# Invariant subspace of differentiation operator

Let $$D \in \mathcal{L}(\mathcal{P}(\mathbb{R}))$$ be the differentiation operator, and let $$U$$ be an invariant subspace of $$D$$. Suppose there exists a $$p \in U$$ with deg $$p = k$$.

a) Show that $$\mathcal{P}_k(\mathbb{R}) \subseteq U$$.

b) Evidently $$\{0\}$$, $$\mathcal{P}(\mathbb{R})$$, and $$\mathcal{P}_k(\mathbb{R})$$ are invariant subspaces of $$D$$ if $$k$$ is a nonnegative integer. Show that there are no others.

For a) I was thinking of showing that all elements of $$\mathcal{P}_k(\mathbb{R})$$ can be written as a basis of $$U$$, but I'm not sure how to show this generally. For b) I was thinking of trying to do a proof by contradiction, but I'm not sure where to get started (also I don't know if this is the best approach).

Hint. As $$U$$ is invariant under $$D$$, $$Dp = p'$$ belongs to $$U$$ and $$\deg p' = k-1$$.
• The first thing that came to mind: suppose $U$ is an invariant subspace, and that $U \neq \{0\}$. Then the set $A = \{k \in \mathbb N : \exists p \in U ( k = \deg p)\}$ is non-empty, and then try to prove that $U = \mathcal P_M(\mathbb R)$, where $M = \max A$. Commented Aug 13, 2020 at 21:13
• I'll try that out for (b). I'm using your hint for (a), and I was thinking that since deg$p' = k - 1$ and there is a $p$ with deg$k$, then any polynomial of degree $k$ exists in the span of $U$. Would that be a valid proof? Commented Aug 13, 2020 at 21:15
• If $p \in U$ and $\deg p = k$, it is not necessarily true that any polynomial of degree $k$ is in $U$. In fact you can only guarantee that any scalar multiple of $p$ is in $U$, since it is a subspace. I think the following can help you: if $n$ is a positive integer and there are polynomials $p_0,p_1,\dots,p_n$ such that $\deg p_i = i$, then $p_0,p_1,\dots,p_n$ is a basis for $\mathcal P_n(\mathbb R)$. Commented Aug 13, 2020 at 22:36
• For your response regarding (b) I understand why I should show that $U=\mathcal{P}_M(\mathbb{R})$ but I am stuck on how to actually do that. Commented Aug 13, 2020 at 22:58