# Problem of the Week #30 - October 22nd, 2012

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#### Chris L T521

##### Well-known member
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Thanks to those who participated in last week's POTW!! Here's this week's problem!

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Problem: Let $V$ be the vector space of continuous functions with basis $\{e^t,e^{-t}\}$. Let $L:V\rightarrow V$ be defined by $L(g(t)) = g^{\prime}(t)$ for $g(t)\in V$. Show that $L$ is diagonalizable.

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#### Chris L T521

##### Well-known member
Staff member
This question was correctly answered by Siron and Sudharaka. You can Sudharaka's solution below:

Take any $$g(t)\in V$$. Then,

$g(t)=\alpha e^{t}+\beta e^{-t}$

$\Rightarrow g'(t)=\alpha e^{t}-\beta e^{-t}$

Hence,

$T\left(\begin{matrix} \alpha \\ \beta \end{matrix}\right)\rightarrow\left(\begin{matrix} \alpha \\ -\beta \end{matrix}\right)$

Note that for each $$v=\left(\begin{matrix} \alpha \\ \beta \end{matrix}\right)\in V$$

$\left(\begin{matrix} 1&0 \\ 0&-1 \end{matrix}\right)\left(\begin{matrix} \alpha \\ \beta \end{matrix}\right)=\left(\begin{matrix} \alpha \\ -\beta \end{matrix}\right)$

The linear transformation $$T$$ can be represented by a diagonal matrix with respect to the basis $$\{e^t,\,e^{-t}\}$$. Hence $$T$$ is diagonalizable.

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