MATH 174A: PROBLEM SET 3

Suggested Solution

Problem 1. (Cf. Taylor I.1.3.) Let Mn×n(C) denote the set of n× n complex matrices.Suppose A ∈ Mn×n(C) is invertible. Using

det(A + tB) = (det A) det(I + tA−1B)

show that

D det(A)B = (det A) Tr(A−1B).

(Hint: you have already shown D det(I)B = Tr B.)

Note: this shows that SLn(C) defined as the set of matrices A ∈ Mn×n with det A = 1 is

a C∞, indeed holomorphic, (hyper)surface in Mn×n = Cn2: take B = A to conclude that

D det(A) is surjective. (The same calculation using Mn×n(R) shows that SLn(R) is realanalytic.)

Solution.

D det(A)B =d

dtdet(A + tB)

∣∣∣∣t=0

= (det(A))d

dtdet(I + tA−1B)

∣∣∣∣t=0

= (det(A))D det(I)(A−1B)

= (det(A)) Tr(A−1B).

Problem 2. Let On(R) denote the set of matrices A ∈ Mn×n(R) with the property thatAAt = I; here At denotes the transpose of A (i.e. the ij entry of At is the ji entry of A).Let Sn×n(R) denote the set of symmetric matrices, i.e. matrices A such that At = A. Note

that Sn×n can be identified with Rn(n+1)

2 as for symmetric matrices A, the below diagonalentries are determined by the above diagonal entries.

Consider the map F : Mn×n → Sn×n given by F (A) = AAt. Show that

(DF )(A)B = ABt + BAt,

and show that for A ∈ On(R), DF (A) : Mn×n → Sn×n is surjective.

Use this to show that On(R) is a compact surface in Mn×n of dimension n(n−1)2

. On(R) iscalled the orthogonal group on Rn.

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Solution.

(DF )(A)(B) =d

dsF (A + sB)

∣∣∣∣s=0

=d

ds(A + sB)(A + sB)t

∣∣∣∣s=0

=d

ds(AAt + sABt + sBAt + s2BBt)

∣∣∣∣s=0

= ABt + BAt.

To show that for any A ∈ On(R), DF (A) : Mn×n → Sn×n is surjective, pick an arbitraryC ∈ Sn×n (i.e. Ct = C), take B = 1

2CA, then

(DF )(A)(B) = ABt + BAt =1

2(AAtCt + CAAt) =

1

2(Ct + C) = C,

where we have used the fact that AAt = I as A ∈ On(R).

For the last statement, first observe that On(R) = {A ∈ Mn×n : AAt = I} = F−1(I). F isclearly continuous map, hence On(R) is closed as the preimage of a closed set. Also, On(R)can be interpreted as those matrices such that its columns (or rows) form an orthonormalbasis of Rn, thus the abolute values of its entries are all no bigger than 1. Hence On(R)

is a closed and bounded set in Rn2, which is compact by Heine-Borel. Since DF (A) is

surjective at each A ∈ On(R), by implicit function theorem, On(R) has dimension equals

to dim(Mn×n)−dim(Sn×n) = n2 − n(n+1)2

= n(n−1)2

.

Problem 3. Suppose that M is a smooth k-dimensional surface in Rn. Show that foreach p ∈ M , the set of vectors tangent to M at p form a k-dimensional linear subspaceof Rn. (Hint: Use the straightening out of the previous problem set.) We denote this byTpM .

Show also that if on a neighborhood O of p in Rn, M is defined by Φ = 0, then TpM isthe nullspace of DΦ(p) : Rn → Rn−k.

Use this to conclude that the disjoint union of tangent spaces TpM , p ∈ M , is a 2k-dimensional surface in R2n: consider the set

TM = {(p, v) ∈ R2n : p ∈ M, v ∈ TpM},and show that the map

F : O × Rn → R2(n−k), F (x, v) = (Φ(x), DΦ(x)v)

defines TM on O × Rn. (That is, TM is given by F = 0, and DF is surjective on TM .)TM is called the tangent bundle of M .

Note: Let (., .) denote the standard inner product on Rn. Every v ∈ Rn defines a linearmap ι(v) : Rn → R by ι(v)w = (v, w). Conversely, for every linear map A : Rn → R there

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is a vector v ∈ Rn such that for all w ∈ Rn, (v, w) = Aw, i.e. ι is surjective. (ι is alsoinjective.)

