the Wronskian of vector solutions
/ VRON-skee-an /
You have several candidate solutions to a system and you need one yes-or-no test: are they genuinely independent, or is one secretly a combination of the others? A single number, computed at any convenient instant, answers it. That number is the Wronskian.
Take your n solution vectors x_1(t), ..., x_n(t) of x' = A(t) x and stand them up as the columns of an n-by-n matrix — this is just the fundamental matrix candidate Psi(t). The Wronskian is its determinant: W(t) = det Psi(t). If W(t0) is not zero at some instant t0, the columns are linearly independent there, so the solutions form a fundamental set; if W(t0) = 0, they are dependent and do not span the solution space. What makes the test painless is Abel's (Liouville's) theorem: W(t) = W(t0) exp(integral from t0 to t of trace A(s) ds). Because the exponential is never zero, W is either nonzero for all t or zero for all t — you only ever need to check one point.
This is the system version of the second-order Wronskian, with two unknowns replaced by n and the two-row determinant by an n-by-n one. The all-or-nothing dichotomy is special to solutions of a linear ODE — for arbitrary vector functions a vanishing Wronskian at one point tells you nothing. It works here only because the vectors are constrained to satisfy the same differential equation.
For solutions (e^t, e^t) and (e^(3t), -e^(3t)) of some 2-by-2 system, W(t) = det [e^t, e^(3t); e^t, -e^(3t)] = -e^t e^(3t) - e^t e^(3t) = -2 e^(4t). Never zero, so the pair is a fundamental set.
Compute the determinant once; by Abel's theorem its never-zero status holds for every t.
Caution: the all-or-nothing law applies only to vectors that solve the same linear system. Two unrelated vector functions can have a Wronskian that is zero at some points and nonzero at others.