Homework set 4 - Solutions

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1 Homework set 4 - Solutions Math 407 Renato Feres 1. Exercise 4.1, page 49 of notes. Let W := T0 m V and denote by GLW the general linear group of W, defined as the group of all linear isomorphisms of W onto itself. For each σ in the symmetric group S m consisting of all the permutations of {1,,...,m} we have defined the multilinear map on V given by Πσv 1,..., v m = v σ 1 1 v σ 1 m for σ S m. In this problem we regard Πσ as a linear map on W. So we write Πσv 1 v m = v σ 1 1 v σ 1 m. Show that Πσ GLW for each σ S m and that Πσσ = Πσ Πσ and Πσ 1 = Πσ 1. This means that Π : S m GLW is a group homomorphism, and that Π defines a linear representation of S m with representation space W. Solution. For σ,σ S m, Πσσ v 1 v m = v σσ 1 1 v σσ 1 m = v σ 1 σ 1 1 v σ 1 σ 1 m = Πσ v σ 1 1 v σ 1 m = Πσ Πσ v 1 v m. It is immediate from the definition that if e is the identity element in the group, then Πev 1 v m = v 1 v m. Therefore, ΠσΠσ 1 = Πσσ 1 = Πe = I, from which we obtain that Πσ 1 = Πσ 1.. Exercise 4., page 50 of notes. Show that the antisymmetrizing operator A, defined by A = 1 signσπσ satisfies A = A.

2 Solution. Keep in mind the following observation: A = 1 signσ σπσ σ for any σ S m. Then A 1 = signσ Πσ 1 signσπσ = 1 1 signσ signσπσ Πσ = 1 1 signσ σπσ σ = 1 A = A. 3. Exercise 4.9, page 55 of notes. Show that if ω is an alternating n-form on the n-dimensional real vector space V and T : V V is any linear map, then ωt v 1,...,T v n = dett ωv 1,..., v n, for arbitrary vectors v 1,..., v n V. Solution. We may suppose that ω 0 since the claim is trivial otherwise. Thinking of ω as a non-zero linear map on V V makes it clear that the set of vectors v 1,..., v n for which ωv 1,..., v n 0 is dense. This is because the complement set of the kernel of a nonzero linear map is a dense set. So, by continuity, it is sufficient to prove the identity under the assumption ωv 1,..., v n 0. In particular, α = {v 1,..., v n } must be a basis of V. Let {v 1,..., v n } be the dual basis. Recall that v i T v j is the i, j -entry of the matrix [T ] representing T in the basis α. So the isomorphism between R n and V obtained under the basis α sends the column vector [T ] 1j,...,[T ] n j t to T v j. This makes it clear that the function F [T ] := ωt v 1,...,T v n ωv 1,..., v n is multilinear when regarded as a function of the columns of [T ]. It is also clear that this is antisymmetric and that F I = 1, where I is the identity matrix. But the determinant is uniquely characterized by these properties. Since the determinant of the matrix of a linear transformation is the same number regardless of choice of basis, we must have Consequently, the claimed identity holds. dett = ωt v 1,...,T v n. ωv 1,..., v n 4. Exercise 4.10, page 56 of notes. If θ is a k-form and ω is an l -form, show that θ ωv 1,..., v k+l = 1 signσθv σ1,..., v σk ωv σk+1,..., v σk+l. k!l! σ S k+l Solution. The wedge product of forms is defined in the notes by k + l θ ω = θ ω A. k

3 The definition of A yields k + l 1 θ ωv 1,..., v k+l = signσθ ωv k k + l! σ 1 1,..., v σ 1 k+l σ S k+l = 1 signσθv σ1,..., v σk ωv σk+1,..., v σk+l. k!l! σ S k+l Note that, summing over all σ 1 gives the same result as summing over all the σ. Thus the stated identity holds. 5. Exercise 5.10, page 65 of notes. If D is the determinant function, show that it is everywhere differentiable and its directional derivative at A in direction W is dd A W = DAtrW A 1 for all A GLn,R and all W Mn,R. Solution. It is clear that DA is differentiable to every order at all points since it is a polynomial function of the entries of A. Note that DA + tw = DI + tw A 1 DA. So dd A W = d DA + tw = DA d DI + tw A 1. Thus it suffices to prove the identity for A = I. That is, it suffices to show that d DI + tw = trw. Expressing the determinant explicitly as a function of the columns of the matrices, we have d DI + tw = d n De 1 + t w 1,...,e n + t w n = De 1,...,e i 1, w i,e i+1,...,e n = trw. Thus the stated identity holds. 6. Let M symm n,r Mn,R be the space of symmetric matrices. Define F : Mn,R M symm n,r by F A = A t A. i=1 a Show that for all A, X Mn, R df A X = A t X + X t A. b Show that if A is invertible, then df A : Mn,R M symm n,r is surjective. Solution. a We have df A = d F A + t X = d A + t X t A + t X = d A t A + ta t X + X t A + t X t X = A t X + X t A. b Let A be invertible. We need to show that for an arbitrary symmetric matrix S, there exists X such that A t X + X t A = S. But this holds for X = 1 At 1 S since A t X + X t A = 1 At A t 1 S + 1 S A 1 A = S. 3

4 7. Exercise 5.13, page 67 of notes. Spectral theorem for symmetric matrices. Let f : U R be a differentiable function defined on an open subset of R n equipped with the standard inner product. The sphere of radius 1 centered at the origin is indicated by S n 1 = {x R n : x = 1}. a Show that grad x f is orthogonal the kernel of d f x. b Show that grad x f is the maximum rate of change of f along any direction: max d f x v = grad x f v =1 and that the maximum is achieved when v = grad x f / grad x f. c Let A Mn,R be a symmetric matrix and define f : R n R by f x = 1 Ax, x. Show that if x S n is a point where f achieves its maximum or its minimum valued, then x is an eigenvector of A. d Show that there exists an orthonormal basis of R n consisting of eigenvectors of A. For this, use a finite induction starting from the existence of one eigenvector, then restricting the function to the intersection of the sphere with the subspace orthogonal to the previously obtained eigenvectors. Solution. a Let u be any vector in the kernel of d f x, so that d f x u = 0. By definition grad x f,u = d f x u = 0 proving the claim. b By the Schwarz inequality, d f x v = grad x f, v grad x f v. On the other hand, grad d f x f x grad x f = grad grad x f, x f = grad grad x f x f. This shows that the maximum of d f x v over vectors v of length 1 is the norm of the gradient, and the maximum is attained when v is the direction of the gradient. c A point x of maximum or minimum is a critical point for f, meaning that d f x v = 0 for all v tangent to S n 1 at x. Note that x is perpendicular to all these tangent vectors. This means that the gradient vector of f at x is parallel to x: grad x f = λx for some λ R n. On the other hand, the gradient of f x satisfies grad x f, w = d f x w = d 1 Ax + t w, x + t w = 1 Ax, w + Aw, x = Ax, w = Ax, w. Therefore, the gradient of f at any point x is grad x f = Ax. Thus if x is a critical point we have Ax = λx for some λ. But this means that x is an eigenvalue of A. 4

5 d Let V be the n 1-dimensional subspace of R n perpendicular to the critical point of R n. Since A is symmetric, 0 = λ x, v = Ax, v = x, Av. This means that Av is also perpendicular to x. So V is invariant under A. We can now repeat the argument of c for the sphere in V. A simple finite induction now gives an orhonormal basis of R n consisting of eigenvectors of A. 5