Let $\lambda_i \in \mathbb C$ $(i=1,2,\cdots,n)$ be $n$ different complex numbers and $p_i \in \mathbb C[t]$ $(i=1,2,\cdots,n)$ polynomials. If we have the relation ${e^{{\lambda _1}t}}{p_1}(t) + {e^{{\lambda _2}t}}{p_2}(t) + \cdots + {e^{{\lambda _n}t}}{p_n}(t) = 0$ for all $t \geq 0$, can we conclude that all $p_i(t)=0$?
The linear dependence of exponential functions
1 Answers
Yes, we can. Assume we have a linear dependance $\tag1\sum_{i=1}^n e^{\lambda_it}p_i(t)=0$ with minimal number $n>0$ of summands. Take derivatives to obtain $\tag2\sum_{i=1}^n \lambda_i e^{\lambda_it}p_i(t)+\sum_{i=1}^n e^{\lambda_it}p_i'(t)=0$ Then $(\lambda_np_n(t)+p_n'(t))\cdot (1)-p_n(t)\cdot(2)$ is also a vanishing linear combination, but it does not involve $e^{\lambda_nt}$. By minimality of $n$, we conclude that this is the zero combination, i.e. $\tag3 (\lambda_np_n(t)+p_n'(t))p_i(t)-p_n(t)(\lambda_ip_i(t)+p_i'(t))=0$ for all $i$. This simplifies to $ (\lambda_n-\lambda_i)p_n(t)p_i(t)=-p_n'(t)p_i(t)+p_n(t)p_i'(t)$ If $\lambda_n\ne\lambda_i$, the polynomial degree on the right hand side is less than on the left hand side, which is impossible. Therefore we must have $n=1$. But if $ e^{\lambda t}p(t)=0$ for all $t$, then $p$ is the zero polynomial because the exponential is never zero.