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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!

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Well, consider the what this linear transformation does to the basis given:

$L(e^t) = \frac{d}{dt}e^t = e^t$

and similarly

$L(e^{-t}) = \frac{d}{dt}e^{-t} = -e^{-t}$

In other words, taking coordinates from the ordered basis $\beta = \{e^{t}, e^{-t}\}$ we have

$L(1,0) = (1,0)$ and $L(0,1) = -(0,1)$

Hence, by definition, the matrix of $L$ with respect to the ordered basis $\beta$ is given by

$[L]_\beta = \begin{pmatrix}1 & 0\\ 0 & -1\\ \end{pmatrix}$

which is diagonal.

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    @KyleR. No problem. Notice by the way that the matrix you first suggested is the correct matrix for the ordered basis $\{e^t, -e^{-t}\}$. So in the sense that a linear transformation, unlike its matrix, is independent of the basis, that was also a correct answer. However, whoever assigned the problem probably meant you should use the basis given. – 2012-07-02