Definition

Let’s first understand how the linear dependence/independence of functions is defined¹.

DEFINITION 1. Linear Dependence/Independence
A set of functions $f_1(x), f_2(x),\cdots f_n(x)$ is said to be linearly dependent on an interval $I$ if there exist constants $c_1, c_2, \cdots c_n$, not all zero, such that $$c_1f_1(x)+c_2f_2(x)+\cdots+c_nf_n(x) = 0$$ for every $x$ in the interval. If the set of functions is not linearly dependent on the interval, it is said to be linearly independent

In other words, a set of $n$ functions is linearly independent if the equation

\[c_1f_1(x) + c_2f_2(x)+\cdots+c_nf_n(x) = 0\]

is satisfied for all $x$ in the interval only when all coefficients $c_i(i=1,2,\cdots,n)$ are zero.

Now let’s look at how the Wronskian is defined.

DEFINITION 2. Wronskian
Suppose each of the functions $f_1(x), f_2(x), \cdots, f_n(x)$ possesses at least $n-1$ derivatives. The determinant $$W(f_1,f_2,\cdots,f_n) = \begin{vmatrix} f_1 & f_2 & \cdots & f_n \\ f_1' & f_2' & \cdots & f_n' \\ \vdots & \vdots & \vdots & \vdots \\ f_1^{(n-1)} &f_2^{(n-1)} & \cdots & f_n^{(n-1)} \\ \end{vmatrix}$$ where the primes denote derivatives, is called the Wronskian of the function.

Note that when the Wronskian $W$ is not equal to zero, we say that the set of functions $f_1(x), f_2(x), \cdots, f_n(x)$ is linearly independent.

It is important to note that just because the Wronskian $W$ is zero does not necessarily mean that the set of functions is always linearly dependent.

Wronskian and Linear Independence Test

To understand the idea of the Wronskian, let’s consider the following process. For a set of functions $f_1(x), f_2(x), \cdots, f_n(x)$ that have at least $n-1$ derivatives, we can think of the following equation:

\[c_1f_1(x)+c_2f_2(x)+\cdots+c_nf_n(x) = 0\]

This equation represents a linear combination of the functions we initially wrote. Depending on the conditions, we can differentiate the above equation with respect to $x$ as follows:

\[c_1f_1(x) + c_2f_2(x)+\cdots+c_nf_n(x)=0\] \[c_1f_1'(x) + c_2f_2'(x)+\cdots+c_nf_n'(x)=0\] \[\vdots \notag\] \[c_1f_1^{(n-1)}(x) + c_2f_2^{(n-1)}(x)+\cdots+c_nf_n^{(n-1)}(x)=0\]

Now we have obtained a system of $n$ simultaneous equations with $n$ unknowns $c_i(i=1,2,\cdots,n)$. We can represent these $n$ equations as a matrix and a vector multiplication.

\[\begin{pmatrix} f_1(x) & f_2(x) & \cdots & f_n(x) \\ f_1'(x) & f_2'(x) & \cdots & f_n'(x) \\ \vdots & \vdots & \vdots & \vdots \\ f_1^{(n-1)}(x) & f_2^{(n-1)}(x) & \cdots & f_n^{(n-1)}(x) \\ \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ \end{pmatrix}\]

Now let’s confirm the fact that we can determine linear independence using the Wronskian through a proof by contradiction.

In equation (8), the determinant of the matrix on the left-hand side is the Wronskian $W$. Let’s assume that the functions are linearly dependent even though $W \neq 0$.

This means that the matrix on the left-hand side of equation (8) has an inverse, and the only solution for $c_i(i=1,2,\cdots,n)$ is the zero vector.

According to the initial Definition 1, when the unique solution for $c_i$ is the zero vector, the function set $f_1(x), f_2(x), \cdots, f_n(x)$ is linearly independent. This is a clear contradiction.

Therefore, if the Wronskian is not equal to zero, the function set $f_1(x), f_2(x), \cdots, f_n(x)$ is always linearly independent.