= Solution
If $f\in V_n\setminus V_{n-1}$, write $f(x)=a_nx^n+\cdots$ with $a_n\ne0$. In the expression $Tf=f''-2xf'-2f$, the first term has <polynomial degree> at most $n-2$, while the other two contribute
$$
-2na_nx^n-2a_nx^n=-2(n+1)a_nx^n.
$$
Thus, if $Tf=\lambda f$, comparison of the leading coefficients gives
$$
\boxed{\lambda=-2(n+1).}
$$
For $n=0$ take $V_{-1}=\{0\}$; the same result follows from $T1=-2$. This identifies the <eigenvalue> conditional on the existence of the <eigenvector>; existence is established next.
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