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.