6.17. THE METHOD OF LAGRANGE MULTIPLIERS 133
If this matrix has rank m+1 then some m+1×m+1 submatrix has nonzero determinant.It follows from the implicit function theorem that there exist m+1 variables, xi1 , · · · ,xim+1such that the system
F(x,a) = 0 (6.17.48)
specifies these m+ 1 variables as a function of the remaining n− (m+1) variables and ain an open set of Rn−m. Thus there is a solution (x,a) to 6.17.48 for some x close to x0whenever a is in some open interval. Therefore, x0 cannot be either a local minimum or alocal maximum. It follows that if x0 is either a local maximum or a local minimum, thenthe above matrix must have rank less than m+ 1 which requires the rows to be linearlydependent. Thus, there exist m scalars,
λ 1, · · · ,λ m,
and a scalar µ, not all zero such that
µ
fx1 (x0)...
fxn (x0)
= λ 1
g1x1 (x0)...
g1xn (x0)
+ · · ·+λ m
gmx1 (x0)...
gmxn (x0)
. (6.17.49)
If the column vectors g1x1 (x0)...
g1xn (x0)
, · · ·
gmx1 (x0)...
gmxn (x0)
(6.17.50)
are linearly independent, then, µ ̸= 0 and dividing by µ yields an expression of the form fx1 (x0)...
fxn (x0)
= λ 1
g1x1 (x0)...
g1xn (x0)
+ · · ·+λ m
gmx1 (x0)...
gmxn (x0)
(6.17.51)
at every point x0 which is either a local maximum or a local minimum. This proves thefollowing theorem.
Theorem 6.17.1 Let U be an open subset ofRn and let f : U→R be a C1 function. Then ifx0 ∈U is either a local maximum or local minimum of f subject to the constraints 6.17.46,then 6.17.49 must hold for some scalars µ,λ 1, · · · ,λ m not all equal to zero. If the vectorsin 6.17.50 are linearly independent, it follows that an equation of the form 6.17.51 holds.