We've been going over diagonalization in my linear algebra class, but we've only been dealing with matrices—nothing too complicated. All of a sudden this problem came along and blindsided me:
Let V be the vector space of continuous functions with basis {$e^t, e^{-t}$}. Let $L:V\rightarrow V$ be defined by $L(g(t)) = g'(t)$ for $g(t)$ in $V$. Show that $L$ is diagonalizable.
I have the feeling that this will somehow involve a matrix similar to \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} because of the basis given, but I don't really know what to do with the derivative or how it begins to fit in. Any ideas? Thanks in advance!