The Wronskian We know that a standard way of testing whether a set of n n-vectors are linearly independent is to see if the n × n determinant having them as its rows or columns is non-zero. This is also an important method when the nvectors are solutions to a system; the determinant is given a special name. (Again, we will assume n = 2, but the deﬁnitions and results generalize to any n.) Deﬁnition 3 Let x1 (t) and x2 (t) be two 2-vector functions. We deﬁne their Wronskian to be the determinant � � � x1 ( t ) x2 ( t ) � � W (x1 , x2 )(t) = �� (1) y1 ( t ) y2 ( t ) � whose columns are the two vector functions. The independence of the two vector functions should be connected with their Wronskian not being zero. For points, the relationship is clear. Using the result mentioned above, we can say � � � x1 ( t0 ) x2 ( t0 ) � � = 0 ⇔ x1 (t0 ) and x2 (t0 ) are dependent. � W (x1 , x2 )(t0 ) = � y1 ( t0 ) y2 ( t0 ) � (2) However for vector functions, the relationship is clear-cut only when x1 and x2 are solutions to a well-behaved ODE system (3). The theorem is: We are still considering the system x� y� = a(t) x + b(t)y = c(t) x + d(t)y � , x y �� = a(t) b(t) c(t) d(t) �� x y � , (3) Theorem 3 Wronskian vanishing theorem. On an interval I where the entries of A(t) are continuous, let x1 and x2 be two solutions to (3) and W (t) their Wronskian (1). Then either a) W (t) ≡ 0 on I, and x1 and x2 are linearly dependent on I, or b) W (t) is never 0 on I, and x1 and x2 are linearly independent on I. Proof. Using (2), there are just two possibilities. a) x1 and x2 are linearly dependent on I; say x2 = c1 x1 . In this case they are dependent at each point of I, and W (t) ≡ 0 on I, by (2). 18.03SC b) x1 and x2 are linearly independent on I, in which case by Theorem 2A they are linearly independent at each point of I, and so W (t) is never zero on I, by (2). �