# Suppose $\lambda_j$'s are distinct for $1\leq j \leq n$. Prove that $e^{\lambda_j x}$ as functions are linearly independent.

Suppose $$\lambda_j \in \mathbb{R}$$ are distinct for $$1\leq j \leq n$$. Let $$V := \{f: \mathbb{R} \to \mathbb{R} \}$$, a collection of functions from $$\mathbb{R}$$ to itself, be a vector space over $$\mathbb{R}$$. Prove or disprove that $$e^{\lambda_1 x}, \cdots, e^{\lambda_n x}$$ are linearly independent.

My attempt: I think they should be linearly independent.

I firstly tried $$n = 2$$ case, and suppose that they are linearly dependent. Suppose $$\lambda_1 > \lambda_2 \geq 0$$. Then we have $$k_1, k_2 \neq 0$$ such that for all $$x$$, $$k_1 e^{\lambda_1 x} + k_2 e^{\lambda_2 x} = 0,$$ and we pass $$x \to \infty$$ (if $$\lambda_1 < \lambda_2 \leq 0$$, then pass $$x \to -\infty$$), then $$e^{\lambda_1 x}$$ term blows up faster than the other. So the equality will fail.

Question: Does my argument above work? Could it be generalized to higher dimensional cases?

• Vandermonde's matrix is lurking in the shadows. Can you find a way to pull it out into the light? – Carl Christian May 4 '19 at 12:02
• Oh yeah, thank you for your idea. I believe you mean that I could write $k_1e^{\lambda_1 x} + \cdots + k_n e^{\lambda_n x} = 0$, and differentiate it both sides with respect to $x$. For the $j$-th derivatives, if we take $x := 0$, then we have $\sum_{i=1}^n k_i \lambda_i^j = 0$. So if we take the derivatives repeatly, and regard $k$'s as our unknowns, then we have a linear system of equations, say $A(k_1, \cdots, k_n) = 0$, and $A$ is a Vandermonde's matrix. The system has unique solution if and only if $\det(A) \neq 0$, i.e., $\lambda_i \neq \lambda_j$ for $i \neq j$, which is our assumption – mathdoge May 4 '19 at 13:02
• @CarlChristian Could you help have a look at my comment above? – mathdoge May 4 '19 at 13:03
• You have done exactly what I hoped you would. – Carl Christian May 4 '19 at 14:24

Argue by induction. Assume that any $$n-1$$ of the functions $$e^{\lambda_k x}$$ are independent and suppose $$\sum\limits_{k=1}^{n} c_ke^{\lambda_k x}=0$$ for all $$x$$. Let $$i$$ be such that $$\lambda_i \geq \lambda_j$$ for all $$j$$. Multiply both sides of the equation by $$e^{-\lambda_i x}$$ and let $$x \to \infty$$. We get $$c_i=0$$ and we are left with $$\sum\limits_{ k\neq i, k=1}^{n} c_ke^{\lambda_k x}=0$$. This implies $$c_k=0$$ for all $$k$$.