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Given the vector space $({\mathbb{R}}^{\mathbb{R}},+,\cdot)$, with ${\mathbb{R}}^{\mathbb{R}}$ defined as the set of all functions $f: \mathbb{R} \rightarrow \mathbb{R}$, we are asked to prove that $f_1(x) = e^x$ and $f_2(x) = e^{-x}$ are linearly independent.

We recall that $f_1(x)$ and $f_2(x)$ are linearly independent if, and only if, $\forall a_1,a_2 \in \mathbb{R}$ $\big(a_1f_1(x) + a_2f_2(x) = 0 \big) \Rightarrow \big( a_1 = a_2 =0 \big).$

Is it enough to do the following?

Proof: Consider the case $x=0$, then clearly we have that $a_1 = -a_2$. Then consider the more general case. We have:

$$a_1e^x - a_2e^{-x} = 0 \iff a_1e^x - a_1e^{-x} = 0$$ from our equality obtained above. This implies that $$a_1(e^x-e^{-x}) = 0,$$ which subsequently implies that either $a_1 = 0$ or $e^x = e^{-x}$.

Thus for all $x$, $a_1 = 0$, and from our earlier equality $a_1=-a_2=0$, which implies $f_1,f_2$ are linearly independent.

In the lecture notes I am referring to, the latter part of the argument wasn't done for general $x$, but rather it was done in particular for $x=1$, and the same conclusion was reached. I am struggling to see how this tells you anything about the functions for all $x$, but perhaps I am just being dense.

As an additional question, the lecture notes also discuss using derivatives to prove linear independence of functions, but this has not been justified anywhere in the notes thus far. Why is this a reasonable thing to do?

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    $\begingroup$ So long as there is one value of $x$ for which the linear combination is not zero, the linear combination of the two functions is not identically zero, whence the functions are linearly independent. To put it the other way, for the two functions to be linearly dependent, there must exist coefficients that work for every value of $x$. $\endgroup$ – Gerry Myerson Feb 18 '19 at 23:27
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This is basically correct, but I would start by supposing the hypothesis: "Suppose $a_1, a_2 \in \mathbb{R}$ so that $a_1e^x + a_2 e^{-x} = 0$ for all $x$."

Then, you come to the conclusion that $$a_1(e^x - e^{-x}) = 0.$$ But remember the definition of $0$ in the space of real-valued functions: it is actually the function that maps every $x \in \mathbb{R}$ to $0$. To prove two functions are equal is to prove they are equal on every input. So there is really a hidden quantifier in this statement. You have found that

$$a_1(e^x - e^{-x}) = 0 \text{ for all $x$. }$$

Thus either $a_1 = 0$, or else $e^x = e^{-x}$ for all $x$. You need to show that the latter is not possible. This might be regarded as obvious, but in general, to disprove a universally quantified statement (a "for all x" statement) you need to furnish a counterexample. Thus, you need to prove there exists an $x$ so that $e^x \neq e^{-x}$. Choosing $x=1$, for example, does the job.

I can't really answer your second question without knowing what the course is. But if it is not a course in calculus/analysis, where you go to the trouble of defining derivatives, it is not unreasonable to assume that all this work has been done and that the basic facts of calculus are known.

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  • $\begingroup$ This answered my question thank you. But RE the derivative question: I don’t understand why the derivative tells us anything about the linear independence of the original function. $\endgroup$ – Benjamin Feb 19 '19 at 8:50
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    $\begingroup$ @Benjamin I can't help without details about how the derivative is being used in that particular case. It is just one of the tools available to you. $\endgroup$ – Jair Taylor Feb 19 '19 at 16:15

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