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Let $W$ be the vector space of functions generated by two functions $e^t,e^{2t}$. Then show $\left\{ e^t,e^{2t}\right\}$ is a basis of $W$.

Proof. Initially, we need to show that $e^t,e^{2t}$ are linearly independent, i.e., for $\alpha$ and $\beta$,

$\alpha e^{t}+\beta e^{2t}=0$

then $\alpha=\beta=0$. Assume not. Assume $\alpha\neq 0$. Then,$\alpha e^{t}+\beta e^{2t}=0$ and we obtain $1=\dfrac {-\beta e^{2t}} {\alpha e^{t}}$. Hece, if $\beta=0$ then $1=0$ for all $t$. Contrudiction.

Now, assume $\beta\neq 0$. This case similar. Therefore, $\alpha=\beta=0$ for all $t$. Then, $e^t,e^{2t}$ are linearly independet.

Now, we will show that $e^t,e^{2t}$ are generate of $W$. BUT, HOW? Can you help?

Also, can you check my proof of first statement (linearly indepent statement)?

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  • $\begingroup$ evaluate the WRONSKIAN of the basis en.wikipedia.org/wiki/Wronskian $\endgroup$ – Jose Garcia Mar 3 '17 at 9:52
  • $\begingroup$ What are $V$ and $W$? Without this information, we cannot say whether the required space is spanning or not, right? $\endgroup$ – астон вілла олоф мэллбэрг Mar 3 '17 at 9:56
  • $\begingroup$ @JoseGarcia The link is only about linearly independent. $\endgroup$ – pozcukushimatostreet Mar 3 '17 at 9:56
  • $\begingroup$ @астонвіллаолофмэллбэрг Yes. Edited. $\endgroup$ – pozcukushimatostreet Mar 3 '17 at 9:59
  • $\begingroup$ Ah, so you only need to show that it is linearly independent, right? (It already spans), but then there are clear mistakes in your proof. You have to assume that at least one of $\alpha,\beta$ is non-zero, and derive a contradiction. $\endgroup$ – астон вілла олоф мэллбэрг Mar 3 '17 at 9:59
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Now, we will show that $e^t,e^{2t}$ are generate of $W$. BUT, HOW? Can you help?

There is nothing to show because $W$ is given as the space generated by these two functions:

Let $W$ be the vector space of functions generated by two functions $e^t,e^{2t}$. Then show $\left\{ e^t,e^{2t}\right\}$ is a basis of $W$.

So if you are given that $W = \mbox{span}\left\{ e^t , e^{2t} \right\}$ and you need to check if $\left\{ e^t , e^{2t} \right\}$ is a basis, then you only need to show that the two functions are linearly independent.

Comments suggest using the Wronskian but an approach like yours works as well. You want the following to hold for all $t$: $$\alpha e^t + \beta e^{2t} = 0$$ But for $t=0$ this means $\alpha + \beta=0$ while for $t=1$ you get $\alpha e + \beta e^{2} = 0$; this system only has the zero solution for $\alpha$ and $\beta$.

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  • $\begingroup$ You are right. Thanks. If writer didn't say $W$ generated by two functions $e^t, e^{2t}$ then how can we show this generates $W$? $\endgroup$ – pozcukushimatostreet Mar 3 '17 at 10:56
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    $\begingroup$ But then $W$ should have been given in some other way; in that case, you should show that any element of $W$ can be written as a linear combination of $e^t$ and $e^{2t}$; i.e. that for an arbitrary $w \in W$, you can always find $\alpha$ and $\beta$ such that $w = \alpha e^t+\beta e^{2t}$. $\endgroup$ – StackTD Mar 3 '17 at 11:12
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The only thing to prove is that $e^t$ and $e^{2t}$ are linearly independent. The other part is already given in the primes.

However I have objections to your proof in the case $\beta\ne0$. It's not that clear how you come to the conclusion that $\alpha=\beta=0$. What you instead can do is that $\alpha e^t + \beta e^{2t} = 0$ must be true for both $t=0$ and $t=\ln 2$. That is:

$$\alpha e^0 + \beta e^{2\cdot0} = \alpha + \beta = 0$$ $$\alpha e^{\ln 2} + \beta e^{2\ln 2} = \alpha\cdot 2 + \beta \cdot 2^2 = 0$$

Then you see that this equation system only has the solution $\alpha=\beta=0$.

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The differentiation operator $(D-2)f=f'-2f$ annihilates $e^{2t}$. Therefore, if $$ \alpha e^{t}+\beta e^{2t} = 0, \;\; t\in\mathbb{R}, $$ you can apply $(D-2)$ to the above in order to obtain $$ (D-2)(\alpha e^{t}+\beta e^{2t}) = -\alpha e^{t} = 0,\;\; t\in\mathbb{R}. $$ Hence $\alpha=0$. And $\alpha=0$ then gives $\beta e^{2t}=0$ for all $t$, which implies that $\beta=0$ also holds. So $\{ e^t,e^{2t} \}$ is a linearly-independent set of functions on $\mathbb{R}$ (or on any finite interval of non-zero length.) You can argue the same for any number of different exponents $r_1,r_2,\cdots, r_n$ by applying an operator that annihilates all but one term; for example, $(D-r_2)(D-r_3)(\cdots)(D-r_n)$ annihilates everything but $e^{r_1 x}$. So $\{ e^{r_1 x}, e^{r_2 x},\cdots, e^{r_n x} \}$ is a linearly independent set of functions on an interval $I$ of non-zero length. The linear span of such functions then has this set of functions as a basis because (a) everything in the span may be written a linear combination of such functions and (b) these functions form a linearly independent set of functions.

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