For each real linear space below, find

[Jamin2112]

Homework Statement

For each real linear space below, find a basis and its dimension.

[...]

(c) {2nd degree polynomials p such that p(2) = 0}

Homework Equations

Definition of basis and dimension.

The Attempt at a Solution

So, we know that

p(x) = ax² + bx + c
and
p(2) = 0 = 4a + 2b + c

so that we have

p(x) = ax² + bx - 4a -2b.

But where do I go from here?

 

[LCKurtz]

Homework Statement

For each real linear space below, find a basis and its dimension.

[...]

(c) {2nd degree polynomials p such that p(2) = 0}

Homework Equations

Definition of basis and dimension.

The Attempt at a Solution

So, we know that

p(x) = ax² + bx + c
and
p(2) = 0 = 4a + 2b + c

so that we have

p(x) = ax² + bx - 4a -2b.

But where do I go from here?

Try writing it as a(x²-4) + b(x-2)

 

[Jamin2112]

Try writing it as a(x²-4) + b(x-2)

Done.

Now what?

 

[wisvuze]

the question is asking you to find a set of vectors ( polynomials) that span all polynomials of degree 2 such that 2 is a root for the polynomial. In other words, you need to find polynomials so that every other polynomial of this kind is a linear combination of those.. in other words, the answer is looking right at you :)

 

[Jamin2112]

the question is asking you to find a set of vectors ( polynomials) that span all polynomials of degree 2 such that 2 is a root for the polynomial. In other words, you need to find polynomials so that every other polynomial of this kind is a linear combination of those.. in other words, the answer is looking right at you :)

I have p(x) = a(x²-4)+b(x-2)

a and b can be any real number, so I really have an infinite set

{p₁(x) = a₁(x²-4)+b₁(x-2), p₂(x) = a₂(x²-4)+b₂(x-2), p₃(x) = a₃(x²-4)+b₃(x-2), ...}

I'm not sure where this is going in terms of basis, span, etc.

 

[HallsofIvy]

Yes, of course. Every vector space contains an infinite number of vectors. The whole point of a basis is to be able to write them, in a unique way, in terms of finite (or at least smaller) set of vectors- the basis.

You now know that any vector in your first set can be written as $ax^2+ bx- 4a- 2b= a(x^2- 4)+ b(x- 2)$. Therefore, every such vector can be written as a linear combination of what two basis vectors? What is the dimension of that space?

 

[Jamin2112]

Yes, of course. Every vector space contains an infinite number of vectors. The whole point of a basis is to be able to write them, in a unique way, in terms of finite (or at least smaller) set of vectors- the basis.

You now know that any vector in your first set can be written as $ax^2+ bx- 4a- 2b= a(x^2- 4)+ b(x- 2)$. Therefore, every such vector can be written as a linear combination of what two basis vectors? What is the dimension of that space?

The basis vectors would be (x² - 4) and (x - 2), the dimension being 2. Right?

 

[LCKurtz]

The basis vectors would be (x² - 4) and (x - 2), the dimension being 2. Right?

It will be 2 if the two basis vectors are independent. Have you checked that?

 

[Jamin2112]

It will be 2 if the two basis vectors are independent. Have you checked that?

They most definitely are independent.

 

[LCKurtz]

The basis vectors would be (x² - 4) and (x - 2), the dimension being 2. Right?

It will be 2 if the two basis vectors are independent. Have you checked that?

They most definitely are independent.

Sure, and it's easy to show. But it is part of what you need to include in your proof.