Thread: vector space

I have to show P1={a0 + a1x|a0 - 2a1 = 0} is a vector space using the 10 parts of the definition such as closed under addition and so on.

is the first statement right?
1. vector addition
Let a0+a1x and b0+b1x be two elements of P1. Then (a0+a1x)+(b0+b1x)=(a0+b0)+(a1+b1)x as long as (a0+b0)-2(a1+b1)=(a0-2a1)+(b0-2b1)=0+0=0.

well you have to show closure, that is if f(x),g(x) is in P1, so is f(x)+g(x).

now (a₀ + a₁x) + (b₀ + b₁x) = (a₀+b₀) + (a₁+b₁)x, no matter if P1 is closed or not.

however, you are correct that if a₀ - 2a₁ = b₀ - 2b₁ = 0, then:

(a₀+b₀) - 2(a₁+b₁) = (a₀-2a₁) + (b₀-2b₁) = 0 + 0 = 0.

if you have already established that V = {a₀ + a₁x : a₀,a₁ in F} (you don't SAY what F is, but presumably it is the field of real numbers, it doesn't HAVE to be, though...there ARE other fields) is a vector space, you don't have to "re-prove" commutativity of vector addition or associativity of vector addition, since these things hold for ALL elements of V (including those that happen to be in P1). we say such properties are "inherited" by any subset.

you WILL have to show that P1 has a 0-vector (or at the very least that at least ONE polynomial of degree 1 or less, or the 0-polynomial, satisfies your membership requirement for P1).

if you show "closure under scalar multiplication", then the two distributive laws, and the unit scalar multiplication law are also "inherited" properties.

Originally Posted by bonfire09

I have to show P1={a0 + a1x|a0 - 2a1 = 0} is a vector space using the 10 parts of the definition such as closed under addition and so on.

is the first statement right?
1. vector addition
Let a0+a1x and b0+b1x be two elements of P1. Then (a0+a1x)+(b0+b1x)=(a0+b0)+(a1+b1)x as long

There is no "as long" here- that is the definition of any the sum of two such polynomials
what you mean is that it satisifies the condition "a0+ b0- 2(a1+ b1)= 0" "as long as"

as (a0+b0)-2(a1+b1)=(a0-2a1)+(b0-2b1)=0+0=0.

although I would have said "because" rather than "as long as". That statement is, by definition, true for functions in the original set.

But, except for inaccuracies in the terminology, yes, your idea is correct.

oh ok thanks. I have a quick question. Since P1={a0 + a1x|a0 - 2a1 = 0} where a0 - 2a1 = 0. Also a0=2a1. Could I rewrite the set as such P1={2a1 + a1x|a0=2a1). I still get the same elements either way.

Yes, that's the same thing.