MHB Q: What is the maximum dimension of [D - λ]^-1(W) for a subspace W of V?

[Chris L T521]

Here's this week's problem!

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Problem: It is a basic result in analysis that for any $f\in V$, and any $\lambda\in\mathbb{C}$, $Df=\lambda\cdot f$ if and only if $f(x)=C\cdot e^{\lambda x}$ for some $C\in\mathbb{C}$. Now assume that $W$ is a subspace of $V$. Define $$[D-\lambda]^{-1}(W)=\{f\in V:Df-\lambda f\in W\}.$$

Prove that $\dim([D-\lambda]^{-1}W)\leq\dim(W)+1$.

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[Chris L T521]

This week's problem was correctly answered by Deveno. You can find his solution below.

[sp]Let $U = [D - \lambda]^{-1}(W)$ and let $T$ be the restriction of $D-\lambda$ to $U$.

Then we have:

$\dim(U) = \dim(\text{ker }T) + \dim(\text{im }T)$

But $\text{im }T = T(U) = (D-\lambda)(U) = (D-\lambda)([D - \lambda]^{-1}(W)) = W$.

Moreover, we have that $\text{ker }T \subseteq \text{ker }(D-\lambda)$, which has dimension 1, since it is spanned by $\{e^{\lambda x}\}$.

Since $\text{ker }T$ is clearly a subspace of $\text{ker }(D - \lambda)$ (namely: $U \cap \text{ker }(D - \lambda)), \dim(\text{ker }T) \leq 1$.

Thus:

$\dim([D - \lambda]^{-1}(W)) = \dim(U) = \dim(\text{ker }T) + \dim(\text{im }T) = \dim(\text{ker }T) + \dim(W) \leq 1 + \dim(W)$[/sp]