Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
3  
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? –  mt_ 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
    
hint: x is of the form Aexp(kt) for non-zero A; why is A non-zero? If A is non-zero what does it have? So why is exp(kt) in U? Does this mean Cexp(kt) is in U for any C? –  Adam Jan 14 '12 at 20:57
1  
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

2 Answers 2

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 $f(t)=e^{kt}$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.