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Here's the statement:

The following set, $V$, only has subspaces $\{0\}$ and $V$. V=\{f(t) \colon \mathbb R \to \mathbb R \mid f'(t) = k\cdot f(t) \text{ where } k \text{ is a constant}\}

I'm having trouble understanding why there are no other subspaces. Why is this the case here? Examples are welcome. I provided a definition of "subspace" in the comments.

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    For example, $f(x) = e^x + e^{2x}$ does not satisfy $f'(x) = k \cdot f(t)$ for any real number $k$, although the equation holds for $e^x$ and $e^{2x}$.2012-01-14

2 Answers 2

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Hint: $V$ is the set $\{ C e^{kt} :C\in \Bbb R\}$. So $V$ is the linear span of the single element $f(t)=e^{kt}$.

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HINT $\rm\displaystyle\ \begin{align} f{\:'} &=\ \rm k\ f \\ \rm \:\ g' &=\ \rm k\ g \end{align}\ \Rightarrow\ \dfrac{f{\:'}}f\: =\: \dfrac{g'}g\: \iff\: \bigg(\!\!\dfrac{g}f\bigg)' =\ 0\ \iff \ g\: =\: c\ f,\ \ \ c'\: =\ 0,\ $ i.e. $\rm\ c\:$ "constant".

This is a special case of the the Wronskian test for linear dependence.