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.

2 Answers

Relevance
  • 8 years ago
    Favorite Answer

    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).

    • Commenter avatarLogin to reply the answers
  • 4 years ago

    nicely, one thank you to teach it fairly is to teach the column vectors are no longer linearly self sustaining. If we improve them out: | a² a²+2a+a million a²+4a+4 a²+6a+9 | | b² b²+2b+a million b²+4b+4 b²+6b+9 | | c² c²+2c+a million b²+4c+4 b²+6c+9 | | d² d²+2d+a million d²+4d+4 d²+6d+9 | If we are able to make some non-0 linear combination of {a², a²+2a+a million, a²+4a+4, a²+6a+9} to be identically 0 for any integer a, then we've discovered a linear combination that proves the column vectors are no longer linearly self sustaining, and hence, the determinant might equivalent 0 for all a, b, c, and d. assume there exists p, q, r, s, such that: pa² + q(a²+2a+a million) + r(a²+4a+4) + s(a²+6a+9) = 0 for all a. Then: (p + q + r + s)a² + (2q + 4r + 6s)a + (q + 4r + 9s) = 0 If we are able to locate values of p, q, r, and s such that the above is genuine, then we've our linear combination. p + q + r + s = 0 2q + 4r + 6s = 0 q + 4r + 9s = 0 This has non-0 recommendations, by using fact it fairly is an equation of much less equations than unknowns. The rank of the coefficient matrix is at maximum 3, and so the nullity would desire to be a minimum of a million. to that end, there exist non-0 p, q, r, s such that: pa² + q(a²+2a+a million) + r(a²+4a+4) + s(a²+6a+9) = 0 for all a, and hence, p, q, r, s grant coefficients for a linear combination of the column vectors of the matrix which grant us the 0 vector. hence the matrix has 0 determinant.

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.