Linear algebra: diagonalization 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?
 A: 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?
A: Hints:


*

*Pick a basis of $P_2$. What is the standard basis?

*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).

*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 $\mathbb{R}^3$?

