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 $\R$.

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Hint.

1. Consider a linear combination \[a_1 e^{c_1 x}+a_2 e^{c_2x}+\cdots + a_ne^{c_nx}=0.\]
2. Differentiate this equality $n-1$ times and you will get $n$ equations.
3. 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 problems are more concrete versions of the current problem.

See the post ↴
Exponential Functions Form a Basis of a Vector Space
for the solution.

The solutions is given in the post ↴
Using the Wronskian for Exponential Functions, Determine Whether the Set is Linearly Independent

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