Linear Algebra problem

[Xehrad]

Let P3(R) designate the R-vector space of real polynomials: p= a + bx +cx^2, where a ,b, c ∈ R, of degree ≤ 2

a) Show that the spanning set S = {p ∈ P3(R) | p(1) = 0}, is a subspace of P3(R)
Tried showing that 0 is an element in S, that addition holds within S and that multiplication holds within S

Letting P(X) = 0, since P(1) = 0

Letting P(X),T(X) be arbitrary. Then: P(3)+T(3) = 0 + 0 = 0, since that (P(x)+T(x))) ∈ S

Letting k be a constant and since that P(x) ∈ S

Then: k*P(3) = k*0 = 0.

S is therefore a subspace.

2) Argue that p1 = 1 − x^2 , p2 = x-x^2 spans S.
Knowing that p1 and p2 only spans S if they are linearly independent:

a1p1 +a2p2 = 0
a1(1-x^2)+a2(x-x^2) = 0

a1-a1x^2+a2x-a2x^2 = 0
-a1x^2-a2x^2+a2x+a1 = 0

Giving us three linear equations:
-a1x^2-a2x^2 = 0
a2x = 0
a1= 0

Therefore only trivial solutions a1=a2=0, therefore p1 and p2 spans S.

Now to the part where i'm stuck.

3) Argue that S cannot be spanned by one element.
I'm not really sure where to start with this problem, do I have to show that it's not linearly independent if I only have one element? And if so, how?

4) Conclude that the dimensions of S is 2.
Do I have to find the number of vector basis' for S?

 

[HallsofIvy]

Let P3(R) designate the R-vector space of real polynomials: p= a + bx +cx^2, where a ,b, c ∈ R, of degree ≤ 2

a) Show that the spanning set S = {p ∈ P3(R) | p(1) = 0}, is a subspace of P3(R)
Tried showing that 0 is an element in S, that addition holds within S and that multiplication holds within S

This is poorly stated. S is not a "spanning set" though it might be a "spanned" set!

Letting P(X) = 0, since P(1) = 0

You should show, or at least state, that p(x)= 0 for all x is in S.

Letting P(X),T(X) be arbitrary. Then: P(3)+T(3) = 0 + 0 = 0, since that (P(x)+T(x))) ∈ S
You should state that P(X), T(X) are arbitrary polynomials in S. And the condition on S is NOT that P(3)= 0 but that P(1)= 0 and T(1)= 0.

Letting k be a constant and since that P(x) ∈ S

Then: k*P(3) = k*0 = 0.

Again, k*P(1)= k*0= 0!

S is therefore a subspace.

2) Argue that p1 = 1 − x^2 , p2 = x-x^2 spans S.
Knowing that p1 and p2 only spans S if they are linearly independent:

That is true IF S is two dimensional. How do you conclude that?

a1p1 +a2p2 = 0
a1(1-x^2)+a2(x-x^2) = 0

a1-a1x^2+a2x-a2x^2 = 0
-a1x^2-a2x^2+a2x+a1 = 0

Giving us three linear equations:
-a1x^2-a2x^2 = 0
a2x = 0
a1= 0

Therefore only trivial solutions a1=a2=0, therefore p1 and p2 spans S.

I think you are confusing a "spanning set" with a "basis"!
A set of vectors spans a subspace if and only if every vector in the subspace can be written as a linear combination of vectors in the set.
Any vector is S is of the form p(x)= ax^2+ bx+ c with the condition that p(1)= a+ b+ c= o. We can write c= - a- b so that p(x)= ax^2+ bx+ - a- b. We need to show that, for any a and b, there exist A and B such that ax^2+ bx+ 1- a- b= A(1- x^2)+ B(x- x^2)= (-A- B)x^2+ Bx+ A.
We need to show that we can solve -A- B= a, B= b and A= 1-a-b. Clearly B= b so -A-B= -A-b= a, A= b+ a. That shows that those two polynomials span S.

Now to the part where i'm stuck.

3) Argue that S cannot be spanned by one element.
I'm not really sure where to start with this problem, do I have to show that it's not linearly independent if I only have one element? And if so, how?

4) Conclude that the dimensions of S is 2.
Do I have to find the number of vector basis' for S?

Yes, you need to find the number of vectors in a basis for S.

As I said before, I think you are confusing "spanning set", "independent vectors", and "basis".

A basis for a finite dimensional vector space has three properties:

1) The vectors span the space.
2) The vectors are independent.
3) The number of vectors is equal to the dimension of the vector space.

Further, any two of those is sufficient to prove the third!

I showed, as part 2, that the given set of two polynomials spans the space. You showed, not that they span the space, but that they are independent. Those together show that the dimension of the space is two. Since the dimension is not 1, a single vector cannot span the set.

 

[Steven G]

Where is P(3) and T(3) coming from? The 3 is from the upper limit for the number of coefficients that a polynomial of degree or less might have. The problem clearly stated to find p(1).