MATH 174A: PROBLEM SET 3. Suggested Solution. det(a + tb) = (det A) det(i + ta 1 B)

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1 MATH 174A: PROBLEM SET 3 Suggested Solution Problem 1. (Cf. Taylor I.1.3.) Let M n n (C) denote the set of n n complex matrices. Suppose A M n n (C) is invertible. Using show that det(a + tb) = (det A) det(i + ta 1 B) D det(a)b = (det A) Tr(A 1 B). (Hint: you have already shown D det(i)b = Tr B.) Note: this shows that SL n (C) defined as the set of matrices A M n n with det A = 1 is a C, indeed holomorphic, (hyper)surface in M n n = C n2 : take B = A to conclude that D det(a) is surjective. (The same calculation using M n n (R) shows that SL n (R) is real analytic.) Solution. D det(a)b = d dt t= det(a + tb) = (det(a)) d dt det(i + ta 1 B) = (det(a))d det(i)(a 1 B) = (det(a)) Tr(A 1 B). Problem 2. Let O n (R) denote the set of matrices A M n n (R) with the property that AA t = I; here A t denotes the transpose of A (i.e. the ij entry of A t is the ji entry of A). Let S n n (R) denote the set of symmetric matrices, i.e. matrices A such that A t = A. Note that S n n can be identified with R n(n+1) 2 as for symmetric matrices A, the below diagonal entries are determined by the above diagonal entries. Consider the map F : M n n S n n given by F (A) = AA t. Show that (DF )(A)B = AB t + BA t, t= and show that for A O n (R), DF (A) : M n n S n n is surjective. Use this to show that O n (R) is a compact surface in M n n of dimension n(n 1) 2. O n (R) is called the orthogonal group on R n.

2 2 Solution. (DF )(A)(B) = d ds F (A + sb) s= = d ds s= (A + sb)(a + sb)t = d ds (AAt + sab t + sba t + s 2 BB t ) = AB t + BA t. To show that for any A O n (R), DF (A) : M n n S n n is surjective, pick an arbitrary C S n n (i.e. C t = C), take B = 1 CA, then 2 s= (DF )(A)(B) = AB t + BA t = 1 2 (AAt C t + CAA t ) = 1 2 (Ct + C) = C, where we have used the fact that AA t = I as A O n (R). For the last statement, first observe that O n (R) = {A M n n : AA t = I} = F 1 (I). F is clearly continuous map, hence O n (R) is closed as the preimage of a closed set. Also, O n (R) can be interpreted as those matrices such that its columns (or rows) form an orthonormal basis of R n, thus the abolute values of its entries are all no bigger than 1. Hence O n (R) is a closed and bounded set in R n2, which is compact by Heine-Borel. Since DF (A) is surjective at each A O n (R), by implicit function theorem, O n (R) has dimension equals to dim(m n n ) dim(s n n ) = n 2 n(n+1) = n(n 1). 2 2 Problem 3. Suppose that M is a smooth k-dimensional surface in R n. Show that for each p M, the set of vectors tangent to M at p form a k-dimensional linear subspace of R n. (Hint: Use the straightening out of the previous problem set.) We denote this by T p M. Show also that if on a neighborhood O of p in R n, M is defined by Φ =, then T p M is the nullspace of DΦ(p) : R n R n k. Use this to conclude that the disjoint union of tangent spaces T p M, p M, is a 2kdimensional surface in R 2n : consider the set and show that the map T M = {(p, v) R 2n : p M, v T p M}, F : O R n R 2(n k), F (x, v) = (Φ(x), DΦ(x)v) defines T M on O R n. (That is, T M is given by F =, and DF is surjective on T M.) T M is called the tangent bundle of M. Note: Let (.,.) denote the standard inner product on R n. Every v R n defines a linear map ι(v) : R n R by ι(v)w = (v, w). Conversely, for every linear map A : R n R there

