I couldn't understand why it says the linear system has a nontrivial solution since the number of rows is :$1$(the original function）$+(n-1)$ (take the derivative $n-1$ times)$=n$. Also there are $n$ columns or unknowns. Why the system is not an independent one?
On
I use [1] as the basis of my proof. As in [1], I employ a proof by contradiction [2].
Step 0. Define the proposition to be proved
By $P$, I denote the proposition to be proved, which is
$P := $ Given $n$ functions $f_1(x), f_2(x), \ldots, f_n(x)$ that are elements of the set $C^{n-1}{(I)}$ [3] (i.e., the space of continuous functions that have continuous first $n-1$ derivatives in the interval $I$), if the Wronskian is not equal to zero for any value $c$ in the interval $I$, then the $n$ functions are linearly independent in the interval $I$ [4].
Step 1. The proposition to be proved, $P$, is assumed to be false
I assume that given $n$ functions $f_1(x), f_2(x), \ldots, f_n(x)$ that are elements of the set $C^{n-1}{(I)}$, if the Wronskian is not equal to zero for any value $c$ in the interval $I$, then the $n$ functions are linearly dependent in the interval $I$.
Step 2. I derive a contradiction
I know that a sequence of functions $f_{1}{(x)},f_{2}{(x)},\dots ,f_{n}{(x)}$ from a function space are said to be linearly dependent, if there exist constant scalars $a_{1},a_{2},\dots ,a_{n}$, which are not all zero, such that $$a_{1}\,f_{1}{(x)}+a_{2}\,f_{2}{(x)}+\cdots +a_{k}\,f_{k}{(x)}= 0_f,\quad\text{for all}~x\in I ;\tag{1}$$ where $ 0_f$ denotes the zero function [5]. Next, I take the derivative of Equation $1$ $n-1$ times. I obtain the following system of $n$ equations in $n$ unknowns: $$ \begin{bmatrix} f_1{(x)} &f_2{(x)} & \cdots & f_n{(x)} \\ f^{'}_1{(x)} &f^{'}_2{(x)} & \cdots & f^{'}_n{(x)} \\ \vdots & \vdots & \vdots & \vdots \\ f^{(n-1)}_1{(x)} &f^{(n-1)}_2{(x)} & \cdots & f^{(n-1)}_n{(x)} \end{bmatrix} \, \begin{bmatrix} a_1\\a_2\\\vdots\\a_n \end{bmatrix} = \begin{bmatrix} 0\\0\\\vdots\\0 \end{bmatrix},\quad\text{for all}~x\in I. $$ This may be re-written in terms of any constant $c\in I$. $$ \begin{bmatrix} f_1{(c)} &f_2{(c)} & \cdots & f_n{(c)} \\ f^{'}_1{(c)} &f^{'}_2{(c)} & \cdots & f^{'}_n{(c)} \\ \vdots & \vdots & \vdots & \vdots \\ f^{(n-1)}_1{(c)} &f^{(n-1)}_2{(c)} & \cdots & f^{(n-1)}_n{(c)} \end{bmatrix} \, \begin{bmatrix} a_1\\a_2\\\vdots\\a_n \end{bmatrix} = \begin{bmatrix} 0\\0\\\vdots\\0 \end{bmatrix}\qquad~\text{for any}~c\in I. $$
According to my proposition, I have $$ \begin{vmatrix} f_1{(c)} &f_2{(c)} & \cdots & f_n{(c)} \\ f^{'}_1{(c)} &f^{'}_2{(c)} & \cdots & f^{'}_n{(c)} \\ \vdots & \vdots & \vdots & \vdots \\ f^{(n-1)}_1{(c)} &f^{(n-1)}_2{(c)} & \cdots & f^{(n-1)}_n{(c)} \end{vmatrix} \neq 0\qquad~\text{for any}~c\in I. $$ I can therefore invert the square matrix of coefficients. I obtain a solution for my unknowns (i.e., $a_1, a_2,\ldots, a_n$). I obtain that $$ \begin{bmatrix} a_1\\a_2\\\vdots\\a_n \end{bmatrix} = \begin{bmatrix} f_1{(c)} &f_2{(c)} & \cdots & f_n{(c)} \\ f^{'}_1{(c)} &f^{'}_2{(c)} & \cdots & f^{'}_n{(c)} \\ \vdots & \vdots & \vdots & \vdots \\ f^{(n-1)}_1{(c)} &f^{(n-1)}_2{(c)} & \cdots & f^{(n-1)}_n{(c)} \end{bmatrix}^{-1} \begin{bmatrix} 0\\0\\\vdots\\0 \end{bmatrix}\qquad~\text{for any}~c\in I. $$ Upon matrix mulitplication, I find $$a_1 = a_2=\cdots = a_n=0.$$
These functions were said to be linearly dependent since there existed scalars $a_1, a_2, \ldots, a_n$ that are not all zero. Yet, I just arrived at the contradictory statement that the scalars $a_1, a_2, \ldots, a_n$ are all zero.
