The Jacobian Matrix in the Implicit Function Theorem
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In the attached problem, I am having trouble showing that the determinants are in the same form in the Lagrangian as the one in the implicit function theorem.
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Solution Summary
In the proof of the Theorem of Lagrange multipliers via the Implicit Function Theorem, we need to verify that a certain Jacobian matrix is non-singular. We explain how to calculate such a Jacobian matrix and why it is non-singular.
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Let us start with a general discussion. Given a function G: R^nto R^m, denote by g_1,...,g_m: R^nto R the component
functions of G, meaning that for X=(x_1,...,x_n)in R^n,
G({X})=(g_1({X}),...,g_m({X})).
Suppose that all the partial derivatives of g_1,...,g_m exist at a point {X}in R^n. We define the {Jacobian
matrix} of G at {X}, denoted by G_{{X}}({X}) and also by DG({X}) or G'({X}), by
the following expression:
G_{{X}}({X})={pmatrix}
pd{g_1({X})}{x_1} ... pd{g_1({X})}{x_n}
vdots ddots vdots
pd{g_m({X})}{x_1} ... pd{g_m({X})}{x_n}
{pmatrix}.
The Jacobian matrix of a function from R^n to R^m is an mtimes n matrix, that is, it has m rows and n columns. The
number of rows equals the number component functions, or equivalently, equals the exponent of R where the image of G is
contained, in this case m; the number of columns equals the number of variables, in this case the vector {X} has n
variables.
Note that the i^{th}-row of the Jacobian matrix corresponds to the (transpose) of the gradient vector nabla g_i,
nabla g_i({X})={pmatrix}
pd{g_i({X})}{x_1}
pd{g_i({X})}{x_2}
vdots
pd{g_1({X})}{x_n}
{pmatrix},
so (nabla g_i)^T=(pd{g_i}{x_1},pd{g_i}{x_2},...,pd{g_i}{x_n}). Here the superscript T denotes transposition. We ...
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