The question I am to answer is as follows.
Consider the Legendre operator $ Φ = (1 − x^2)\frac{d^2}{dx^2} - 2x\frac{d}{dx}$ acting on vector space $R_2[x]$ of polynomials of degree $≤ 2$. Find its eigenvalues and eigenvectors.
In my lecture notes Legendre Operators are only swiftly covered and glossed over, and from the notes I can't work out what I'm supposed to do with it in order to find the characteristic polynomial I need to start the process to find the eigenvalues and vectors.
Even when I have the eigenvalues, I won't have a matrix to substitute the values back into (as I see it) and will struggle to find the eigenvectors.
A helping hand as to how I'm supposed to approach and use the operator would be appreciated.
As a basis of $R_2[x]$ You may take $\{1,x,x^2\}$ for simplicity and compute $$\Phi(1)=(1-x^2)\underbrace{\frac{d^2}{dx^2}1}_{=0}-2x\underbrace{\frac{d}{dx}1}_{=0}=0,$$ $$\Phi(x)=(1-x^2)\underbrace{\frac{d^2}{dx^2}x}_{=0}-2x\frac{d}{dx}x=-2x,$$ $$\Phi(x^2)=(1-x^2)\frac{d^2}{dx^2}x^2-2x\frac{d}{dx}x^2=2-6x^2.$$ So $\Phi$ is represented by the matrix $$A_{\Phi}=\begin{pmatrix}0&0&2\\0&-2&0\\0&0&-6\end{pmatrix}$$ and the characteristic matrix is $$\lambda I-A_{\Phi}=\begin{pmatrix}\lambda&0&-2\\0&\lambda+2&0\\0&0&\lambda+6\end{pmatrix}$$ and the characteristic polynomial $$\chi_{\Phi}(\lambda)=\lambda(\lambda+2)(\lambda+6).$$ For the eigenvalue $\lambda=0$ You already have the eigenvector $1$, for $\lambda=-2$ You already have the eigenvector $x$ and for the eigenvalue $\lambda=-6$ You can get from the singular matrix $$\begin{pmatrix}-6&0&-2\\0&-4&0\\0&0&0\end{pmatrix}$$ the eigenvector $1-3x^2$. These eigenvectors span the corresponding onedimensional eigenspaces.