Finding a Basis for S: Polynomials in P3 with Specific Form

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SUMMARY

The discussion focuses on finding a basis for the subspace S of P3, defined by polynomials of the form ax2 + bx + 2a + 3b. The basis is determined to be the set {(x2 + 2), (x + 3)}. The Wronskian of these polynomials, calculated as x2 + 6x - 2, confirms their linear independence, establishing that they form a valid basis for the subspace S.

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  • Familiarity with the Wronskian determinant
  • Basic algebraic manipulation and factoring techniques
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Dustinsfl
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My mind is shot.

Let S be a subspace of P3 consisting of all polynomials of the form ax2+bx+2a+3b. Find a basis for S.

I am not sure where to start.
 
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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?
 
Try "factoring out a". It can't factor out of everything, but try factoring it out of all the terms you can.

--------

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?
 
Last edited:
Then set that equal to 0? Also factor out b the same way?
 
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.
 
a(x2+2)+b(x+3)
 
I can then set up the Wronskian.
 
Do you know what to do now, or do you still have questions?
 
I am still confused but I evaluated the Wronskian. How can I come up with the basis now?
W=x^2+6x-2
 
  • #10
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.
 
  • #11
That is it?
Thanks.
 

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