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?

  • $\begingroup$ Vandermonde's matrix is lurking in the shadows. Can you find a way to pull it out into the light? $\endgroup$ – Carl Christian May 4 '19 at 12:02
  • $\begingroup$ 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 $\endgroup$ – mathdoge May 4 '19 at 13:02
  • $\begingroup$ @CarlChristian Could you help have a look at my comment above? $\endgroup$ – mathdoge May 4 '19 at 13:03
  • $\begingroup$ You have done exactly what I hoped you would. $\endgroup$ – 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 $k$.


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