= Solution
A <linear operator> $\alpha$ on a real inner-product space is <self-adjoint> when
$$
\langle\alpha v,w\rangle=\langle v,\alpha w\rangle
$$
for all $v,w$.
To prove the <finite-dimensional spectral theorem>, first note that the continuous <quadratic function> $\langle\alpha v,v\rangle$ has a maximum on the unit sphere. The <Lagrange multiplier> equation at a maximizing <vector> $v$ gives $\alpha v=\lambda v$, so $\alpha$ has a real unit <eigenvector>. Its orthogonal complement is invariant because
$$
\langle\alpha w,v\rangle=\langle w,\alpha v\rangle
=\lambda\langle w,v\rangle=0.
$$
Induction on the dimension supplies an orthonormal eigenbasis of that complement and hence of $V$.
For $f\ne0$,
$$
\langle f,f\rangle=\int_0^\infty f(x)^2e^{-x}\,dx>0,
$$
while symmetry and bilinearity are immediate, so the displayed formula defines an <inner product> on $P_n$. Since
$$
e^{-x}\alpha(f)
=\bigl(xe^{-x}f'(x))',
$$
<integration by parts> gives
$$
\begin{aligned}
\langle\alpha f,g\rangle
&=\left[gxe^{-x}f'\right]_0^\infty
-\int_0^\infty x e^{-x}f'g'\,dx\\
&=\langle f,\alpha g\rangle.
\end{aligned}
$$
The boundary term vanishes for <polynomials>, proving that $\alpha$ is self-adjoint.
On the monomial $x^k$,
$$
\alpha(x^k)=-kx^k+k^2x^{k-1}.
$$
Thus the <matrix> of $\alpha$ in the monomial <basis> is triangular with diagonal
$$
0,-1,-2,\ldots,-n.
$$
These are therefore its <eigenvalues>. For $n=2$, corresponding <eigenvectors> are
$$
\boxed{
1,\qquad x-1,\qquad x^2-4x+2
}
$$
for <eigenvalues> $0,-1,-2$, respectively. They are the first <Laguerre polynomials> up to normalization, as described by the <Laguerre differential operator on polynomials>.
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