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Are the set of functions $(1,e^{x},e^{-x})$ linearly independent?

I wrote it as an augmented matrix but it brought me to nowhere. Can somebody help me?

Kerem
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    evaluate the equation $a+be^x+ce^{-x}=0$ for three different $x$ and show that then $a=b=c=0$. – Surb Dec 25 '14 at 20:06
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    Start with a putative linear dependency relation $c_1+c_2e^x+c_3e^{-x}=0$. Remember that equality in the space of functions means that this holds for all $x$. Your goal is to prove that $c_1=c_2=c_3=0$. You can get helpful equations in many ways. Plug in some values for $x$. Differentiate. Repeat. You will get homogeneous equations. It should not be too hard to get enough of them to conclude that the constants must all be zero. – Jyrki Lahtonen Dec 25 '14 at 20:10
  • @Kerem: See this technique. – Mhenni Benghorbal Dec 26 '14 at 02:54

4 Answers4

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Yes, they are independent. Fix any $a,b,c\in\mathbb{R}$ such that $$\forall x\in\mathbb{R},\qquad a+be^x + ce^{-x} = 0\ . $$ Now:

  • Taking $x=0$ in the above, $a+b+c=0$, or equivalently $a=-(b+c)$.
  • By differentiating, you get that for all $x$, $be^x - ce^{-x}=0$. Applying to $x=0$, this gives $b=c$.

Summing up, right now we get $$ \forall x\in\mathbb{R},\qquad b(e^x + e^{-x}) = 2b\ . $$ But as if $b\neq 0$ the LHS would go to $\pm\infty$ when $x\to\infty$ (while the RHS is constant), we must have $b=0$. Thus, $a=0$ and $c=0$ as well.

Clement C.
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Hint: show that the equation $$a+be^x+ce^{-x}=0,\qquad \forall x \in \Bbb R$$ implies $a=b=c=0$ by evaluating at $x = 0,1,2$ for example.

Surb
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In general, eigenvectors of distinct eigenvalues are linearly independent.

For the case at hand, the operator is differentiation.

lhf
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I believe this question has missing background information: it doesn't specify the vector space.

Let thus the vector space be $\mathcal{F}(\mathbb{R})$, the space of all real-valued functions. In this case, 1 is the identity function. One way to show that $1,e^x,e^{-x}$ are linearly independent is the following.

Let $c_1,c_2,c_3\in\mathbb{R}$, and assume that \begin{equation} c_1 + c_2 e^{x} + c_3 e^{-x} = 0 \,\,\,\,\,(*) \end{equation} for all $x\in\mathbb{R}.$ Differentiate both sides of $(*)$ to get the equation \begin{equation} c_2 e^x + c_3^\prime e^{-x}=0, \end{equation} where $|c_3^\prime| = |c_3|$. If $c_2,c_3^\prime$ were to be different from zero we would get a contradiction; check, for instance, what happens when $x = 0$ and then when $x=1$. So we must have $c_2=c_3=0$ and thus $c_1=0$ as well.

utobi
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