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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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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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