T(f) = f + f', show that T is not diagonalizable

[L.Wil]

Homework Statement

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

T(f) = f + f'

show that T is not diagonalizable for n >= 1

Homework Equations

The Attempt at a Solution

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

 

[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

 

[Dick]

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

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

 

[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?

 

[Dick]

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

So would the ODE be λf = f + f'

And then I solve for λ?

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

 

[L.Wil]

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

 

[Dick]

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

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?

 

[L.Wil]

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

 

[Dick]

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

What do you conclude from that?

 

[L.Wil]

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

I'm not really sure!

 

[Dick]

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

I'm not really sure!

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?

 

[L.Wil]

No I don't think so?

 

[L.Wil]

I solved it to get f(x) = ke^((λ-1)x)

Should i rearrange for λ?

 

[Dick]

I solved it to get f(x) = ke^((λ-1)x)

Should i rearrange for λ?

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?

 

[L.Wil]

Thanks so much for all of your help.

If λ=1, f'(x)=0

and so f(x) = constant?

 

[Dick]

Thanks so much for all of your help.

If λ=1, f'(x)=0

and so f(x) = constant?

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?