Proving a linear transform is diagonalizable

1. The problem statement, all variables and given/known data

Let U be the R-vector space consisting of all polynomials of degree at most n with
coefficients that are real.

The Derivative Map
F : U [tex]\rightarrow[/tex] U
f(x) [tex]\rightarrow[/tex] f'(x)

Is the derivative function F diagonalizable?


3. The attempt at a solution

My instinct says yes but I'm not too sure why. Am I right in saying that eigenvalues are invariant after a linear transform. I'm pretty sure that before the function is applied you can have a matrix that has n+1 distinct eignenvalues but not sure if this is the right lines or if it is how to set about writing it down in a proof format.

Any help would be great thanks !

 

The linear transform is differentiation function.

The eigenvalues I'm not so sure of , the main reason I say it is because in my notes it seems to be the only thing remotely to do with this question.

In the following deffinitions I have , I assume he means A is the vector space you start with

Let A be an (nxn) matrix, A is diagonalisable by similarity transform if and only if there is a basis for R^n consisting of entirely eigenvectors for A.

and

If A is diagonisible then

the sum of the distinct eigenvalues for dimension of the eigenspace = n

Oh and I also know that eigenvalues are similarity invariant but their eigenvector aren't (altho they have a 1 to 1 correspondance after the function is applied)

Sorry I'm pretty confused at how to go about this, so I'm sorry if it doesn't make sense