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Show that $e^x$ and $e^{-x}$ are linearly independent in C$(-\infty,\infty).$

In order to solve this one must use the Wronskian of $f_1,f_2..f_n$

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Using this we show

$$W[e^x,e^{-x}] = \begin{vmatrix}e^x & e^{-x} \\e^x& -e^{-x} \end{vmatrix} = -2$$

Can anyone explain why this matrix is equal to $-2$?

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    That just follows from how to compute determinants. In your case, $W[e^x,e^{-x}]=e^{x}\cdot -e^{-x}-e^{x}\cdot e^{-x}=-1-1=-2$. Since $W[e^{x},e^{-x}]\neq 0$, this gives you the linear independence of the two functions that you seek. – 2017-01-17
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    The determinant $\left|\begin{array}{rr}a&b\\c&d\end{array}\right|=ad-bc$ – 2017-01-17

2 Answers 2

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First of all it is not the matrix equal -2, but its determinant.

$$\begin{vmatrix}e^x & e^{-x} \\e^x& -e^{-x} \end{vmatrix} = e^x(-e^{-x}) - e^{-x}e^x = -e^{x-x}-e^{x-x} = -e^0-e^0 = -1-1=-2$$

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The elementary solution is also possible in this simple case of two exponential functions. If $e^x$ and $e^{-x}$ were linearly dependent, then they were proportional: $$e^x=\alpha e^{-x}$$ for some $\alpha\ne 0$ and for all $x\in\Bbb R$. For $x=0$ we get $\alpha=1$ and $e^x=e^{-x}$ for all $x$, which is the nonsense.

This argument works also for $e^{ax}$ and $e^{bx}$ with $a\ne b$.