Finding eigenvalues of the linear map of differentiation on polynomial sets I have the following question:

Let $\mathbf{P_2}$ denote the vector space of real polynomials of degree at most two, and let $\mathbf{L}:\mathbf{P_2}\rightarrow\mathbf{P_2}$ be the linear map $\mathbf{L}f=f', $(the derivative of $f$). What are the eigenvalues of $\mathbf{L}$?

I got since the linear map takes $$\begin{pmatrix} \alpha \\ \beta\\ \gamma \end{pmatrix}\rightarrow \begin{pmatrix} 0 \\ 2\alpha \\\beta \end{pmatrix}$$We have $$\begin{pmatrix} 0 \\ 2\alpha \\\beta \end{pmatrix}=\lambda\begin{pmatrix} \alpha \\ \beta\\ \gamma \end{pmatrix}$$
This only will have a solution without $\alpha=\beta=\gamma=0$, when $\lambda=0$ 
Is this correct? Just making sure my reasoning is sound.
 A: Hint: A base for $\mathbf{P_2}$ is $\{1,x,x^2\}$ and the derivatives are
$$1\longmapsto 0,$$
$$x\longmapsto 1,$$
$$x^2\longmapsto 2x.$$
Then the matrix of the transformation is
$\left(\begin{array}{ccc}
0&1&0\\
0&0&2\\
0&0&0
\end{array}\right).$
The characteristic matrix is
$\left(\begin{array}{ccc}
x&-1&0\\
0&x&-2\\
0&0&x
\end{array}\right)$, whose determinant is the characteristic polynomial
$$\chi(x)=x^3,$$ 
with a  triple zero $\lambda=0$.
These are the matrix's eigenvalues.
The only non trivial eigenvector is 
$\left(\begin{array}{c}
1\\
0\\
0
\end{array}\right)$, since
$$\left(\begin{array}{ccc}
0&1&0\\
0&0&2\\
0&0&0
\end{array}\right)
\left(\begin{array}{c}
1\\
0\\
0
\end{array}\right)=0
\left(\begin{array}{c}
1\\
0\\
0
\end{array}\right)
.$$
Remark:
Here a polynomial $a+bx+cx^2$ is manipulated by is components conventionally as  $\left(\begin{array}{c}a\\ b\\ c\end{array}\right)$.
A: Yes, correct. 
$0=\lambda\alpha\ \implies\ $ either $\lambda=0$ or $\alpha=0$. 
If $\lambda=0$, we get $2\alpha=0$ and $\beta=0$. 
If $\lambda\ne 0$ then $\alpha=0$ and hence $2\alpha=\lambda\beta \implies \beta=0$ and also $\beta=\lambda\gamma\implies \gamma=0$. So this case won't give an eigenvector.
