# Vector space

1. Oct 7, 2007

### eyehategod

15. Determine wheter the set is a vector space.
The set of all fifth-degree polynomials with the standard operations.
AXIOMS
1.u+v is in V
2.u+v=v+u
3.u+(v+w)=(u+v)+w
4.u+0=u
5.u+(-u)=0
6. cu is in V
7.c(u+v)=cu+cv
8.(c+d)u=cu+cd
9.c(du)=(cd)u
10.1(u)=u

the back of my book says that axioms 1,4,5, and 6 fail. I dont know why 4,5,and 6 fail. Can anyone help me?

Last edited: Oct 7, 2007
2. Oct 7, 2007

### bob1182006

hm..if it fails 1 and 6 shouldn't it also fail 7? o.o...Sorry I can't see a counterexample that 1/4/5/6 fails.

say you have
$$u=ax^5+bx^4y+cx^3y^2+dx^2y^3+exy^4+fy^5$$
$$v=gx^5+hx^4y+ix^3y^2+jx^2y^3+kxy^4+ly^5$$
where a,b,...,l are arbitrary constants where at least 1 of a,b,...,f and g,h,...,l doesn't equal 0.

does:
f(u+v)=f(u)+f(v)?
f(u+0)=f(u)?
f(u+(-u))=0?
f(cu)=cf(u)?

you can write it them out and I think they do hold, or can you give 1 counter-example that proves it doesn't?

Last edited: Oct 8, 2007
3. Oct 8, 2007

### matt grime

Oh boy is that too complicated. 4,5,6 simply fail because 0 is not a 5th degree polynomial, thus there is no 0 to add to u in 4, no 0 for u+(-u) to equal, and 0*u is not a 5th degree poly

4. Oct 8, 2007

### HallsofIvy

Staff Emeritus
Notice that the set of all polynomials with degree less than or equal to 5 is a vector space. But here, you must have polynomials of precisely degree 5. The "0" polynomial is not in that set.

5. Oct 8, 2007

### eyehategod

to matts response:
im still getting use to this abstract way of thinking and am not sure if this is stupid or not. For axiom 4 to pass, does the zero vector have to be a 5th degree polynomial? cant the zero vector just be 0? b/c lets say (x^5+x)+(0)=(x^5+x). Whats wrong with the way im approaching this?

6. Oct 8, 2007