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

1. ## Dimension of the kernal of F, and determining the basis for the image.

Im struggling with the following Let F: $\P_2(R)$ $\rightarrow$ $\P_3(R)$ be the linear transformation determined by: F(f)(x)=6 $\int_{0}^{2x-1} f(t) dt$

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.

2. ## Re: Dimension of the kernal of F, and determining the basis for the image.

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

$\mathrm{ker}(F) = \{f \in P_2|F(f) = 0\}$

now F(f)(x) is defined to be $F(f)(x) = 6\int_0^{2x - 1} f(t)\ dt$.

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

if $6\int_0^{2x - 1} f(t)\ dt = 0$ then for $f(x) = ax^2 + bx + c$ we have:

$6\int_0^{2x-1} at^2 + bt + c\ dt = 6(\frac{a}{3}(2x-1)^3 + \frac{b}{2}(2x - 1)^2 + c(2x - 1)) = 0$ so that:

$16ax^3 - 24ax^2 + 12ax - 2a + 12bx^2 - 12bx + 3b + 12cx - 6c$

$= 16ax^3 + (12b - 24a)x^2 + (12a - 12b + 12c)x + 3b - 2a - 6c = 0$ for all x.

hence:

$16a = 0 \implies a = 0$
$12b - 24a = 0 \implies 12b = 0 \implies b = 0$
$12a - 12b + 12c = 0 \implies 12c = 0 \implies c = 0$

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 $\{1,x,x^2\}$.

so $F(1)(x) = 6\int_0^{2x-1} dt = 6(2x - 1) = 12x - 6$

and $F(x)(x) = 6\int_0^{2x-1} t\ dt = 3(2x - 1)^2 = 12x^2 - 12x + 3$

and finally, $F(x^2)(x) = 6\int_0^{2x-1} t^2\ dt = 2(2x - 1)^3 = 16x^3 - 24x^2 + 12x - 2$ so a basis for Im(F) is:

$\{12x - 6, 12x^2 - 12x + 3, 16x^3 - 24x^2 + 12x - 2\}$