0.1 Tangent Spaces and Lagrange Multipliers

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1 01 TANGENT SPACES AND LAGRANGE MULTIPLIERS 1 01 Tangent Spaces and Lagrange Multipliers If a differentiable function G = (G 1,, G k ) : E n+k E k then the surface S defined by S = { x G( x) = v} is called the level surface for G( x) = v Note that each of the functions G i : E n+k R If we denote by S i the level surface for the equation G i ( x) = v i, then S = k i=1 S i Suppose that x 0 = (x 0 1,, x 0 n+k ) S and that G ( x 0 ) L(E n+k, E k ) has rank k Let δ i,j = 1 if i = j and 0 if i j With respect to the standard basis { e j = (δ 1,j,, δ n+k,j ) j = 1, 2,, n + k} for E n+k and the analogous smaller basis for E k, we note that the matrix [ G ( x 0 )] k (n+k) has its whole set of k row vectors linearly independent, and these row vectors are the gradient vectors G 1 ( x 0 ), G 2 ( x 0 ),, G k ( x 0 ) Let φ : R S be a differentiable function for which φ(0) = x 0 Then we call the vector v = φ (0) a tangent vector to S at x 0 Definition 011 The tangent space T x 0(S) at x 0 S is the set of all tangent vectors to S at x 0 The translate x 0 + T x 0 = { x 0 + v v T x 0(S)} is called the tangent plane to the surface S, with point of tangency at x 0 A translate a + V = { a + x x V } of a vector subspace V of E n is called an affine subspace of E n An affine subspace is a vector subspace if and only if a V (See Exercise 1) Theorem 011 Let G : E n+k E k be a differentiable function Let x 0 S = { x G( x) = v} Suppose G ( x 0 ) has rank k Then the tangent space T x 0(S) is the vector subspace T x 0(S) = ( span R { G 1 ( x 0 ),, G k ( x 0 )} ) of dimension n In words, T x 0(S) is the orthogonal complement of the span of the k gradient vectors G 1 ( x 0 ),, G k ( x 0 )

2 2 Proof Suppose first that v = φ (0) T x 0(S) This implies that φ maps into each level surface G i ( x) = v i We will show that v G i ( x 0 ) for each i = 1,, k In fact, G i ( φ(t)) v i, a real constant We differentiate using the Chain Rule to find that G i( φ(0)) φ (0) = 0 In terms of the matrix representation of the left side of the latter equation, we have G i ( x 0 ) φ (0) = 0, so that v G i ( x 0 ) This shows that T x 0(S) G i ( x 0 ) for each i This implies that T x 0(S) ( span R { G 1 ( x 0 ),, G k ( x 0 )} ) The hypothesis that rank( G ( x 0 )) = k implies that dimt x 0(S) n If we can show that the tangent space is at least n-dimensional, then it will have to be the entire orthogonal complement of the span of the gradient vectors as claimed Thus it will suffice to produce a linearly independent set of n vectors in the tangent space Because the rank of a matrix is also the number of linearly independent column vectors, it follows that the matrix [ G ( x 0 )] has k independent columns We can rearrange the order of the n elements of the standard basis of E n+k to arrange that the first k columns are linearly independent By the Implicit Function Theorem, there exists an open set U E k containing (x 0 1,, x 0 k ) and an open set V En containing (x 0 k+1,, x0 k+n ) such that there are unique differentiable functions solving the equation x 1 = ψ 1 (x k+1,, x k+n ) x k = ψ k (x k+1,, x k+n ) G(ψ 1 (x k+1,, x k+n ),, ψ k (x k+1,, x k+n ), x k+1,, x k+n ) = v Next we define n differentiable curves on S by the equations φ 1 (t) = (ψ 1 (x 0 k+1 + t, x0 k+2,, x0 k+n ),, ψ k(x 0 k+1 + t, x0 k+2,, x0 k+n ), x0 k+1 + t, x0 k+2,, x0 k+n ) φ n (t) = (ψ 1 (x 0 k+1,, x0 n+k 1, x0 k+n + t),, ψ k(x 0 k+1,, x0 k+n 1, x0 k+n + t), x0 k+1, x0 k+n 1, x0 k+n + t) In comparing the vectors φ i(0) for i = 1,, n, observe that for each of these vectors the final n entries are all 0 except for a single entry which is 1 The location of the 1 is different for each of these vectors Thus the n vectors are independent and the theorem is proved

