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Given functions
 , then the Wronskian determinant (or simply the Wronskian)
 is the determinant of the square matrix
where $f^{(k)}$ indicates the $k$ th derivative of $f$ (not exponentiation).
The Wronskian of a set of functions $F$ is another function, which is zero over any interval where $F$ is linearly dependent. Just as a set of vectors is said to be linearly dependent when there exists a non-trivial linear relation between them, a set of functions
 is also said to be dependent over an interval $I$ when there exists a non-trivial linear relation between them, i.e.,
for some
 , not all zero, at any $t \in I$ .
Therefore the Wronskian can be used to determine if functions are independent. This is useful in many situations. For example, if we wish to determine if two solutions of a second-order differential equation are independent, we may use the Wronskian.
Consider the functions $x^2$ , $x$ , and $1$ . Take the Wronskian:
Note that $W$ is always non-zero, so these functions are independent everywhere. Consider, however, $x^2$ and $x$ :
Here $W = 0$ only when $x = 0$ . Therefore $x^2$ and $x$ are independent except at $x = 0$ .
Consider $2x^2+3$ , $x^2$ , and $1$ :
Here $W$ is always zero, so these functions are always dependent. This is intuitively obvious, of course, since $$ 2x^2 + 3 = 2(x^2) + 3(1) $$
Given $n$ linearly independant functions
 , we can use the Wronskian to construct a linear differential equation whose solution space is exactly the span of these functions. Namely, if $g$ satisfies the equation
then
 for some choice of
 .
As a simple illustration of this, let us consider polynomials of at most second order. Such a polynomial is a linear combination of $1$ , $x$ , and $x^2$ . We have $$ W (1, x, x^2, g(x)) = \left| \begin{matrix} 1 & x & x^2 & g(x) \\ 0 & 1 & 2x & g'(x) \\ 0 & 0 & 2 & g''(x) \\ 0 & 0 & 0 & g'''(x) \end{matrix} \right| = 2 g''' (x) $$ Hence, the equation is $g''' (x) = 0$ which indeed has exactly polynomials of degree at most two as
solutions.
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