# Homework Help: T(f) = f + f', show that T is not diagonalizable

1. Oct 30, 2012

### L.Wil

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

T: k[x]n -> k[x]n

T(f) = f + f'

show that T is not diagonalizable for n >= 1

2. Relevant equations

3. The attempt at a solution

I would usually start by getting a characteristic polynomial but I don't know how to do that here?

2. Oct 30, 2012

### L.Wil

I also thought I could make it into an ODE and solve for the eigenvalue but i couldn't get that to work

3. Oct 30, 2012

### Dick

k[x] are polynomials? If so look at the ODE an eigenvector has to satisfy and think about degrees of polynomials.

4. Oct 31, 2012

### L.Wil

It doesn't say whether k[x] are polynomials.

So would the ODE be λf = f + f'

And then I solve for λ?

5. Oct 31, 2012

### Dick

The definition of k[x] is pretty important to how you solve this problem. You need to find out what it is.

6. Oct 31, 2012

### L.Wil

Looking at my notes I think that k[x] are polynomials

7. Oct 31, 2012

### Dick

Write your equation as (λ-1)f = f'. If λ isn't 1, what can you say about the degrees of the polynomials on each side?

8. Oct 31, 2012

### L.Wil

That the degree of the polynomial on the right is one less than on the left?

9. Oct 31, 2012

### Dick

What do you conclude from that?

10. Oct 31, 2012

### L.Wil

That there aren't enough eigenvalues for it to be diagonalizable?

I'm not really sure!

11. Oct 31, 2012

### Dick

You going to try to show there aren't enough eigenvectors or functions. To do that you have to figure out all the possible ways (λ-1)f = f' can be solved. Go back to your comment about the λ≠1 case. Can two polynomials of different degree be equal?

12. Oct 31, 2012

### L.Wil

No I don't think so?

13. Oct 31, 2012

### L.Wil

I solved it to get f(x) = ke^((λ-1)x)

Should i rearrange for λ?

14. Oct 31, 2012

### Dick

That doesn't work because the exponential isn't generally a polynomial. You only want polynomial solutions. You've ruled out λ≠1. Are there polynomial solutions if λ=1? What do they look like?

15. Oct 31, 2012

### L.Wil

Thanks so much for all of your help.

If λ=1, f'(x)=0

and so f(x) = constant?

16. Oct 31, 2012

### Dick

Very welcome! Ok, so your only possible eigenvalue is 1 and all of the eigenfunctions are constants. If T were diagonalizable in k[x]_n how many linearly independent eigenfunctions would you need? What's the dimension of k[x]_n?