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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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Subspace: U is a subspace of V if U is a subset of V and: 1) U is nonempty 2) U is closed under addition 3) U is closed under scalar multiplication – Casey Patton Jan 14 '12 at 20:36
You need two things. Firstly, that a one-dimensional vector space has no subspaces except {0} and itself. Secondly, that your V is a one-dimensional vector space. Can you do either of these? – m_t_ Jan 14 '12 at 20:47
While it may seem 'obvious' a subspace of (some finite dimensional space) V has to have dimension less than or equal to V, it is actually not an entirely trivial proof - while simple it is rather tedious. For instance, you actually have to prove any subspace does have a basis (via explicit construction) then prove that a set of linearly independent vectors can not have greater cardinilty than a set of spanning vectors. So, for this example, it's best to spot V = <exp(kt)> and say if x =/= 0 is in U then ... – Adam Jan 14 '12 at 20:55
By the way, we need to assume that the constant $k$ is the same for each element of $V$, otherwise $V$ is not a vector space. It is misleading, I think $k$ should be declared outside the set. – Mikko Korhonen Jan 14 '12 at 20:59
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 $e^{2x}$. – Mikko Korhonen Jan 14 '12 at 21:06
up vote 4 down vote accepted

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.

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