# linear independence of exponentials

Let $X$ be the set of functions $e^{p(x)}$ of the real vector $x$, where $p$ is a multivariate polynomial with $p(0)=0$.

Is any finite subset of $X$ linearly independent? If yes, why? If no, is the answer true for other, restricted choices of $p$? (The answer is yes when the polynomials are restricted to have degree 1.)

• What does applying a polynomial to a vector mean in this context? Does it just mean applying a multivariate polynomial to a given set of values? Feb 27, 2018 at 13:24
• If these are univariate polynomials, then the set is indeed linearly independent, which is easily seen by considering the behaviour of a linear combination as $x\to\infty$ (one of the polynomials will dominate). Feb 27, 2018 at 14:31
• @DanielMcLaury: yes; corrected Feb 27, 2018 at 15:01
• Of course, as a minor matter of terminology, "$X$ is linearly independent" (for $X$ a subset of an abstract vector space) means "every finite subset of $X$ is linearly independent". Feb 27, 2018 at 19:18
• For an elementary proof in the univariate case look here Oct 28, 2018 at 8:01

Yes. If you require that no difference $p_j-p_k$ is constant (which follows from your assumption $p(0)=0$), then $c_1e^{p_1}+\ldots+c_me^{p_m}=0$ implies that all $c_j=0$.
For polynomials in one variable, this is easy: look at the asymptotics, for example when $z$ tends to infinity on an appropriate ray, and use induction in degree of the polynomials. To obtain the same for several variables, restrict your identity onto lines in $C^n$.