Basis for polynomials of the form ax² + bx + 2a + 3b in P₃

[Dustinsfl]

My mind is shot.

Let S be a subspace of P<sub>3 consisting of all polynomials of the form ax²+bx+2a+3b. Find a basis for S.

I am not sure where to start.</sub>

 

[VeeEight]

What are the coefficients for P₃ in? Are a, b elements of some field?
Find a generating set for this subspace. How about {x², x, ...}? What else should be in there to generate S?

 

[Mathnerdmo]

Try "factoring out a". It can't factor out of everything, but try factoring it out of all the terms you can.

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If that doesn't make sense, try to get a feel for what the subspace is by choosing values for a and b.
For instance, which element does a = 2, b = 0 give you?

 

[Dustinsfl]

Then set that equal to 0? Also factor out b the same way?

 

[Mathnerdmo]

My hint about factoring is purposefully vague. What did you get when you tried factoring out the a?

However, yes, if you correctly factor out the "a"s and the "b"s, you should start to see the answer.

 

[Dustinsfl]

a(x²+2)+b(x+3)

 

[Dustinsfl]

I can then set up the Wronskian.

 

[Mathnerdmo]

Do you know what to do now, or do you still have questions?

 

[Dustinsfl]

I am still confused but I evaluated the Wronskian. How can I come up with the basis now?
W=x^2+6x-2

 

[Mathnerdmo]

We have all the information now. We just need to realize what we have.

(1)
Every element of S can be written as
a(x²+2) + b(x+3)
This shows that (x²+2) and (x+3) generate S (also, setting a=1,b=0 or vice versa shows that they're in S).

(2)
The Wronskian of these two elements is x²+6x-2, which is not identically zero. This shows that these two elements are linearly independent.

A linearly independent generating set is precisely a basis, so these two polynomials form your basis.

 

[Dustinsfl]

That is it?
Thanks.