Anonymous
Anonymous asked in Science & MathematicsMathematics · 8 years ago

# Prove determinant of the Jacobian is non-zero?

Suppose f: V -> W is a bijection from V = B_r (a) in R^n to W in R^n, and f is differentiable at a. Suppose also that f^(-1): W -> V (inverse) is defferentiable at f(a). Prove that the determinant of the Jacobian of f at a is non-zero, that is, det J_f(a) is non-zero.

Relevance
• 8 years ago

Let M = J_f(a) and N = J_{f^{-1}}(f(a)). Note that M and N are nxn matrices and that they exist by the given hypotheses.

The chain rule says that whenever a map h is differentiable at a point x, and a map g is differentiable at h(x), then the composite function g o h is differentiable at x, and that J_{g o h}(x) = J_g(h(x)) J_h(x). If we apply this fact here to h = f, x = a, and g = f^{-1} we learn that

NM = J_{f^{-1}}(f(a)) J_f(a) = J_{f^{-1} o f} (a)

But f^{-1} o f is the identity map from V to V, and clearly the Jacobian of the identity map, at any point of V, is just the nxn identity matrix I_n. So

NM = I_n.

Taking the determinant of both sides and applying well known facts about the determinant you deduce that det(N) det(M) = 1. This implies that det(M) is nonzero (if it were zero, then for all real numbers y, the product y det(M) would be 0).