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I'm trying to prove commutativity of addition for vector spaces, using the axioms for vector spaces. Apparently commutativity can be proven! Im having trouble getting a good feel for what is allowed and what is not. Here's my work so far:

$u+v+u+v = 2(u+v) = 2u + 2v = u+u+v+v = u+(u+v)+v$

Here I just wanna claim that $u+(v+u)+v = u+(u+v)+v$

$\Rightarrow -u+u+(v+u)+v+(-v) = -u+u+(u+v)+v+(-v)$ : here im just adding -u to the right, and -v to the left. Question: is this "adding to both sides" really legit in this context? Why?

Quick help proof: $-v+v = (-1)v+(1)v = (-1+1)v = 0v = 0 = v-v$

And another: $ 0+v = v+(-v) + v = (1)v + (-1)v + v = (1-1)v + v = v = v+0$

We have $0 + (u+v) + 0 = 0+(v+u)+0 \Rightarrow u+v = v+u$

This feels ugly and not at all elegant, especially the great leap "add -u to both sides" feels completely out of place. Do I need more lemmas? Is there a more elegant way?

//not homework or anything, just for my own pleasure, feel free to provide theory, as it is more insightful than solutions. :) Thanks!

EDIT: corrected notation a little.

JuliusL33t
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5 Answers5

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Normally commutativity is taken as an axiom, but you can deduce it from associativity, distributivity and from the existence of inverses as follows:

$(u + v) - (v + u) = (u + v) - v - u$ (by distributivity)
$ = u + (v - v) - u$ (by associativity)
$ = u + 0 - u = (u + 0) - u = u - u = 0$

So $u + v = v + u$

Mathmo123
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  • This I like! But from what I understand, u+0 = u is an axiom but 0+u = u is not, but follows from commutativity, which itself is provable from the other axioms. I'm trying to be as careful as possible to make sure I use the axioms and NOTHING else, so I understand what can be assumed and what has to be proven. Even though the commutativity is often taken as an axiom, the whole point is to prove commutativity as carefully as possible from the others, and to show that it can be done! And your proof is really nice, much more elegant and concise than my mess of characters! :) – JuliusL33t Jun 20 '14 at 13:09
  • I've edited it slightly - it uses that u+0 = u rather than 0+u = u – Mathmo123 Jun 20 '14 at 13:13
  • Actually, $-(v+u) = -v - u$ needs $1v = v$ to prove, not sole distributivity. If there’s no 1, we can only prove $-(v+u) = -u - v$ which wouldn’t help. – arseniiv May 03 '20 at 18:44
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Your first equation essentially answers the question: $$u+v+u+v=2(u+v)=2u+2v=u+u+v+v$$ From here, because we know that a vector space is a group under addition, add on the left by $-u$ and on the right by $-v$ to get $$-u+u+v+u+v-v=-u+u+u+v+v-v$$ $$v+u=u+v$$

This is legitimate because by definition a vector space is a group under addition. If your definition doesn't have this as part of it, I'd recommend adding your definition.

Hayden
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The proof of this fact almost goes as follows (after generalizing associativity to omit parentheses):

$$u + u + v + v = (u + v)(1 + 1) = u + v + u + v$$

Then cancel $u$ on the left, and cancel $v$ on the right, to obtain $u + v = v + u$, "QED."

The issue, though, is that left and right cancellation laws, although not so tough to prove, are generally done by assuming commutativity. Let us suppose we have only proved right cancellation, and now we will prove left cancellation without overassuming, i.e., without assuming $0$ is a left identity and without assuming that the inverse on the right also works on the left.

Proposition 1. Given that $x + y = 0$ we can conclude that $y + x = 0$.

Proof. $y + x + y + x = y + 0 + x = y + x$. Right cancellation on $y+x$ yields the desired result.

Proposition 2. Given that $x + 0 = x$ we can conclude that $0 + x = x$.

Proof. Suppose $y$ is the (right and left, cf. Prop 1) inverse of $x$: $0 + x = x + y + x = x + 0 = x$.

Now that we have right and left cancellation laws, the scare quotation marks can be removed from the QED above.

Benjamin Dickman
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I hate to be a bearer of bad news but you cannot prove commutativity of addition using the other axioms.

Proving $(u+v) - (v+u) = 0$ does not necessarily mean $-(v + u)$ is the additive inverse of $u+v$
unless you assume the uniqueness of the additive inverse.

To prove uniqueness of the additive inverse you need commutativity. Not one axiom in the definition of a vector space is redundant.

Anindya Prithvi
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gummychaos
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  • Vector space is, in particular, a group under addition, where we have uniqueness of inverse (even for non-Abelian groups). – Sahiba Arora Jun 10 '22 at 07:36
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I agree that commutativity is an axiom for Vector Spaces, so what you might need to show is that the ordinary vector spaces you know about satisfy the axioms of Vector Spaces. Then you can forget about my capital letters.

You know numbers commute. Use that to show vectors commute.

$\begin{align}u+v&=&(u_1,u_2,...,v_n)+(v_1,v_2,...,v_n)\\ &=&(u_1+v_1,u_2+v_2,...,u_n+v_n)\\ &=&(v_1+u_1,v_2+u_2,...,v_n+u_n)\\ &=&v+u \end{align}$

Empy2
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    Doesn't this rely on the fact what we already know a little something about some representation of vectors as n-tuples etc etc? Should I feel free to use that definition? Because it feels like if we just define a vector as a element of a set and impose the axioms on the set, and say nothing else, then we can prove it in a much more general sense. Is this not true? – JuliusL33t Jun 20 '14 at 13:14
  • One of the great things about vector spaces is that lots of things are vector spaces - not only the ordinary vectors, but polynomials, $2\times 3$ matrices and the set of functions. It's nice to have concrete examples before getting abstract. Then, when you prove something about orthogonal projection, say, you know why that is useful to all these things. – Empy2 Jun 20 '14 at 13:53
  • Just to finish my comment: But first you have to show these things are vector spaces. – Empy2 Jun 20 '14 at 14:05