📂Linear Algebra

Wronskian Definition and Determination of Linear Independence

Definition¹

Let us consider a set of functions that are differentiable up to n times, denoted by $S=\left\{ f_{1}, f_{2}, \dots, f_{n} \right\}$. The Wronskian $W$ of this set is defined by the following determinant.

$$W(x) = W(f_{1}, f_{2}, \dots, f_{n}) := \begin{vmatrix} f_{1} & f_{2} & \cdots & f_{n} \\ f_{1}^{\prime} & f_2^{\prime} & \cdots & f_{n}^{\prime} \\ \vdots & \vdots & \ddots & \vdots \\ f_{1}^{(n-1)} & f_{2}^{(n-1)} & \cdots & f_{n}^{(n-1)} \end{vmatrix}$$

Explanation

The set of differentiable functions constitutes a vector space, known as the function space. While there is no general method to determine the linear independence/dependence of a set of functions, the Wronskian can be used to ascertain linear independence for differentiable functions.

It is crucial to understand that the theorem introduced below is not bidirectional. It is essential to realize that the converse does not hold. If $W(x) \ne 0$, it signifies linear independence, but if $W(x)=0$, it does not determine whether the set is independent or dependent.

Theorem

Let $S$ be the set as defined here. If the Wronskian of $S$ is not $0$, then $S$ is linearly independent.

Proof

The proof is by contrapositive. In other words, if $S$ is linearly dependent, then the Wronskian $W$ is always $0$.

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Assume $S=\left\{ f_{1},\ f_{2},\ \cdots,\ f_{n} \right\}$ to be linearly dependent. Then, by definition, there exists a non-$0$ $k_{i}(i=1,2,\dots,n)$ satisfying the equation below.

$$\begin{equation} k_{1} f_{1} + k_{2} f_{2} + \cdots + k_{n} f_{n} = 0 \label{eq1} \end{equation}$$

Differentiating the above equation yields:

$$\begin{align*} k_{1} f_{1}^{\prime} + k_{2} f_{2}^{\prime} + \cdots + k_{n} f_{n}’&=0 \\ k_{1} f_{1}^{(2)} + k_{2} f_{2}^{(2)} + \cdots + k_{n} f_{n}^{(2)}&=0 \\ \vdots& \\ k_{1} f_{1}^{(n-1)} + k_{2} f_{2}^{(n-1)} + \cdots + k_{n} f_{n}^{(n-1)} &=0 \end{align*}$$

Converting this system of equations into matrix representation results in:

$$\begin{align*} \begin{pmatrix} f_{1} & f_{2} & \cdots & f_{n} \\ f_{1}^{\prime} & f_2^{\prime} & \cdots & f_{n}^{\prime} \\ \vdots & \vdots & \ddots & \vdots \\ f_{1}^{(n-1)} & f_{2}^{(n-1)} & \cdots & f_{n}^{(n-1)} \end{pmatrix} \begin{pmatrix} k_{1} \\ k_{2} \\ \vdots \\ k_{n} \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \\ \mathbf{F} \mathbf{k} &= \mathbf{0} \end{align*}$$

In this case, the above equation has a nontrivial solution $\mathbf{k} \ne \mathbf{0}$. Therefore, by equivalence, $\mathbf{F}$ is not an invertible matrix, and its determinant is $0$. Since the determinant of $\mathbf{F}$ is the Wronskian,

$$W(x) = 0,\quad \forall x \in \mathbb{R}.$$

Hence, if $S$ is linearly dependent, the Wronskian $W$ is always $0$.

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