Verify that F^n is a vector space over F

[Saladsamurai]

I that this is probably another really simple question, but I would like some help learning how one starts a 'verification problem.'

Verify that Fⁿ is a vector space over F.

I know that I just have to show that commutativity and scalar multiplication etc. are satisfied. But I am not used to approaching problems with such generality.

I am thinking that I should first define the elements of Fⁿ; maybe like:

Fⁿ={(x₁,...,x_n) : x_j $\epsilon$ F for j=1,2,...,n}.

Now what? This seems kind of redundant... for commutativity I would just show that

$$(x_1,x_2,x_3,...,x_n)+(y_1,y_2,y_3,...,y_n)=(x_1+y_1,x_2+y_2,x_3+y_3,...x_n+y_n)=(y_1+x_1,y_2+x_2,y_3+x_3,...,y_n+x_n)$$

Right? Is this the right approach?

 

[snipez90]

Looks fine to me. Sooner or later you'll realize that the only step that requires any care at all in these axiom verifications is the step where you justify with "this follows from [some property of the reals]".

 

[CompuChip]

Yep, that's about it. By definition, Fⁿ consists of n-tuples of elements of F, as you wrote. The addition is defined as stated in the first equality of your last equation. The second equality of that equation holds because F is a field, hence its addition is commutative. And again by the definition of addition of elements in Fⁿ, the last thing you wrote is again $(y_1,y_2,y_3,...,y_n)+(x_1,x_2,x_3,...,x_n)$.

Of course the reason that the proof is not complicated, is that the definition of a vector space is based on that of Rⁿ and R is just a particular field F.

 

[Saladsamurai]

Yep, that's about it. By definition, Fⁿ consists of n-tuples of elements of F, as you wrote. The addition is defined as stated in the first equality of your last equation. The second equality of that equation holds because F is a field, hence its addition is commutative. And again by the definition of addition of elements in Fⁿ, the last thing you wrote is again $(y_1,y_2,y_3,...,y_n)+(x_1,x_2,x_3,...,x_n)$.

Of course the reason that the proof is not complicated, is that the definition of a vector space is based on that of Rⁿ and R is just a particular field F.

Okay, this makes sense. I was wondering where the separation of definition and verification occurred. It was in my lack of understanding (or even knowing) the proper definition of a field. But after reading this, I looked it up and this whole verification business seems a whole lot clearer.

Thanks guys

 

[HallsofIvy]

Verification of "A is B" simply means showing that A satisifies the definition of B.

Snipez said "the only step that requires any care at all in these axiom verifications is the step where you justify with "this follows from [some property of the reals]". What he meant was that you have to use the fact that "some property of the reals" as used here is just a part of the definition of "field".

The fact, for example, that all Cauchy sequences of real numbers converge to a real number is a property of real numbers that is not part of the definition of "field" but is not used in defining a vector space over a field.