# Eigenvalues and diagonalisation of differentiation as a linear transformation

1. Feb 13, 2010

### Kate2010

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

Let V be the space of polynomials with degree $$\leq$$ n (dimV=n+1)
i. Let D:V->V be differentiation, i.e. D: f(x) -> f'(x)
What are the eigenvalues of D? Is D diagonalisable?

ii. Let T be the endomorphism T:f(x) -> (1-x)2 f''(x).
What are the eigenvalues of T? Is T diagonalisable?

2. Relevant equations

3. The attempt at a solution

i. I have constructed the matrix of D with respect to the basis {1,x,x2,...,x2}
C = $$\left[ \begin{array}{ccccc} 0 & 1 & 0 & ... & 0 \\ 0 & 0 & 2 & ... & 0 \\ : & : & : & : & : \\ 0 & 0 & 0 & 0 & n \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]$$

The characteristic polynomial of this (I think) is xn+1 + (-1)n(n!)

I have no idea how to solve this to get eigenvalues?

Or is there a better way to approach this question?

2. Feb 13, 2010

### vela

Staff Emeritus
Your characteristic polynomial is wrong. Try expanding the determinant along the first column.

3. Feb 13, 2010

### Kate2010

Oh yes so it is, the characteristic polynomial is just xn+1 I think, this equals zero when $$\lambda$$ = 0, so $$\lambda$$ = 0 is the only eignvalue, and D is not diagonalisable?

4. Feb 13, 2010

### vela

Staff Emeritus
Right. It has only one eigenvector. The problem is basically finding polynomials that satisfy $p'(x)=\lambda p(x)$. If p(x) is degree n>0, the LHS is degree n-1, so there is no solution for n>0. The only solution is p(x)=constant.

5. Feb 13, 2010

### Kate2010

Any ideas on part ii? I tried to express T as a matrix but it seems a lot more complicated.

So I tried to solve (1-x2) f''(x) = $$\lambda$$x and got a value for f(x) involving logs. Does this mean $$\lambda$$ = 0 is the only eigenvalue again?

6. Feb 14, 2010

### vela

Staff Emeritus
Were you able to express T as a matrix? One way to do it is to apply T to the nth basis vector to get the nth column of T. For example, the third column would be given by:

$$T\begin{bmatrix}0\\0\\1\\0\\ \vdots \\ 0\end{bmatrix}=(1-x)^2 (x^2)'' = 2-4x+2x^2 = \begin{bmatrix}2\\-4\\2\\0\\ \vdots \\ 0\end{bmatrix}$$

You'll need to figure out what pattern it follows.

7. Feb 14, 2010

### Kate2010

Would you not also square the x in the first bracket, so it would be (1-x2)2f''(x2)

Then when I try to create the matrix I think the onlu number in the diagonal is a -4 from this column, and the rest of the matrix contains a lot of 0s, so just guessing would the characteristic polynomial be xn(x+4) in which case we have eigenvalues 0 and -4? Is there a better way to express this pattern?

Last edited: Feb 14, 2010
8. Feb 14, 2010

### Kate2010

If so, considering 0, f(x) = px + q would work, however I can't think of a polynomial that would work for -4.

9. Feb 14, 2010

### vela

Staff Emeritus
What exactly is T? In the original post, you wrote (1-x)2f''; in your later post, you have (1-x2)f''. I used the former in my example.

I'd try finding the eigenvalues for the n=3 or n=4 case. See if you can spot a pattern.

Another approach you could take is finding a series solution to the differential equation. See if there's a way to get the series to terminate after a finite number of terms so you get a polynomial solution. By the way, in the differential equation you wrote down, the RHS should have f(x), not just x.

10. Feb 16, 2010

### vela

Staff Emeritus
No, f(x)=x2 is the vector. T takes its second derivative and multiplies it by (1-x)2.

You're off by a column. The bottom 2 would be the element in the diagonal.

Try working out the n=3 case. You'll get a 4x4 matrix that you can read the eigenvalues off of. When you solve for the eigenvectors, you'll (hopefully) see the pattern, from which you can deduce the solution to the differential equation.

11. Feb 16, 2010

### Kate2010

Thanks for all your help, I think I got there in the end. I had misunderstood the question. When I made a matrix of T I found the eigenvectors to be 2,6,...,n(n-1) and the eigenvalues to be 1, x, (1-x)^n :)