Question

Let P(lowercase 3) be the set of all real polynomials of degree or less. Show {2x^3+x+1,x-2,x^3-x^2} is a linearly independent set and find a basis for P(3) which includes these three polynomials.

Answer 1

let \(\mathbf{P}_3\) be the vector space of real univariate polynomials of degree 3 or less. We need to prove that \(\{2x^3+x+1,x-2,x^3-x^2\}\) is a linearly independent set:$$a(2x^3+x+1)+b(x-2)+c(x^3-x^2)=0\\(2a+c)x^3-cx^2+(a+b)x+(a-2b)=0$$To prove this set is linearly independent, then, we solve for \(a,b,c\) by equating coefficients on both sides:$$2a+c=0\\-c=0\\a+b=0\\a-2b=0$$Immediately we see that \(c=0\) hence \(a=0\), and therefore \(b=0\); given this system is overdetermined we know \(a=b=c=0\) is the only solution and they are indeed linearly independent

Answer 2

To find a basis, first realize our vector space's dimension is \(4\) hence we need another vector; the easiest method is to pick a random vector and decompose it into sums of the ones available -- how about \(x^3-1\)?$$x^3-1=\frac12(2x^3+x+1)-\frac12(x+1)-1=\frac12(2x^3+x+1)-\frac12(x-2)-\frac52$$since we're left with a constant term \(-5/2\), we can pick any other constant as our last basis vector -- so to be clean, how about \(1\)?

Answer 3

$$a(2x^3+x+1)+b(x-2)+c(x^3-x^2)+d=0\\(2a+c)x^3-cx^2+(a+b)x+(a-2b+d)=0$$Just to check, let's verify this is still linearly independent (so that it's actually a basis):$$2a+c=0\\-c=0\\a+b=0\\a-2b+d=0$$Once again we immediately conclude \(c=0\) hence \(a=0\) and \(b=0\) and therefore it follows \(d=0\) -- we're good!

Answer 4

thats pretty interesting how we can apply polynomials to this

Answer 5

hmm seems like u have to be clear with what u have to try and solve. but yea fully understnad that now

Answer 6

good :-) and yes this shows that the theory of vector spaces is more powerful than just 'arrows in the plane'

Answer 7

exactly

Answer 8

thats preety interesting when u think about it