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Given $ \alpha_1 > \alpha_2 > \cdots > \alpha_n \geq 0$, $f_1(x)= e^{-\alpha_1x},..., f_n(x)=e^{-\alpha_nx}$, prove that they are linearly independent . Hint: don’t forget the limit $ x \rightarrow \infty $.

I have already seen proofs about similar questions. What you have to do is to set the combination of those functions (or polynomials) equal to 0 and show that all coefficients have to be 0. But here, the coefficients are in the exponent, they can’t be equal to each other, and the negative exponent disturbs me a bit.

Thanks for your help.

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    $\begingroup$ Use Vandermonde determinants. $\endgroup$ – kimchi lover Jan 1 '18 at 14:42
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As usual, we will consider $$\beta_1 e^{-\alpha_1 x}+\beta_2 e^{-\alpha_2 x}=0$$

Dividing by $e^{-\alpha_2 x}$ gives $$\beta_1 e^{-(\alpha_1-\alpha_2)x}+\beta_2=0$$ Letting $x\to\infty$ we see that $e^{-(\alpha_1-\alpha_2)x}\to 0$ since $\alpha_1>\alpha_2\geqslant 0$. Therefore $\beta_2=0$. Since $e^{-\alpha_1x}>0$ for all $x$, we also have $\beta_1=0$. Therefore $\{e^{-\alpha_1 x}, e^{-\alpha_2 x}\}$ is linearly independent.

I am sure that you can prove it now for the more general case.

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  • $\begingroup$ Thanks a lot ! Just to be sure, why aren’t the alphas (in the exponents) the coefficients that have to be 0 ? $\endgroup$ – Poujh Jan 1 '18 at 14:48
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    $\begingroup$ Linear independence for a set of vectors $\{v_1,v_2\}$ means exactly: $\beta_1 v_1+\beta_2 v_2=0 \implies \beta_1=\beta_2=0$. Therefore, the coefficients in the exponents have nothing to do with the linear independence. $\endgroup$ – rae306 Jan 1 '18 at 14:55
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You can use the Vandermonde determinant.

Let $$\beta_1e^{-\alpha_1x} + \cdots + \beta_ne^{-\alpha_1x} = 0$$

Plugging in $x = 0, 1, \ldots, n-1$ yields:

$$\beta_1 + \cdots + \beta_n = 0$$ $$\beta_1e^{-\alpha_1} + \cdots + \beta_ne^{-\alpha_n} = 0$$ $$\beta_1e^{-2\alpha_1} + \cdots + \beta_ne^{-2\alpha_n} = 0$$ $$\vdots$$ $$\beta_1e^{-(n-1)\alpha_1} + \cdots + \beta_ne^{-(n-1)\alpha_n} = 0$$ The determinant of this linear system is the Vandermonde determinant:

$$\begin{vmatrix} 1 & 1 & \cdots & 1\\ e^{-\alpha_1} & e^{-\alpha_2} & \cdots & e^{-\alpha_n}\\ e^{-2\alpha_1} & e^{-2\alpha_2} & \cdots & e^{-2\alpha_n}\\ \vdots & \vdots & \ddots & \vdots\\ e^{-(n-1)\alpha_1} & e^{-(n-1)\alpha_2} & \cdots & e^{-(n-1)\alpha_n}\\ \end{vmatrix} = \prod_{1 \le i < j \le n} (e^{-\alpha_j} - e^{-\alpha_i}) \ne 0$$

because $e^{-\alpha_i} \ne e^{-\alpha_j}$ for all $i \ne j$. Hence, the system has the unique solution $\beta_1= \beta_2 = \cdots = \beta_n = 0$, which implies linear independence.


Another solution using Vandermonde.

Assume

$$\beta_1e^{-\alpha_1x} + \cdots + \beta_ne^{-\alpha_1x} = 0$$

Taking the derivative $n-1$ times yields:

$$\beta_1e^{-\alpha_1x} + \cdots + \beta_ne^{-\alpha_nx} = 0$$ $$-\alpha_1\beta_1e^{-\alpha_1x} - \cdots - \alpha_n\beta_ne^{-\alpha_nx} = 0$$ $$\alpha_1^2\beta_1e^{-\alpha_1x} + \cdots + \alpha_n^2\beta_ne^{-\alpha_nx} = 0$$ $$\vdots$$ $$\alpha_1^{n-1}(-1)^{n-1}\beta_1e^{-\alpha_1x} + \cdots - \alpha_n^{n-1}(-1)^{n-1}\beta_ne^{-\alpha_nx} = 0$$

