# Proving a linear transform is diagonalizable

1. May 10, 2009

### dargar

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

2. May 10, 2009

### Hurkyl

Staff Emeritus
Can you be more precise in what you mean? Eigenvalues of what? What linear transform?

3. May 10, 2009

### dargar

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

4. May 10, 2009

### Hurkyl

Staff Emeritus
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)

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.

It's often useful to start with special cases. Try working through the n=0 case. Then the n=1 case. Mabe even the n=2 case. Only then worry about the general case.