Step 3. Conclusion
Since the statements that, the scalars $a_1, a_2, \ldots, a_n$ are not all zero, and the statement that, the scalars $a_1, a_2, \ldots, a_n$ are all zero, cannot both be true, the assumption that $P$ is false must be wrong. I conclude that $P$ must be true.
BIBLIOGRAPHY
[1] https://math.vanderbilt.edu/sapirmv/msapir/prwronskian.html
[3] https://en.wikipedia.org/wiki/Function_space
[4] https://mathworld.wolfram.com/LinearlyDependentFunctions.html
[5] https://en.wikipedia.org/wiki/Linear_independence#Definition
First of all consider the $n$ functions $f_1,...,f_n \in C^{n-1}(\mathbb{R})$ and suppose there exists $k_1,...k_n$ real numbers such that
$$g=\sum_{i=1}^n k_if_i(x)=0$$
for every $x\in\mathbb{R}$; that is $g$ is the zero-function. The set $C^{n-1}(\mathbb{R})$ is a real vector space of infinite dimension, so $g$ is still a function belong to it. Consider now the first derivative of $g$
$$g'(x)=\sum_{i=1}^n k_if_i'(x), $$
because the derivative of the sum is the sum of the derivative, on the other hand $g\equiv0$, so $g'(x)\equiv0$; then$$g'(x)=\sum_{i=1}^n k_if_i'(x)=0. $$
This means that if $f_1,...,f_n$ are linearity dependent, then $f'_1,...,f_n'$ needs to be linearity dependent in $C^{n-1}(\mathbb{R})$. If you repeat for all order of derivation, you obtain the same result
$f_1,...,f_n$ are linearity dependent iff $f_1^{(i)},...,f_n^{(i)}$ are linearity dependent for every $i=1,...,n-1$.
Finally, consider the functions $f_1,...,f_n \in C^{n-1}(\mathbb{R})$ and take the following matrix $$A=\begin{pmatrix} f_1(x)& ... & f_n(x)\\ ... & ... & ...\\ f_1^{(n-1)} & ... & f_n^{(n-1)} \end{pmatrix}$$
the Wronskian $W(x)$ of $A$ is its the determinant, which clearly it depend on $x$. If there exists a non-zero vector column of real numbers such that
$$A=\begin{pmatrix} f_1(x)& ... & f_n(x)\\ ... & ... & ...\\ f_1^{(n-1)} & ... & f_n^{(n-1)} \end{pmatrix} \begin{pmatrix} k_1\\ ...\\ k_n \end{pmatrix}= \begin{pmatrix} 0\\ ..\\ 0 \end{pmatrix},$$
for every $x\in\mathbb{R}$, this mean that the function are linearity dependent by the discussion above. Viceversa if there exists a real number $\overline{x}$ such that $W(\overline{x})$ is not zero, then the function are linearity independent.