Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
cross posted mathoverflow.net/questions/118225 –  Willie Wong Jan 8 '13 at 8:40

1 Answer 1

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 $\,1\leq t\leq n\,$:


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

But then


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

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

share|improve this answer
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
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.