In P3, let W = {f ∈ P3 | f (3) = f (2) + f (1)}. Find a basis for W and dim(W )

Answer 1

A good rule of thumb when thinking about linear algebra problems (and other mathematics, honestly) is to put all of your constants on one side and your variables on the other. In this case, this means rewriting the definition of W to be:

W = { f ∈ P₃ | f(3) - f(2) - f(1) = 0 }

This helps, because now we see that W is the kernel of a transformation T defined by:

T(f) = f(3) - f(2) - f(1)

which is a linear transformation (check this!) from P₃ to R. It's also surjective since, for example, T(x-1) = 2-1-0 = 1 is nonzero. So, by rank-nullity, the kernel (W) has dimension 4-1 = 3.

To actually find a basis, then, we are looking for 3 independent polynomials lying in W. A good first guess might be to pick different degrees but adjust the constant to lie in W, so we'll try a basis of the form {x^3+a, x^2+b, x+c}. To solve for a,b,c, we plug these into T:

0 = T(x^3+a) = (27+a)-(8+a)-(1+a) = 18-a, so we get a = 18

0 = T(x^2+b) = (9+b)-(4+b)-(1+b) = 4-b, so we get b = 4, and

0 = T(x+c) = (3+c)-(2+c)-(1+c) = -c, so we get c = 0.

So, a basis for W is {x^3+18, x^2+4, x}.