Eigenvalues and eigenspaces of differential operator on finitely generated vector space

Question

Let $V$ be a vector space generated by functions $e^{\lambda_1x},\dots, e^{\lambda_nx}$, where $\lambda_1, \dots,\lambda_n$ are pairwise distinct numbers. Define the differential operator $$\dfrac{d^2}{dx^2}:V\to V.$$ Find the eigenvalues and eigenspaces of this operator.

My approach: I know how to show that $e^{\lambda_1x},\dots, e^{\lambda_nx}$ are linearly independent functions. Hence they will be a basis for $V$. It's easy to show that the matrix of this operator in this basis is the following: $$\begin{bmatrix} \lambda^2_1 & 0 & \cdots & 0 \\ 0 & \lambda^2_2 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda^2_n \\ \end{bmatrix}$$

Then it follows easily that eigenvalues are $\{\lambda^2_1,\dots, \lambda^2_n\}$.

Let's try to find eigenspace corresponding to $\lambda^2_k$, i.e. $V_{\lambda^2_k}:=\ker(\dfrac{d^2}{dx^2}-\lambda^2_k\cdot \text{id})$

If $f\in V_{\lambda^2_k}$ then $\dfrac{d^2f}{dx^2}-\lambda^2_kf=0$. Suppose $f=\beta_1e^{\lambda_1 x}+\dots_+\beta_ne^{\lambda_n x}$ and since those exponents are basis it means that: $\beta_j\lambda_j^2-\beta_j\lambda_k^2=0$ for $j=1,\dots,n$. It implies that $\beta_j(\lambda_j+\lambda_k)=0$. Then I stuck here.

But after some moment I've realized that $\dim V_{\lambda^2_k} \leq$ multiplicity of $\lambda^2_k$ which is 1. Then I have checked that $\langle e^{\lambda_k x}\rangle\subseteq V_{\lambda^2_k}$ which imples that $\dim V_{\lambda^2_k}=1$ and $V_{\lambda^2_k}=\langle e^{\lambda_k x}\rangle$.

I am sure that this is correct solution but one moments which confuses is that I was not able to deduce it from those above equations.

Can anyone show how to deduce it from those equations?

Answer 1

It follows from your equation $$\beta_j (\lambda_j + \lambda_k) = 0$$ that for each $j \neq k$ we have $\beta_j = 0$ or $\lambda_k = -\lambda_j$.

• If the former is true for all $j \neq k$, then we have computed that the eigenspace $V_{\lambda_k^2}$ is spanned by $e^{\lambda_k x}$.
• On the other hand, if the latter is true for some $j \neq k$ (by hypothesis it can only be true for one $j$), then we have computer that the eigenspace $V_{\lambda_k^2}$ is spanned by $e^{\lambda_k x}$ and $e^{\lambda_j x} = e^{-\lambda_k x}$.

In particular, this shows that the multiplicity of $\lambda_k^2$ (and the dimension of $V_{\lambda_k^2}$) need not be $1$, and indeed that it is not precisely when $\lambda_j = -\lambda_k$ for some $j \neq k$.