3 is a vector v R n such that for all w R n, (v, w) = Aw, i.e. ι is surjective. (ι is also injective.) Now, if Φ = (Φ 1,... Φ n k ), then DΦ j (p) is a linear map from R n to R, Φ j (p) denotes the image of DΦ j (p) under ι 1 ; it is of course just ( 1 Φ j,..., n Φ j ). Thus, we can reinterpret the result above: T p M is the orthocomplement of the span of Φ 1 (p),..., Φ n k (p). 3 Solution. Using the straightening out techinque in our previous problem set, we know that around a small neighborhood of p, the surface can be parametrized by m coordinates x 1,..., x m by the map g(x 1,..., x m ) = (x 1,..., x m, f 1 (x 1,..., x m ),..., f n k (x 1,..., x m )). Then, it is easy to see that the vectors g x 1,..., g x k forms a basis for the tangent space of M at p. Assume that around a small neighborhood of p, U, M is defined by the equation Φ =. Then any curve lying on O through p is constantly zero under the map, this shows that the differential map DΦ vanishes on every tangent vectors of M at p. Since DΦ is surjective as a defining function of M, be considering dimensions, we conclude that T p M is exactly the kernel of DΦ. Define F : O R n R 2(n k) as F (x, v) = (Φ(x), DΦ(x)v). Then F (x, v) = if and only if Φ(x) = and DΦ(x)(v) = if and only if x O and v T p M. Hence, T M is given by F =. Note that [ ] DΦ O DF =, DΦ which is surjective since DΦ is surjetive. Problem 4. (Taylor I.4.9.) Given X M n, define ad X End(M n ), that is, ad X : M n M n by ad X(Y ) = XY Y X. Show that e tx Y e tx = e t ad X Y. (Hint: If V (t) denotes either side, show that dv/dt = (ad X)(V ), V () = Y.) Solution. Follow the hint, it is clear that both sides satisty the initial condition V () = Y. Moreover, d dt e tx Y e tx = Xe tx Y e tx + e tx Y Xe tx = X(e tx Y e tx ) + (e tx Y e tx )X, d dt e t ad X Y = ad Xe t ad X Y. Hence both sides satisfies the ODE dv/dt = (ad X)(V ), V () = Y, by uniqueness theorem for this kind of ODE with constant coefficients, we know that both sides agree on all t R.

4 4 Problem 5. (Taylor I.5.1.) Let A(t) and X(t) be n n matrices satisfying dx dt = A(t)X. We form the Wronskian W (t) = det X(t). Show that W satisfies the ODE dw = a(t)w, a(t) = Tr A(t). dt (Use the alternative hint: Write X(t + h) = e ha(t) X(t) + O(h 2 ) and use Exercise 3 of section 4 to write det e ha(t) = e ha(t), hence W (t + h) = e ha(t) W (t) + O(h 2 ).) Solution. For h small enough, we have from the Taylor expansion X(t + h) = X(t) + h dx dt (t) + O(h2 ) = X(t) + ha(t)x + O(h 2 ) = e ha(t) X + O(h 2 ). Taking determinant on both sides, using Exericse 3 or section 4, W (t+h) = det(e ha(t) ) det(x(t))+o(h 2 ) = e Tr ha(t) det(x(t))+o(h 2 ) = e ha(t) W (t)+o(h 2 ), hence dw/dt = a(t)w (t). Problem 6. (Taylor I.5.2.) Let u(t) = y(t) 2, for a solution y to (5.1). Show that u M(t)u(t), provided A(t) M(t)/2. Such a differential inequality implies the integral inequality u(t) A + M(s)u(s) ds, t, with A = u(). The following is a Gronwall inequality; namely, if (5.17) holds for a real-valued function u, then provided M(s), we have, for t, u(t) Ae N(t), N(t) = M(s)ds. Prove this. Note that the quantity dominating u(t) in (5.18) is equal to U, solving U() = A, du/dt = M(t)U(t). Solution. Write u(t) = (y(t), y(t)) where (.,.) denotes the standard inner product in R n, we have u (t) = 2(y(t), y (t)) 2 y(t) y (t) = 2 y(t) A(t)y(t) M(t)u(t). Direct integration on both sides give the integral form of this inquality. Let N(t) = M(s)ds. Rearranging (5.17) gives (when A =, there is nothing to prove, so we can assume A > ) u(t) A + 1. M(s)u(s)ds

5 Multiplying both sides by M(t) (we use the non-negativity of M here) then integrate ( ) ln A + M(s)u(s)ds ln(a) N(t), Taking exponential and apply (5.17) again u(t) A + M(s)u(s)ds Ae N(t). 5