# How to prove $\det(e^{\lambda_ix_j})\not=0$ where $\lambda_i\not=\lambda_j$ and $x_i\not=x_j$ if $i\not=j$

In try to figure out the exercise:

Let $$f(x)=\sum_{k=1}^{n}c_ke^{\lambda_kx}$$where $\lambda_i \not=\lambda_j,i\not=j$,and $c_1^2+c_2^2+\dots+c_n^2\not=0$, then the number of $f(x)$'s roots is strictly less than $n$.

My approach(this way can't deal with $f(x)$ has repeated root):

assume $x_1,x_2,\dots,x_n$ are $f(x)$'s roots,and $x_i\not=x_j$ if $i\not=j$. then I get a linear equations about $c_1,c_2,\dots,c_n$: $$e^{\lambda_1x_1}c_1+e^{\lambda_2x_1}c_2+\dots+e^{\lambda_nx_1}c_n=0$$ $$e^{\lambda_1x_2}c_1+e^{\lambda_2x_2}c_2+\dots+e^{\lambda_nx_2}c_n=0$$ $$\dots\dots\dots\dots\dots\dots$$ $$e^{\lambda_1x_n}c_1+e^{\lambda_2x_n}c_2+\dots+e^{\lambda_nx_n}c_n=0$$ I want to show that the solution to this linear equations are $0$,it will be a contradiction.but i can't figure out its determinant of coefficient: $$\begin{vmatrix} e^{\lambda_1x_1}& e^{\lambda_2x_1} &\dots &e^{\lambda_nx_1}\\ e^{\lambda_1x_2}& e^{\lambda_2x_2} &\dots&e^{\lambda_nx_2} \\ \dots&\dots &\dots&\dots \\ e^{\lambda_1x_n}&e^{\lambda_2x_n} &\dots &e^{\lambda_nx_n} \end{vmatrix}\not=0$$

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Maybe related to vandermonde determinants, which see. – Gerry Myerson Nov 4 '12 at 2:04
I think everything is real, and maybe you want to argue that i) you can choose $\lambda_1 = 0$, ii) if that has n zeros it's derivative has n-1 real zeros iii) it's derivative is of the same form but with one fewer $\lambda$. – mike Nov 4 '12 at 11:50
if $f(x)$ has n zeros, $f'(x)$ maybe not has n-1 zeros. for instance: $f(x) = x^2+1$. or maybe I misunderstand you :) – Laura Nov 7 '12 at 5:43

Note that the zeroes of $f$ agree with the zeroes of $$\tag{1} g(x)=e^{-\lambda_1 x}\,\sum_{k=1}^nc_ke^{\lambda_kx_k}=c_1+\sum_{k=2}^nc_ke^{(\lambda_k-\lambda_1)x}.$$ Now we proceed by induction on $n$. If $n=1$, then $g(x)=c_1$ has no zeroes (recall that $c_1\ne0$).
Assume as inductive hypothesis that functions of the form ($1$) have at most $n$ zeroes. Then if $$h(x)=d_1+\sum_{k=2}^{n+1}d_ke^{(\lambda_k-\lambda_1)x},$$ its derivative is $$h'(x)=\sum_{k=2}^{n+1}d_k(\lambda_k-\lambda_1)e^{(\lambda_k-\lambda_1)x}.$$ So $h'$ satisfies the inductive hypothesis and thus has at most $n-1$ zeroes. But then $h$ can have at most $n$ zeroes (as between any two zeroes of $h$ there is a zero of $h'$).