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Linear transformations, eigenvalues, diagonalisation

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?


i. I have constructed the matrix of D with respect to the basis {1,x,x2,...,x2}
C = [010...0002...0:::::0000n00000]\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

So the only eigenvalue is 0 and D is not diagonalisable?

How would I go about part ii?
Reply 1
Avatar for Hathlan
Hathlan
2
Kate2010
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?


i. I have constructed the matrix of D with respect to the basis {1,x,x2,...,x2}
C = [010...0002...0:::::0000n00000]\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

So the only eigenvalue is 0 and D is not diagonalisable?

How would I go about part ii?


I think your first part is right, although it's more because there's only one eigenvector for that eigenvalue.

With the second bit, just try to think what eigenvectors could be. To start with, you get two eigenvectors which'll have eigenvalue zero. What other eigenvectors can you get? Look carefully at f(x) to see if you can find 'em and if not, they're all quite similar to

MASSIVE HINT:

Spoiler



So how many are there? What eigenvalues will we get?
Constructing the matrix of T would be a start.

As an aside, the linear algebra lecturer this term was pretty good.
Reply 3
Avatar for Kate2010
Kate2010
OP
Do we get the eigenvalues 0, 2, 6, 12 ,..., n(n-1) i.e. -n(n-1) for n>0. (From considering the matrix)

For the eigenvectors, we could have g(x) = constant (eigenvalue 0), g(x) = px (eigenvalue 0), g(x) = (x-1)^2 (eigenvalue 2), then could we have (1-x)^n?

So we have n linearly independent eigenvectors, so T is diagonalisable?
Reply 4
Avatar for Hathlan
Hathlan
2
Kate2010
Do we get the eigenvalues 0, 2, 6, 12 ,..., n(n-1) i.e. -n(n-1) for n>0. (From considering the matrix)

For the eigenvectors, we could have g(x) = constant (eigenvalue 0), g(x) = px (eigenvalue 0), g(x) = (x-1)^2 (eigenvalue 2), then could we have (1-x)^n?

So we have n linearly independent eigenvectors, so T is diagonalisable?


Yup, I think that's right.

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