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For any $n$ distinct points $x_1,x_2 , \ldots , x_n$ on the real line show that the matrix $M$ where $M(i,j) = e^{\lambda_j x_i} $ has non-zero determinant where $\lambda_1 \lt \lambda_2 \lt \ldots \lt \lambda_n \in \mathbb{R}$ are fixed constants.

I tried to prove this by induction and I am able to show this for $n=1$(duh...) and $n=2$. but do not know how to proceed. Any hints?

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cross posted mathoverflow.net/questions/118225 –  Willie Wong Jan 8 '13 at 8:40
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1 Answer

Suppose the determinant is zero, then one of its rows is linearly dependent on the preceeding ones, say:

$$\left(e^{\lambda_1x_i}\,,\,e^{\lambda_2x_i}\,,\ldots,e^{\lambda_nx_i}\right)=\sum_{k=1}^{i-1}a_k\left(e^{\lambda_1x_k}\,,\,e^{\lambda_2x_k}\,,\ldots,e^{\lambda_nx_k}\right)\,\,,\,\,a_k\in\Bbb R$$

From the above, we get for each $\,1\leq t\leq n\,$:

$$\sum_{k=1}^{i-1}a_ke^{\lambda_tx_k}=e^{\lambda_tx_i}\Longrightarrow\Phi(\lambda_t):=\sum_{k=1}^{i-1}a_ke^{\lambda_tx_k}-e^{\lambda_tx_i}=0\Longrightarrow$$

$$ \Phi'(\lambda_t)=\sum_{k=1}^{i-1}a_kx_ke^{\lambda_tx_k}-x_ie^{\lambda_tx_i}=0$$

But then

$$\sum_{k=1}^{i-1}a_kx_ke^{\lambda_tx_k}=x_ie^{\lambda_tx_i}=x_i\sum_{k=1}^{i-1}a_ke^{\lambda_tx_k}\Longrightarrow$$

$$\sum_{k=1}^{i-1}a_k(x_k-x_i)e^{\lambda_tx_k}=0\Longrightarrow a_1=a_2=...=a_{i-1}=0$$

since $\,(x_k-x_i)e^{\lambda_tx_k}\neq 0\,\,\,,\,\forall\,\,1\leq k\leq i-1$ ...

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thank you for the reply. I don't get the last implication. how does one conclude the $a_i$'s are zero? I mean you can have non-zero numbers adding up to zero right? –  smilingbuddha Jan 6 '13 at 23:16
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If you are permuting the rows of the matrix such that $x_i$'s are in increasing order then, i think the last statement would make sense, since $(x_k - x_i)$ will all be of the same sign for $1 \leq k \leq i-1 $ which would imply that all the $a_i$'s are zero. –  smilingbuddha Jan 6 '13 at 23:19
    
@smilingbuddha, way to go: you caught that slip. Indeed, as the $\,x_i'$s are different we can, from the beginning, permute the matrix rows as tto get those differences positive. After all, the determinant's is the same up to sign. Thanks. –  DonAntonio Jan 6 '13 at 23:55
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Todd Trimble points out on MO that you don't justify the claim that $\Phi(\lambda_t)=0\implies \Phi'(\lambda_t)=0$. So far as I see, you only have $\Phi(\lambda_t)=0$ for a particular value of $\lambda_t$, not that $\Phi$ is the zero function. –  mt_ Jan 7 '13 at 17:29
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