Thread: Dimension of the kernal of F, and determining the basis for the image.

Im struggling with the following Let F: be the linear transformation determined by: F(f)(x)=6

compute the dimension of the kernel of F. Determine a basis for the image of F.

Firstly is the dimension of ker(F)=1,

Then I have found F(f)(x) and have that it equals: 16ax^3+(12b-24a)x^2 +(12a-12b+12c)x + (-2a+3b-6c), from this is our basis given by [(-2a+3b-6c),(12a-12b+12c),(-24a+12b),16a] ?

Many thanks in advance, any help would be most appreciated.

perhaps we should attempt to find ker(F) first.



now F(f)(x) is defined to be .

so when is this integral identically 0 (for all x)?

if then for we have:

so that:



for all x.

hence:





so the only f with this property is the 0-polynomial. this means that ker(F) = {0}

by the rank-nullity theorem, this means dim(F(P₂)) = 3. thus it suffices to find images of basis elements for P₂, to construct a basis for F(P₂).

so we do that: we will use the basis .

so

and

and finally, so a basis for Im(F) is: