1. Linear Operator Question

I'm working on a practice final exam, and I'm stuck on a problem:

Let $T: \mathcal{P}_2 \to \mathcal{P}_2$ be a linear operator defined by $T(f) = f' + f''$. ( $\mathcal{P}_2$ is the set of all polynomials of order $\leq 2$).

(Note that $T$ is corresponding to a matrix)
(1) Find the matrix $A$.

For this, I'm not sure if I need to use a basis as a reference or not. Is the matrix representation of $T$ supposed to operate on coordinates of $f$? Could I not just assume the standard basis for this, $B = \{1,x,x^2\}$, and use that to find $A$?

I imagine I'd be solving the problem $A[a \ b \ c]^T = [0 \ 2a \ (2a + b)]^T$ since the general polynomial of order two has the coordinates $[a \ b \ c]^T$ in $B$ and, after being operated on by $T$, has coordinates $[0 \ 2a \ (2a + b)]^T$.

Can you tell me if I'm on the right track?

(2) Is $A$ diagonalizable?

If I continue with the method from (1), then $A$ is NOT diagonalizable, but, of course, I'm not certain that my answer in (1) is correct or not.

(3) Find a basis, $B$ of $\mathcal{P}_2$ such that $[T]_B$ is a diagonal matrix.

In this context, is $[T]_B$ just my $A$ from part (1) just with a different basis? I figured that it would be, but, if that were the case, couldn't I just work backwards from (3) to (1) and everything is obvious?

2. Re: Linear Operator Question

Originally Posted by Aryth
I'm working on a practice final exam, and I'm stuck on a problem:

Let $T: \mathcal{P}_2 \to \mathcal{P}_2$ be a linear operator defined by $T(f) = f' + f''$. ( $\mathcal{P}_2$ is the set of all polynomials of order $\leq 2$).

(Note that $T$ is corresponding to a matrix)
(1) Find the matrix $A$.

For this, I'm not sure if I need to use a basis as a reference or not. Is the matrix representation of $T$ supposed to operate on coordinates of $f$? Could I not just assume the standard basis for this, $B = \{1,x,x^2\}$, and use that to find $A$?

I imagine I'd be solving the problem $A[a \ b \ c]^T = [0 \ 2a \ (2a + b)]^T$ since the general polynomial of order two has the coordinates $[a \ b \ c]^T$ in $B$ and, after being operated on by $T$, has coordinates $[0 \ 2a \ (2a + b)]^T$.

Can you tell me if I'm on the right track?

(2) Is $A$ diagonalizable?

If I continue with the method from (1), then $A$ is NOT diagonalizable, but, of course, I'm not certain that my answer in (1) is correct or not.

(3) Find a basis, $B$ of $\mathcal{P}_2$ such that $[T]_B$ is a diagonal matrix.

In this context, is $[T]_B$ just my $A$ from part (1) just with a different basis? I figured that it would be, but, if that were the case, couldn't I just work backwards from (3) to (1) and everything is obvious?
I just read a pretty good treatment of this. You do use the basis you mentioned above and you note

T{1 0 0} = {0 0 0}, T{0 1 0} = {1 0 0}, T{0 0 1} = {2 2 0}

in other words T(1) = 0, T(x) = 1, T(x^2) = 2+2x

 0 1 2 0 0 2 0 0 0

It's pretty clearly not diagonizable.

3) is above my head atm but you might try the Fourier basis, I saw that done in another article I just read trying to come up to speed on this.

3. Re: Linear Operator Question

Originally Posted by romsek
I just read a pretty good treatment of this. You do use the basis you mentioned above and you note

T{1 0 0} = {0 0 0}, T{0 1 0} = {1 0 0}, T{0 0 1} = {2 2 0}

in other words T(1) = 0, T(x) = 1, T(x^2) = 2+2x

 0 1 2 0 0 2 0 0 0

It's pretty clearly not diagonizable.

3) is above my head atm but you might try the Fourier basis, I saw that done in another article I just read trying to come up to speed on this.
Yeah, I'm glad I was on the right track. For (3), I have a few bases I can try. If those don't work, I'll certainly give the Fourier basis a try. Thanks for your help.

4. Re: Linear Operator Question

Originally Posted by Aryth
Yeah, I'm glad I was on the right track. For (3), I have a few bases I can try. If those don't work, I'll certainly give the Fourier basis a try. Thanks for your help.
a little toying with this convinced me Fourier is not the way to go. It looks like minimal polynomials are going to play a role in this.

This may be of interest.

5. Re: Linear Operator Question

I'm not sure why you talk about "trying" a particular basis. A linear transformation, from vector space U to vector V, can only be written as a matrix in given bases for U and V. Are you not told what basis to use? Yes, "1", "x", and " $x^2$". Then T(1)= 0, T(x)= 0+ 1, and [tex]T(x^2)= 2x so, yes, the matrix, in that (ordered) basis is $\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & 1 & 2 \end{bmatrix}$.

Now, a 3 by 3 matrix is invertible if and only if it has three independent eigenvectors. So, yes, find the "characteristic polynomial", set it equal to 0, and solve for the eigenvalues. If there are three distinct eigenvalues ,there definitely are three independent eigenvectors and the matrix is diagonalizable. If there are only one or two distinct eigenvalues there still may be three independent eigenvalues, but you would have to check that.

(3) kind of gives the answer to (2) away- it wouldn't make sense to ask for a basis in which A is diagonal if A were NOT "diagonalizable". That basis is the basis consisting of the eigenvectors of A.

6. Re: Linear Operator Question

Halls point is pretty obvious if you think about it. If T were diagonalizable you could recover f from f' + f''. This is obviously not possible since you lose any information on the 0th order coefficients when you differentiate.