cdot$$) are different for the vector space itself, on the one hand, and the field it is a vector space over, on the other. However in the familiar case of the real numbers over themself, or $$\Bbb R^n$$ over $$\Bbb R$$, for that matter, the operations are indeed defined in terms of the operations in the field.

Closure is indeed fairly obvious in the case of $$\Bbb R^n$$ over $$\Bbb R$$. Or in the case of coordinate spaces. But in general you need to check. As for instance, when you check if you have a subspace (which is a special case of the problem you are considering) , closure is all you have to check (the other properties are inherited).

I would be prepared to check closure: (it seems like) there are some funky vector spaces out there. For one thing, you can tweak the familiar operations in various ways.

Of course, this all kind of sounds like hot air without some examples. For the finite dimensional case, Halmos' Finite Dimensional Vector Spaces is supposed to be good. And don't forget the function spaces and other different examples of infinite dimensional spaces. This list includes Banach spaces, and Hilbert spaces.

Then there are the examples of field extensions, such as $$\Bbb Q(i\sqrt7)$$, that one encounters in algebraic number theory.

For a vector space one needs an abelian group, which also has to be an $$\Bbb F$$-module.