# Checking axioms of Vector Spaces

Currently I am studying a section from my book on vector spaces. I'm having issues in understanding how I am supposed to prove some of the questions in the Exercises section, such as:

In each of the following, determine whether the set, together with the indicated operations, is a vector space. If it is not, identify at least one of the ten vector space axioms that fails.

13. $M_{4,6}$ with the standard operations.

14. $M_{1,1}$ with the standard operations.

15. The set of all third-degree polynomials with the standard operations.

16. The set of all fifth-degree polynomials with the standard operations.

17. The set of all first-degree polynomial functions $ax+b$, $a\neq 0$, whose graphs pass through the origin with the standard operations.

18. The set of all quadratic functions whose graphs pass through the origin with the standard operations.

I don't know how exactly to identify which axiom fails. Here are the axioms:

1. $\mathbf{u}+\mathbf{v}$ is in $V$. Closure under addition.
2. $\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}$. Commutative property.
3. $\mathbf{u}+(\mathbf{v}+\mathbf{w}) = (\mathbf{u}+\mathbf{v})+\mathbf{w}$. Associative property.
4. $V$ has a zero vector $\mathbf{0}$ such that for every $\mathbf{u}\in V$, $\mathbf{u}+\mathbf{0}=\mathbf{u}$. Additive identity.
5. For every $\mathbf{u}\in V$, there is a vector in $V$ denoted by $-\mathbf{u}$ such that $\mathbf{u}+(-\mathbf{u}) = \mathbf{0}$. Additive inverse.
6. $c\mathbf{u}$ is in $V$. Closure under scalar multiplication.
7. $c(\mathbf{u}+\mathbf{v}) = c\mathbf{u}+c\mathbf{v}$. Distributive property.
8. $(c+d)\mathbf{u}=c\mathbf{u}+d\mathbf{u}$. Distributive property.
9. $c(d\mathbf{u})= (cd)\mathbf{u}$. Associative property.
10. $1(\mathbf{u}) =\mathbf{u}$. Scalar identity.

Say we look at question 17: In the answer book it says that axiom 4 fails, but I don't see how that is possible.

Say if you have 4x+1, by axiom 4 you are supposed to add 0 to the vector u, so you would get 4x+1+0=4x+1, which is true....

I'm really not sure how to get my head around these type of problems. Can someone give me a coherent explanation? Am I supposed to test by hand each axiom, or just in my head?

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What is $M_{4,6}$ Is this notation standard? – kuch nahi Jun 23 '11 at 2:57
@kuch: If I had to guess, $M_{n,m}$ refers to the set of $n\times m$ matrices (presumably over $\mathbb{R}$), but I agree Virtuoso should specify. – Zev Chonoles Jun 23 '11 at 3:02
Axiom 4 fails in Problem 17. Can you see why? – GEdgar Jun 23 '11 at 3:12
I guess if you're not getting "your head around these type of problems", it would be a good idea to do some of these exercises in full detail. Write up everything and test all the axioms (even if one has failed already). After doing this several times, the answers to these questions will become completely obvious. By the way: Checking whether $0$ belongs to some putative subspace is the very first thing I always do. It is usually very easy to check and of the answer is no, you know you don't have a linear subspace. – t.b. Jun 23 '11 at 3:12
@Virtuoso: Let's deal with Question $15$. I expect here "standard operations" are supposed to be addition of polynomials, and multiplication of a polynomial by a constant. A polynomial is of third degree if it looks like $ax^3+bx^2+cx+d$, where $a \ne 0$. Then $(-2x^3+5x)+(2x^3+12x^2)=12x^2+5x$, not a degree $3$ polynomial, so (1.) fails. Also, (4.) fails because the zero polynomial does not have degree $3$. If we had instead "polynomials of degree $\le 3$", that would be a vector space. (I don't like the question, it is kind of intended to trick people, which is not nice.) – André Nicolas Jun 23 '11 at 3:23

With practice, one learns to recognize the sort of things that may go wrong with potential "vector spaces", and quickly zoom in on those. But, the thing is, it takes practice to figure this out.

