Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $L: P_2 \longrightarrow P_2$ be the linear operator defined by $L(p(t)) = p'(t)$ for $p(t) \in P_2$, the space of real polynomials of degree at most $2$. Is $L$ diagonalizable? If it is, find a basis $S$ for $P_2$ with respect to which $L$ is represented by a diagonal matrix.

Answer: $L$ is not diagonalizable. The eigenvalues of $L$ are $\lambda_1 = \lambda_2 = \lambda_3 = 0$. The set of associated eigenvectors does not form a basis for $P_2$.

I don't how you solve this problem. How do you transform the polynomial into a matrix so that I can find the eigenvalues? I can solve these questions if they state the matrix but this one concerns polynomials. Can someone please help me?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

Hints:

  1. Pick a basis of $P_2$. What is the standard basis?
  2. Write the matrix of $L$ with respect to that basis (you'll have to apply $L$ to each element of the basis of $P_2$, and calculate its coordinates in that basis; that'll give you the columns of the matrix).
  3. Work with the matrix now: is it diagonalizable?

1.

We can choose the basis $B = \{1, x, x^2\}$.

2.

$L(1) = 1' = 0 = 0 \cdot 1 + 0 \cdot x + 0 \cdot x^2 \\ L(x) = x' = 1 = 1 \cdot 1 + 0 \cdot x + 0 \cdot x^2 \\ L(x^2) = (x^2)' = 2x = 0 \cdot 1 + 2 \cdot x + 0 \cdot x^2 \\ \text{Then} \\ \left[L\right]_B = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 &0 \end{pmatrix}$

3.

Find the roots of $\det(xI - \left[L\right]_B)$, which will be the eigenvalues, and the corresponding eigenvectors. Do you end up with a basis of $\mathbb{R}^3$?

share|improve this answer
    
Would it be the same if I had the basis in the reverse order, {$t^{2}$, t, 1} and the matrix ended up being $\begin{matrix} 0 & 0 & 0\\ 2 & 0&0 \\ 0&1&0\end{matrix}$ ? –  Ashley Jun 24 '12 at 5:27
    
@Ashley You would have a different matrix representation of $L$, but the matrix representation is just that, a representation. It temporarily enables you to work with real numbers, as if you had a linear transformation $\Bbb R^3 \to \Bbb R^3$, forgetting for a moment that in fact it was $P_2 \to P_2$. But it doesn't matter which basis you pick; as long as you do your calculations correctly, you'll end up with the same answer. –  talmid Jun 24 '12 at 5:29
    
I see. But what does it mean in the answer that "The set of associated eigenvectors does not form a basis for P2"? Didn't I just form a basis for P2? Sorry for asking a lot of questions –  Ashley Jun 24 '12 at 5:31
    
@Ashley Yes, you formed a basis for $P_2$, which is completely necessary to solve the problem. $P_2$ certainly has a basis (in fact, lots of them). However, the linear transformation will be diagonalizable if and only if its eigenvectors also contain a basis of $P_2$. You now have to calculate the eigenvalues of the matrix, and see what their associated eigenvectors are; and then see if these eigenvectors are enough to form a basis (independently of the basis you considered earlier, which was just used to do the calculations, and has no inference in whether $L$ is diagonalizable). –  talmid Jun 24 '12 at 5:34
    
Oh, I get it now. Thanks so much for your help talmid :) –  Ashley Jun 24 '12 at 5:39

The first step is to pick your favorite basis for $P_2$; this will allow you to find a matrix for $L$. Then you can use the standard techniques once you have a matrix representation for $L$.

Now, my favorite basis for $P_2$ (which I am guessing from context is all polynomials of degree at most $2$; careful, as sometimes it means the set of polynomial of degree less than $2$), absent countervailing influences, is $\beta=\{1,x,x^2\}$. How does $L$ behave relative to this basis?

$$\begin{align*} L(1) &= (1)' = 0\\ &= 0(1) + 0(x) + 0(x^2);\\ L(x) &= (x)' = 1\\ &= 1(1) + 0(x) + 0(x^2);\\ L(x^2) &= (x^2)' = 2x\\ &= 0(1) + 2(x) + 0(x^2). \end{align*}$$ So the matrix that represents $L$ with respect to the basis $\beta$, $[L]_{\beta}^{\beta}$, is $$[L]_{\beta}^{\beta} = \left(\begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 2\\ 0 & 0 & 0 \end{array}\right)$$ (the first column is the $\beta$-coordinate vector of $L(1)$; the second column is the $\beta$-coordinate vector of $L(x)$; and the third column is the $\beta$-coordinate vector of $L(x^2)$).

The characteristic polynomial of $L$ is the same as the characteristic polynomial of any of its matrix representations, and the eigenvalues of $L$ are the eigenvalues of $[L]_{\beta}^{\beta}$.

Can you go from here?

share|improve this answer
    
Thanks a lot Arturo! :) –  Ashley Jun 24 '12 at 5:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.