Now, if Φ = (Φ1, . . . Φn−k), then DΦj(p) is a linear map from Rn to R, ∇Φj(p) denotesthe image of DΦj(p) under ι−1; it is of course just (∂1Φj, . . . , ∂nΦj).

Thus, we can reinterpret the result above: TpM is the orthocomplement of the span of∇Φ1(p), . . . ,∇Φn−k(p).

Solution. Using the straightening out techinque in our previous problem set, we knowthat around a small neighborhood of p, the surface can be parametrized by m coordinatesx1, . . . , xm by the map g(x1, . . . , xm) = (x1, . . . , xm, f1(x1, . . . , xm), . . . , fn−k(x1, . . . , xm)).Then, it is easy to see that the vectors ∂g

∂x1, . . . , ∂g

∂xkforms 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 = 0.Then any curve lying on O through p is constantly zero under the map, this shows that thedifferential map DΦ vanishes on every tangent vectors of M at p. Since DΦ is surjectiveas a defining function of M , be considering dimensions, we conclude that TpM is exactlythe kernel of DΦ.

Define F : O × Rn → R2(n−k) as F (x, v) = (Φ(x), DΦ(x)v). Then F (x, v) = 0 if and onlyif Φ(x) = 0 and DΦ(x)(v) = 0 if and only if x ∈ O and v ∈ TpM . Hence, TM is given byF = 0. Note that

DF =

[DΦ O∗ DΦ

],

which is surjective since DΦ is surjetive.

Problem 4. (Taylor I.4.9.) Given X ∈ Mn, define ad X ∈ End(Mn), that is, ad X :Mn → Mn by ad X(Y ) = XY − Y X. Show that

e−tXY etX = e−t ad XY.

(Hint: If V (t) denotes either side, show that dV/dt = −(ad X)(V ), V (0) = Y .)

Solution. Follow the hint, it is clear that both sides satisty the initial condition V (0) = Y .Moreover,

d

dte−tXY etX = −Xe−tXY etX + e−tXY XetX = −X(e−tXY etX) + (e−tXY etX)X,

d

dte−t ad XY = − ad Xe−t ad XY.

Hence both sides satisfies the ODE dV/dt = −(ad X)(V ), V (0) = Y , by uniquenesstheorem for this kind of ODE with constant coefficients, we know that both sides agreeon all t ∈ R.

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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

dt= a(t)W, a(t) = Tr A(t).

(Use the alternative hint: Write X(t + h) = ehA(t)X(t) + O(h2) and use Exercise 3 ofsection 4 to write det ehA(t) = eha(t), hence W (t + h) = eha(t)W (t) + O(h2).)

Solution. For h small enough, we have from the Taylor expansion

X(t + h) = X(t) + hdX

dt(t) + O(h2) = X(t) + hA(t)X + O(h2) = ehA(t)X + O(h2).

Taking determinant on both sides, using Exericse 3 or section 4,

W (t+h) = det(ehA(t)) det(X(t))+O(h2) = eTr hA(t) det(X(t))+O(h2) = eha(t)W (t)+O(h2),

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 +

∫ t

0

M(s)u(s) ds, t ≥ 0,

with A = u(0). The following is a Gronwall inequality ; namely, if (5.17) holds for areal-valued function u, then provided M(s) ≥ 0, we have, for t ≥ 0,

u(t) ≤ AeN(t), N(t) =

∫ t

0

M(s)ds.

Prove this. Note that the quantity dominating u(t) in (5.18) is equal to U , solvingU(0) = A, dU/dt = M(t)U(t).

Solution. Write u(t) = (y(t), y(t)) where (., .) denotes the standard inner product in Rn,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) =

∫ t

0

M(s)ds.

Rearranging (5.17) gives (when A = 0, there is nothing to prove, so we can assume A > 0)

u(t)

A +∫ t

0M(s)u(s)ds

≤ 1.

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Multiplying both sides by M(t) (we use the non-negativity of M here) then integrate

ln

(A +

∫ t

0

M(s)u(s)ds

)− ln(A) ≤ N(t),

Taking exponential and apply (5.17) again

u(t) ≤ A +

∫ t

0

M(s)u(s)ds ≤ AeN(t).