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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?

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up vote 1 down vote accepted


  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?


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


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


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 $\mathbb{R}^3$?

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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?

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Thanks a lot Arturo! :) – Ashley Jun 24 '12 at 5:40

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