This is the theorem that we are proving.

Theorem. Let f1, f2,...,fn be functions in C[0,1] each of which has first n-1 derivatives. If the Wronskian of this set of functions is not identically zero then the set of functions is linearly independent.

Proof. By contradiction, suppose that the Wronskian W of this set of functions is not identically zero but the functions are linearly dependant. By the theorem about linearly independent sets this means that there exist numbers a1,...,an, not all equal to zero and such that

a1f1(x)+a2f2(x)+...anfn(x)=0

The 0 in this formula is the zero function. Taking n-1 derivatives of this equality gives us the following system of equalities:

```                           a1f1(x)+a2f2(x)+...anfn(x)=0
a1f1'(x)+a2f2'(x)+...anfn'(x)=0
.....................................
a1f1n-1(x)+a2f2n-1(x)+...anfnn-1(x)=0
```

Thus a1, ..., an form a solution of the homogeneous system of linear equations with the following matrix A(x):

```
f1(x)      f2(x)   .... fn(x)
f1'(x)     f2'(x)  .... fn'(x)
f1''(x)    f2''(x) .... fn''(x)
.........................
f1n-1(x) f2n-1(x) ..fnn-1(x)
```

Thus this system has n unknowns and n equations and has a non-trivial solution. So by the second theorem about invertible matrices, the matrix A(x) is not invertible for any x. Now by the third theorem about determinants, the determinant of A(x) is 0 for every x. But the determinant of this matrix is the Wronskian of our set of functions, and we supposed that this Wronskian is not identically zero. This contradiction completes the proof of the theorem.