Wronskian

Definition
Let $ϕ_{1}, \dots, ϕ_{n} \in C^{\infty} (I)$ , and set

W (ϕ_{1}, \dots, ϕ_{n}) (t) = (ϕ_{i}^{(j - 1)} (t)) .

We define the Wronskian of $ϕ_{1}, \dots, ϕ_{n}$ as $w (t) = det W (t)$ .
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Theorem. If $w (t)$ is not identically zero in $I$ then $ϕ_{1}, \dots, ϕ_{n}$ are linearly independent in $I$ .

Proof: Suppose $c_{1} ϕ_{1} + c_{2} ϕ_{2} + c_{3} ϕ_{3} = 0$ , with $c_{1}, c_{2}, c_{3} \in R$ not all zero. Then, if we differentiate we obtain

c_{1} ϕ_{1}^{'} + c_{2} ϕ_{2}^{'} + c_{3} ϕ_{3}^{'} = 0,

and differentiate again, we obtain

c_{1} ϕ_{1}^{″} + c_{2} ϕ_{2}^{″} + c_{3} ϕ_{3}^{″} = 0.

Therefore, we have

c_{1} \cdot (\begin{matrix} ϕ_{1} \\ ϕ_{1}^{'} \\ ϕ_{1}^{″} \end{matrix}) + c_{2} \cdot (\begin{matrix} ϕ_{2} \\ ϕ_{2}^{'} \\ ϕ_{2}^{″} \end{matrix}) + c_{3} \cdot (\begin{matrix} ϕ_{3} \\ ϕ_{3}^{'} \\ ϕ_{3}^{″} \end{matrix}) = 0.

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Remark.
The converse is not true. The functions $t^{2}$ and $| t | \cdot t$ are linearly independent but their Wronskian is 0 everywhere. So non-null Wronskian implies independence but null Wronskian DO NOT implies linear independence.
On the other hand, a converse can be proven by requiring stronger assumptions on $ϕ_{1}, \dots, ϕ_{n}$ , for example analyticity or if they are known to be the solutions of the same linear ODE of the form $y^{n)} + L y = 0$ , where $L$ is a linear differential operator with respect to $x$ of order less than $n$ . See Wikipedia.