This is the theorem that we are proving.

Theorem. Let *f _{1}, f_{2},...,f_{n}* be functions in

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

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

a_{1}f_{1}(x)+a_{2}f_{2}(x)+...a_{n}f_{n}(x)=0 a_{1}f_{1}'(x)+a_{2}f_{2}'(x)+...a_{n}f_{n}'(x)=0 ..................................... a_{1}f_{1}^{n-1}(x)+a_{2}f_{2}^{n-1}(x)+...a_{n}f_{n}^{n-1}(x)=0

Thus *a _{1}, ..., a_{n}* form a solution of the homogeneous system of
linear equations with the following matrix

f_{1}(x) f_{2}(x) .... f_{n}(x) f_{1}'(x) f_{2}'(x) .... f_{n}'(x) f_{1}''(x) f_{2}''(x) .... f_{n}''(x) ......................... f_{1}^{n-1}(x) f_{2}^{n-1}(x) ..f_{n}^{n-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.