Often, if one thing goes wrong, lots of things will go wrong; sometimes, it is one and only one thing that goes wrong (and it may be hard to spot). At this stage, it might actually be a good idea for you to check each axiom and see whether it is met or not met, because it will afford you a lot of practice. Even though it's enough to find one axiom that fails for something to not be a vector space, finding all the ways in which things go wrong is likely good practice at this stage.

For example, you don't say which problem "says the answer is Axiom 4", and in fact I see no problem, among the ones listed, in which $4x+1$ is even a vector! It's not a $4\times 6$ matrix, it's not a $1\times 1$ matrix, it's not a degree 3 polynomial, it's not a degree 5 polynomial, it's not a first degree polynomial whose graph passes through the origin, and it's not a quadratic function whose graph passes through the origin...

Since user6312 already got you started with Question 15, let's continue: you know it fails Axiom 1. It is not hard to verify that it satisfies axioms 2 and 3. Axiom 4 fails because the zero vector (the polynomial 0) is not in your set....

Axiom 5 is a bit tricky: strictly speaking, Axiom 5 does not even make sense if Axiom 4 fails, because there is no $\mathbf{0}$ in the first place. I would certainly score such a statement as correct. On the other hand, if you have a polynomial of degree exactly 3, $ax^3+bx^2+cx+d$, with $a\neq 0$, then you can find a polynomial of degree exactly 3 that added to it will give you the zero polynomial (which is not in the set). So you might also say Axiom 5 is "sort of" satisfied.

Axiom 6 fails: for example, $x^3$ is in your set, $c=0$ is a scalar, but $0(x^3)$ is not in your set.

It's not hard to verify that Axioms 7, 8, 9, and 10 do hold.

So for 15, the axioms that fail are Axioms 1, 4, 6, and possibly 5 (depending how you interpret it).

You'll find similar problems with 16. There's a bit more to do with 17, because you also have the condition "and passes through the origin"; be sure to take that into account. Similar with 18. As for 13 and 14, I'll spill the beans and tell you that they are vector spaces: you should verify that all the axioms hold, one by one. Be sure to not verify them "by example": it's not enough to show that for particular $4\times 6$ matrices $\mathbf{u}$ and $\mathbf{v}$ you have $\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}$: you must verify it works for all possible choices of $\mathbf{u}$ and $\mathbf{v}$. If you find yourself saying "since, for example..." chances are you're doing it wrong.

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fixed, I was referring to problem 17, and I made a mistake since it isn't passing through the origin. But say if you had 4x, I still don't understand why axiom 4 is not satisfied, because if you add a 0 vector to 4x it still comes out as 4x. – chaver Jun 23 '11 at 5:50
@Virtuoso: Because Axiom 4 says "$V$ has a vector $\mathbf{0}$ such that...". What is the zero vector you are proposing? It's the polynomial $0$. But the polynomial $0$ is not in $V$. In order for something to be in $V$, it must get past the bouncer at the door. And the bouncer checks two pieces of ID: your graph has to go through the origin, and you have to be a polynomial of degree exactly 1. What is the degree of the polynomial $0$? Does it have the required ID to get through the door, or is it standing in the alley outside cooling its heels? – Arturo Magidin Jun 23 '11 at 5:53
@Virtuoso: Also: don't accept an answer until you are satisfied. Here you are telling me you still don't understand something, but you've gone ahead an accepted my answer (which did not explain things fully to you) anyway. That defeats the purpose of marking an answer as accepted. – Arturo Magidin Jun 23 '11 at 5:55

For Ex. 17, ask yourself if zero is in the set. This should point you in the right direction.

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what do you mean if zero is in the set? I thought the axiom wants you to add a zero vector to the vector and see if it comes out the same... – chaver Jun 23 '11 at 5:41
@Virtuoso, the axiom asserts that $V$ has a zero vector, so, if some set $S$ doesn't have a zero vector, then $S$ is not a vector space. In question 17, the point is that the 0 you want to add to, say, $4x+1$ isn't in the set: 0 is not "a polynomial $ax+b$, $a\ne0$". – Gerry Myerson Jun 23 '11 at 5:56