# Exponential Functions are Linearly Independent

## Problem 73

Let $c_1, c_2,\dots, c_n$ be mutually distinct real numbers.

Show that exponential functions

\[e^{c_1x}, e^{c_2x}, \dots, e^{c_nx}\]
are linearly independent over $\R$.

Add to solve later

Sponsored Links

Contents

## Hint.

- Consider a linear combination \[a_1 e^{c_1 x}+a_2 e^{c_2x}+\cdots + a_ne^{c_nx}=0.\]
- Differentiate this equality $n-1$ times and you will get $n$ equations.
- Write a matrix equation for the system. You will see the Vandermonde matrix.

## Proof.

Suppose that we have a linear combination of these functions that is zero.

Namely, suppose we have

\[a_1 e^{c_1 x}+a_2 e^{c_2x}+\cdots + a_ne^{c_nx}=0\]
for some real numbers $a_1, a_2, \dots, a_n$.

We want to show that the coefficients $a_1, a_2, \dots, a_n$ are all zero.

By differentiating the equation, we obtain

\[a_1c_1e^{c_1x}+a_2c_2e^{c_2x}+\cdots +a_n c_n e^{c_n x}=0.\] Differentiating repeatedly we further obtain the equalities

\begin{align*}

& a_1c_1^2e^{c_1x}+a_2c_2^2e^{c_2x}+\cdots +a_n c_n^2 e^{c_n x}=0\\

& a_1c_1^3e^{c_1x}+a_2c_2^3e^{c_2x}+\cdots +a_n c_n^3 e^{c_n x}=0\\

& \dots \\

& a_1c_1^{n-1}e^{c_1x}+a_2c_2^{n-1}e^{c_2x}+\cdots +a_n c_n^{n-1}

e^{c_n x}=0\\

\end{align*}

We rewrite these $n$ equations into the following matrix equation.

\[\begin{bmatrix}

1 & 1 & \dots &1 \\

c_1 & c_2 & \dots & c_n \\[3pt]
c_1^2 & c_2^2 & \dots & c_n^2 \\[3pt]
\vdots & \vdots & \vdots & \vdots \\[3pt]
c_1^{n-1} & c_2^{n-1} & \dots & c_n^{n-1}

\end{bmatrix}

\begin{bmatrix}

a_1 e^{c_1x} \\

a_2 e^{c_2x} \\

\vdots \\

a_n e^{c^nx}

\end{bmatrix}=\begin{bmatrix}

0 \\

\vdots \\

0

\end{bmatrix} \tag{*}

\]

The determinant of the left matrix is

\[\det \begin{bmatrix}

1 & 1 & \dots &1 \\

c_1 & c_2 & \dots & c_n \\[3pt] c_1^2 & c_2^2 & \dots & c_n^2 \\[3pt] \vdots & \vdots & \vdots & \vdots \\[3pt] c_1^{n-1} & c_2^{n-1} & \dots & c_n^{n-1}

\end{bmatrix}=\prod_{i<j}(c_j-c_i)\] by the Vandermonde determinant.

Since by assumption $c_1,\dots, c_n$ are distinct, the determinant is not zero.

Therefore by multiplying equality (*) by the inverse on the left, we obtain

\[\begin{bmatrix}

a_1 e^{c_1x} \\

a_2 e^{c_2x} \\

\vdots \\

a_n e^{c^nx}

\end{bmatrix}=\begin{bmatrix}

0 \\

\vdots \\

0

\end{bmatrix}. \]
Since the functions $e^{c_i x}$ are always positive, we must have $a_1=a_2=\cdots=a_n=0$ as required.

Therefore the functions $e^{c_1 x}, \dots, e^{c_n x}$ are linearly independent.

## Comment.

The determinant that we considered above is called the Wronskian for the set of functions $\{e^{c_1x}, e^{c_2x}, \dots, e^{c_nx}\}$.

## Related Question.

The following problem is a more concrete version of the current problem.

\[\{e^x, e^{2x}, e^{3x}\}\] is linearly independent on the interval $[-1, 1]$.

The solutions is given in the post ↴

Using the Wronskian for Exponential Functions, Determine Whether the Set is Linearly Independent

Add to solve later

Sponsored Links

## 1 Response

[…] The solution is given by Exponential Functions are Linearly Independent. […]