3 01 TANGENT SPACES AND LAGRANGE MULTIPLIERS 3 Corollary 011 Let G : E k+n E k be a differentiable function and let x 0 S = { x G( x) = v} Suppose x 0 is a local extreme point of a differentiable function f : S R and that G ( x 0 ) has rank k Then there exist numbers λ 1,, λ k such that f( x 0 ) = λ 1 G 1 ( x 0 ) + + λ k G k ( x 0 ) (1) The numbers λ 1,, λ k are called Lagrange multipliers Proof If φ : R S is a differentiable curve on S with φ(0) = x 0, let ψ(t) = f( φ(t)) Since this function has an extreme point at 0, we have ψ (0) = f( φ(0)) φ (0) = 0 It follows from Theorem 011 that f( x 0 ) is orthogonal to the tangent space T x 0(S) Since the co-dimension of T x 0(S) is k, it follows that f( x 0 ) lies in the span of the k vectors G 1 ( x 0 ),, G k ( x 0 ) This proves the corollary The method of Lagrange multipliers permits an optimization problem to be replaced by a problem of solving a system of equations From the k + n components of the vectors in Equation 1, we obtain a system of k + n equations in the n+2k unknowns x 1,, x k+n, λ 1,, λ k We get k additional equations from the k components of the equation G( x) = v Thus we obtain a system of n+2k equations in n+2k unknowns Although we have replaced a calculus problem with an algebraic problem, the algebraic problem can be challenging Nevertheless, the method of Lagrange multipliers is a powerful tool for optimization problems Example 011 We will begin with a three-dimensional example Consider the surface S defined by the equation x 4 + y 4 + z 4 = 1 in E 3, shown in Figure 1 We will find both the maximum and the minimum values of the function f( x) = x 2 +y 2 +z 2 on S (In effect, we are determining the closest and furthest distances from the origin on S) In this example, we denote x = (x, y, z) Observe that if we define G( x) = x 4 + y 4 + z 4 then S = G 1 ({1}) Hence S is closed because G is continuous S is also bounded (Why?) Hence the function f must achieve both a maximum and a minimum value somewhere on S Since S is smooth at all points and since G is non-vanishing on S,

4 4 y z x Figure 1: x 4 + y 4 + z 4 = 1 the extreme points must occur at those points for which f( x) = λ G( x) This yields the following system of equations x(1 2λx 2 ) = 0 y(1 2λy 2 ) = 0 z(1 2λz 2 ) = 0 x 4 + y 4 + z 4 = 1 The reader should check the following by making the necessary calculations If none of the three variables is zero, then x 2 = y 2 = z 2 = 1 2λ showing that λ = ± 3 2 This implies that f(x, y, z) = 3 If exactly one of the three variables is zero, then at a point satisfying the system of equations we must have f(x, y, z) = 2 If exactly two of the variables are zero, then at a point satisfying the system we must have f(x, y, z) = 1

5 01 TANGENT SPACES AND LAGRANGE MULTIPLIERS 5 It follows that the maximum value of f on S is 3 But the reader should be able to explain why at least one of the variables must be non-zero Thus the minimum value is 1 There is also an easy way to explain even from the outset why f(x, y, z) 1 everywhere on S Exercises 01 1 Prove that the tangent plane x 0 + T x 0(S) is a vector subspace of E n if and only if x 0 T x 0(S) 2 Describe both the tangent space and the tangent ( plane to the sphere S n 1 = { x E n x = 1} at the point x 0 1 = 1 n, n 1,, n ) 3 The sphere S 3 E 4 is defined by { S 3 = x } 4 x 2 i = 1 i=1 Define f : S 3 R by f( x) = 4 i=1 a ix i where a i is a constant for each i {1, 2, 3, 4} Show that the maximum and minimum values of f on S 3 are ± 4 i=1 a2 i 4 The group SL(2, R) of matrices was defined in Exercise???? Let f : SL(2, R) R be defined by f( x) = x x x x 2 4, where we identify the matrix ( ) x1 x X = 2 SL(2, R) x 3 x 4 with the vector x = (x 1, x 2, x 3, x 4 ) constrained to the surface S in E 4 defined by the equation x 1 x 4 x 2 x 3 = 1 (a) Prove that f achieves a minimum value on S but that it has no maximum (b) Use the method of Lagrange multipliers to find the minimum value of f on S

6 6 5 Let f( x) = 4 i=1 x2 i for all x E 4 Let S 1 be the surface in determined by x 1 x 4 x 2 x 3 = 1 and let S 2 be the surface defined by 4 i=1 x i = 2 Let S = S 1 S 2 (a) Prove that f( x) has a minimum value on S but no maximum (b) Find the minimum value of f( x) on S