The determinant of this linear system is again the Vandermonde determinant:

$$\begin{vmatrix} e^{-\alpha_1x} & e^{-\alpha_2x} & \cdots & e^{-\alpha_nx}\\ (-\alpha_1)e^{-\alpha_1x} & (-\alpha_2)e^{-\alpha_2x} & \cdots & (-\alpha_n)e^{-\alpha_nx}\\ (-\alpha_1)^2e^{-\alpha_1x} & (-\alpha_2)^2e^{-\alpha_2x} & \cdots & (-\alpha_n)e^{-\alpha_nx}\\ \vdots & \vdots & \ddots & \vdots\\ (-\alpha_1)^{n-1}e^{-\alpha_1x} & (-\alpha_2)^{n-1}e^{-\alpha_2x} & \cdots & (-\alpha_n)^{n-1}e^{-\alpha_nx}\\ \end{vmatrix}$$ $$ = e^{-\alpha_1x}\cdots e^{-\alpha_nx} \begin{vmatrix} 1 & 1 & \cdots & 1\\ -\alpha_1 & -\alpha_2 & \cdots & -\alpha_n\\ (-\alpha_1)^2 & (-\alpha_2)^2 & \cdots & (-\alpha_n)^2\\ \vdots & \vdots & \ddots & \vdots\\ (-\alpha_1)^{n-1} & (-\alpha_2)^{n-1} & \cdots & (-\alpha_n)^{n-1}\\ \end{vmatrix} =e^{-\alpha_1x}\cdots e^{-\alpha_nx} \prod_{1 \le i < j \le n} (\alpha_i - \alpha_j) \ne 0$$

because $\alpha_i \ne \alpha_j$ for all $i \ne j$. Hence, the system has the unique solution $\beta_1= \beta_2 = \cdots = \beta_n = 0$, which implies linear independence.

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  • $\begingroup$ Wow this is nice! (+1) $\endgroup$ – rae306 Jan 1 '18 at 15:59
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Hint. If they are linearly dependent then there exist $c_1,\dots,c_n$ not all zero such that for $x\in \mathbb{R}$, $$c_1e^{-\alpha_1 x}+\dots +c_ne^{-\alpha_n x}=0.$$ We may assume that $0\not=c_i=c_{i+1}=\dots =c_n=0$, then $$c_1e^{-\alpha_1 x}+\dots+c_{i-1}e^{-\alpha_{i-1} x}+c_ie^{-\alpha_{i} x}=0,$$ and after multiplying both sides by $e^{\alpha_{i} x}$ we get $$c_1e^{(\alpha_i-\alpha_1) x}+\dots+c_{i-1}e^{(\alpha_i-\alpha_{i-1}) x}+c_i=0.$$ Now note that $\alpha_i-\alpha_k<0$ for $k=1,\dots, i-1$, and take the limit as $x\to +\infty$. What may we conclude about $c_i$?

P.S. Note that the hint works also without the condition $\alpha_n\geq 0$.

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Actually, the proof given by mechanodroid allows us to prove a more general theorem :

Eigenvectors associated with different eigenvalues of a linear operator are linearly independent.

Let's assume it for now, and note $D$ the linear function $D : f \rightarrow f'$ that returns the derivative of a function.

Now each $x\rightarrow \exp(\alpha_ix)$ is an eigenvector of $D$ (with eigenvalue $\alpha_i$) and are therefore independant.

Another consequence could be to prove that $x\rightarrow x^{\alpha_i}$ are independant as eigenvectors of $D_1:f\rightarrow(x\rightarrow x f'(x))$.


Now the proof itself consists in taking $n$ eigenvectors of $D$. Assume a linear combination of $n$ eigenvectors of $D$ is $0$. Calling $\alpha_i,f_i$ a pair eigenvalue, eigenvector, we have:

$a_1 f_1 + ... + a_n f_n = 0$

Applying $D$ once :

$\alpha_1 a_1 f_1 + ... + \alpha_n a_n f_n = 0$

Iterating, for all $i$ :

$\alpha_1^i a_1 f_1 + ... + \alpha_n^i a_n f_n = 0$

We have the following system in matrix form :

$$\left(\begin{array}{ccc} \alpha_{1} & \ldots & \alpha_{n}\\ \vdots & & \vdots\\ \alpha_{1}^{n} & \ldots & \alpha_{n}^{n} \end{array}\right)\left(\begin{array}{c} a_{1}f_{1}\\ \\ a_{n}f_{n} \end{array}\right)=\left(\begin{array}{c} 0\\ \\ 0 \end{array}\right)$$

Now the key is to prove that the matrix is invertible, so that $a_i f_i=0$ for every $i$. But its determinant is:

$$\alpha_1 ... \alpha_n \prod_{1 \le i < j \le n} (\alpha_i - \alpha_j) $$

Which is non zero if the $\alpha$'s are different, and non zero.

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What you have to do is to set the combination of those functions (or polynomials) equal to 0 and show that all coefficients have to be 0.

You can take the Taylor Series and do the same thing. Another tactic is to note that they're orthogonal when restricted to the imaginary axis.

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Note that the Vandermonde proofs, though longer, also works when the $\alpha$'s are complex not necessarily real numbers.

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