# Proving a linear transform is diagonalizable

## Homework Statement

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 $$\rightarrow$$ U
f(x) $$\rightarrow$$ f'(x)

Is the derivative function F diagonalizable?

## 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 !

Hurkyl
Staff Emeritus
Gold Member
Am I right in saying that eigenvalues are invariant after a linear transform.
Can you be more precise in what you mean? Eigenvalues of what? What linear transform?

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

Hurkyl
Staff Emeritus
Gold Member
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.
No, he meant what he said: A is an nxn matrix. Diagonalizability is not a property of vector spaces; it is a property of matrices (and of linear transformations).

(This n, of course, does not necessarily have anything to do with the n in your original problem)

the sum of the distinct eigenvalues for dimension of the eigenspace = n
This doesn't make sense. I know the theorem you're trying to state, but the idea you have in your head doesn't make sense.

If a matrix is diagonalizable, then:

It have a set of (distinct) eigenvalues.
Each of those eigenvalues has a corresponding eigenspace.
Each of those eigenspaces has a dimension.
The sum of all of those dimensions is n.