Linear Transformation Exercise: Determining Kernel and Image

[Tranquillity]

Homework Statement

Let f:R[X] -> R[X] be the linear transformation sending a polynomial P(X) to f(P(X))= P(X+1) - P(X).

a) Let f4: R4[X] -> R[X] be the linear transformation induced by restriction of f to the R-vector space of polynomial of degree at most 4. Determine the kernel and the image of f4. (4 is a subscript)

b)Answer part (a) again with f4 replaced by fn (n any non-negative integer).
Deduce that for any P in R[X], there exists Q in R[X] such that f(Q)=P

c) Let Q in R[X] and let S in R[X] be a polynomial such that f(S)=Q. Show that any other solutions of the equation f(P)=Q can be written P=S+S' with S' in ker(f)."

Homework Equations

I have problem understanding the concepts of Linear Transformations and I would appreciate if you can provide me the full solution of this exercise with baby steps, since I most care about the steps that guide to the solution so I can handle such types of exercises.

The Attempt at a Solution

Let {1, x^2, x^3, x^4} be the basis of R4[x]

Then f(1)=1-1=0
f(x)=(x+1)-x=1
f(x^2)=(x+1)^2-x^2=2x+1=(2C1)*x+(2C2)*1
f(x^3)=(x+1)^3-x^3=3(x^2)+3x+1=(3C1)(x^2)+(3C2) x+(3C3)
f(x^4)=(x+1)^4-x^4=(4C1)(x^3)+(4C2)(x^2)+(4C3)x +(4C4)

where 2C1 means two-choose one notation

Then kerf4=[0 1 1 1 1; 0 0 2 3 4; 0 0 0 3 6; 0 0 0 0 4; 0 0 0 0 0] *[x0 x1 x2 x3 x4]=[0 0 0 0 0]

then [x0 x1 x2 x3 x4]=[c 0 0 0 0] where c exists in R so kerf4=R
since the polynomials in the kernel are P(x)=c

Then we would say that any vector y=[y0 y1 y2 y3 y4] of the image has the form:

[y0 y1 y2 y3 y4]= [0 1 1 1 1; 0 02 3 4; 0 0 0 3 6; 0 0 0 0 4; 0 0 0 0 0]* [x0 x1 x2 x3 x4]
so Imf4=[x1+x2+x3+x4; 2*x2+3*x3+4*x4; 3*x3+6*x4; 4*x4; 0] ?

For b the thing I observed is that f(x^n)=Sum from k=1 to n (n choose k)* x^(n-k). After that how do I continue to find the kernel and the image?

I am not pretty sure for my approach, if it's possible to check it and to help me to go a step further from where I am now, if I am correct!

 

[lanedance]

for a) your reasoning is correct, however for the image is given by span{1, x^2, x^3)

 

[lanedance]

then in b) this will generalise to ker = c, image one less dimension than the original space

 

[Tranquillity]

So kernel for part b is R again and image is {1,x,x^2,...,x^(n-1)}?
How did you find the image in the two cases? And then how do I proceed to finish part b and c? Thanks again!

 

[lanedance]

based on you calc of the action on each of the vectors in the restricted space, you end up with a set of basis vectors for the image

 

[Tranquillity]

Can you give me more analytical steps if it's possible?

 

[lanedance]

i'm not too sure what you want? however generally the philiosphy on this forum is to help lead you in the right direction rather than do the whole problem

 

[Tranquillity]

How did you find that the image it's spanned by {1,x,x^2,x^3} in the first case?
I mean we know that the linear transformation reduces the degree of the polynomial by one, but the how did we arrived there by the definition of image?
Secondly you told me my reasoning for kernel in part a was right, the same reasoning holds for part b? And for part b how do I deduce that there exists Q in R[X] such that f(Q)=P?

 

[lanedance]

you take the basis and get
f(1)=0
f(x)=1
f(x^2)=2x+1
f(x^3)=3x^2+3x+1
f(x^4)= 4x^3+6x^2+4x +1
any vector in the image can be written as a combination of the image of the basis vectors

 

[Tranquillity]

Thanks for the help. I have another two final questions:

a)So for part a kernel is R, image is {1,x,x^2,x^3} and in part b kernel is R and image is {1,x,x^2,...,x^(n-1)}?

b) How do I deduce that for any P in R[X], there exists Q in R[X] such that f(Q)=P?

 

[lanedance]

try induction

 

[lanedance]

also i would refer to the kernal as the space given by span{1}, as it its a little clearer in the current context to speak of basis vectors

 

[Tranquillity]

Ok, thanks for all!