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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.

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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 $\left(\begin{array}{c}a\\ b\\ c\end{array}\right)$.

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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.

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