Finding eigenvalue based on vector space and linear operator?

Question

So given that $L$ is a linear operator from a vector space $V$ to itself, with $V=\mathbb R_2[t]$ and $Lf = f'' - f$ I need to show that $\lambda = -1$ is the only eigenvalue.

I know how to do this with standard matrices but it has been a couple of years since I've taken differential equations so I'm very rusty and unsure of how to convert this information to a matrix I can use the standard $det(A - \lambda I)$ equation on. Any tips?

Answer 1

Basis of $V$ is $\{1,t,t^2\}$

Write the matrix with respect to this basis

$L(1)=0-1=-1\cdot1+0\cdot t+0\cdot t^2$

$L(t)=0-t=0\cdot 1-1\cdot t+0\cdot t^2$

$L(t^2)=2-t^2=2\cdot 1+0\cdot t-1\cdot t^2$

So matrix is $$\begin{bmatrix}-1 & 0&2\\ 0&-1&0\\0&0&-1\end{bmatrix}$$

Now you can clearly see $\lambda =-1$ is the only eigen value!

Answer 2

The easiest approach, I think, is to note that $$ [L + \operatorname{id}]^2f = 0 $$ for all polynomials $f \in \Bbb R_2[t]$. So, if $f$ is an eigenvector associated with eigenvalue $\lambda$, then $[L + \operatorname{id}]^2f = (\lambda + 1)^2f = 0$