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Show that the set consisting of the functions $$x, e^x, e^{-x}$$ on $\mathbb R$ is linearly independent.

So I have the equation $$ax + be^x + ce^{-x} = 0$$ and I want to show that this is only satisfied when $a = b = c = 0$

Letting x = 0, $b + c = 0$

Letting x = 1, $a + be + ce^{-1} = 0$

Letting x = -1, $-a + be^{-1} + ce = 0$

Using these equations as columns of a matrix I have

$\begin{bmatrix} 0 & 1 & 1 \\ 1 & e & e^{-1} \\ -1 & e^{-1} & e \end{bmatrix}$

$\begin{bmatrix} 1 & e & e^{-1} \\ 0 & 1 & 1 \\ -1 & e^{-1} & e \end{bmatrix}$

$\begin{bmatrix} 1 & e & e^{-1} \\ 0 & 1 & 1 \\ 0 & e^{-1}+e & e^{-1}+e \end{bmatrix}$

But if I now subtract (e^{-1}+e) times row 2 from row 3 I will get all zeros in the last row meaning linear dependence. So what have I done wrong?

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Write c=-b in your second and third equations and compare them. You've just used the wrong three values of $x$. –  Mark Bennet Apr 28 '12 at 20:41

5 Answers 5

up vote 2 down vote accepted

You have shown that the vectors of values of these three functions at these particular three points are linearly dependent. That does not show that the functions are linearly dependent. You will need a different choice of points.

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By choice of points I mean $x = -1, 0, 1$. You got really unlucky. A "generic" choice of three points should work. Try $0, 1, 2$. –  user29743 Apr 28 '12 at 20:39
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Alternative proof: differentiate your relationship twice. You get that $a$ is 0 for free and you're reduced to showing that $e^x$ and $e^{-x}$ are linearly independent, which is easy because you can evaluate at 0 and at 1. –  user29743 Apr 28 '12 at 20:40
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And if we are willing to also differentiate $3$ times, we don't have to worry about evaluating at $1$. –  André Nicolas Apr 28 '12 at 21:04

Several people have posted rather general methods for attacking this problem. Here is an alternative approach that is rather specific to this case. Suppose that $a$, $b$, and $c$ are numbers and that $$ ax + b e^x + c e^{-x} = 0, \qquad x \in \mathbb{R}. $$ Divide both sides of this by $e^x$ and take the limit as $x \to \infty$ to deduce $b = 0$. Once you know that, dividing both sides of the above by $x$ and taking the limit as $x \to \infty$, you deduce $a = 0$. And $c=0$ clearly follows from there.

The same basic idea can be used slightly more generally, to prove linear independence of functions of different "growth rates" without solving systems of equations.

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This is most naturally attacked using the Wronskian, $$W(f_1,\ldots,f_n)(x) = \det\left( \begin{array}{ccc} f_1(x) & \cdots & f_n(x) \\ f_1'(x) & \cdots & f'_n(x) \\ \vdots & \vdots & \vdots \\ f_1^{(n-1)}(x) & \cdots & f_n^{(n-1)}(x) \end{array}\right),$$ where $f_k^{(m)}(x)$ is the $m$th derivative of $f_k(x)$. If the Wronskian does not vanish, the functions $f_k(x)$ are linearly independent.

We find $$W(x,e^x,e^{-x}) = \det\left( \begin{array}{ccc} x & e^x & e^{-x} \\ 1 & e^x & -e^{-x} \\ 0 & e^x & e^{-x} \end{array}\right) = 2x.$$ Therefore, the functions are linearly independent on $\mathbb{C}\setminus\{0\}$. (And therefore on $\mathbb{R}\setminus\{0\}$.)

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You got that for those three values of x there are infinite solutions a,b,c, but it does not mean that those solutions verify the equation for all x. If you had that for three values of x only the trivial solution, it verifies for all x too then it is the unique solution for all x.

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You just need to pick different values. For $x=0,1,2$ you get: $$\begin{bmatrix} 0 & 1 & 1 \\ 1 & e & e^{-1} \\ 2 & e^2 & e^{-2} \end{bmatrix}$$

This row reduces to: $$\begin{bmatrix} 0 & 1 & 1 \\ 1 & e & e^{-1} \\ 0 & e(e-2) & e^{-2}(1-2e) \end{bmatrix}$$

Now take the determinant: you get $e(e-2)-e^{-2}(1-2e)\neq0$ because $e$ is transcendental.

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