that we cannot expect a matrix A to have a logarithm whenever det A =

0. Thus, in the following we restrict to invertible matrices A. The most

elegant de¬nition of logarithms of matrices is by means of Cauchy™s integral

formula: Let an invertible matrix A be given, whose eigenvalues »k , 1 ¤

k ¤ µ, necessarily are all nonzero. We de¬ne

log A = (2πi)’1 log z (zI ’ A)’1 dz, (C.1)

γ

where γ is not a single curve, but rather a collection of small circles, posi-

tively oriented, one about each eigenvalue »k , and none of them encircling

the origin. Hence in fact, we do not have one integral but a sum of ¬nitely

many, one for each such circle. The radii of these circles should be taken

so small that the corresponding closed discs do not intersect, and we may

make arbitrary choices for a branch of log z along each one of the circles.

Every matrix log A computed according to this formula will be called a

branch of the logarithm of A. Note that we cannot hope for a single-valued

function, since the logarithm already is multivalued in the scalar case.

From the above de¬nition one can deduce a list of rules that enable us, in

principle, to compute the branches for log A. Note, however, that formulas

containing more than one matrix logarithm hold only if, after choosing

branches for all but one logarithm, we choose the correct branch for the

remaining one.

• For a diagonal invertible matrix Λ = diag [»1 , . . . , »ν ], we have

log Λ = diag [log »1 , . . . , log »ν ],

where on the right we choose branches of the logarithms such that

»j = »k implies log »j = log »k .

• For a nilpotent matrix N , one (the main) branch of the logarithm is

(’1)n+1 n

log(I + N ) = N.

n

n≥1

Observe that the series terminates. Every other branch is obtained

from the main one by adding 2kπiI, k ∈ Z.

• For a Jordan matrix J = Λ + N = Λ(I + Λ’1 N ), with invertible

diagonal Λ and nilpotent N , commuting with Λ, we have

log J = log Λ + log(I + Λ’1 N ),

C.2 Logarithms of a Matrix 241

provided that the logarithms on the right are computed according to

the rules stated above; observe that then the matrices on the right

commute!

• For A = T ’1 J T , with J in Jordan form, we have

log A = T ’1 log(J) T.

It can be shown that the following identities for logarithms of matrices hold;

observe, however, what was said above about formulas containing several

logarithms.

1. For every invertible A we have

exp[log A] = A.

However, for certain A there are X with exp[X] = A so that no

branch of log A equals X.

2. For B = T ’1 A T we have

log B = T ’1 (log A) T.

3. For A, B commuting and invertible, the rule

log(A B) = log A + log B

holds modulo the following restriction: If branches for any two out of

the three logarithms are selected, then the formula is correct provided

that we select the branch for the third accordingly.

4. For triangularly blocked invertible matrices A with diagonal blocks

Ak , all branches of log A are likewise blocked and have diagonal blocks

log Ak in the same order.

For our purposes it is important to be able to solve equations of the form

exp[2πiX] = C, for given invertible C. From what we have said above

about logarithms of matrices, one can show that the following is correct:

Theorem 71 Given any invertible matrix C, there exist matrices M such

that exp[2πiM ] = C; namely, M = (2πi)’1 log C. If C is triangularly

blocked of some type, then so are the matrices M . The eigenvalues of any

two such M can only di¬er by integers, and there is a unique matrix M

whose eigenvalues have real parts in the half-open interval [0, 1).

242 Appendix C. Functions of a Matrix

Exercises: In what follows, consider the matrices

0 0 1 0

A= , J= .

1 2πi 1 1

1. Show that eA = I.

2. Show that the above matrix A is not a branch of log I.

3. For A = I + N , N nilpotent, show

(zI ’ A)’1 = (z ’ 1)’n’1 N n

n≥0

(observe that the series terminates). Use this to obtain the formula

given for the main branch of log(I + N ) from (C.1).

Solutions to the Exercises

Here we give some detailed hints toward the solution of the exercises, in

particular to those that are used in later chapters of the book. Exercises

left out here require only straightforward computation.

Section 1.1:

1. Since H(G, C ) is a vector space over C , it su¬ces to verify the

subspace properties.

z

2. De¬ne x0 (z) ≡ x0 , xk+1 (z) = A(u) xk (u) du, k ≥ 0, and show the

z0

following:

(a) Every xk (z) is holomorphic in D, k ≥ 0.

(b) For every ρ < ρ, there exists K = K(˜) such that xk (z) ¤

˜ ρ

K |z ’ z0 | x0 /k!, |z ’ z0 | ¤ ρ, k ≥ 0.

k k

˜

∞

(c) The series x(z) = 0 xk (z) converges compactly on D, i.e.,

uniformly on compact subsets of D, and represents a solution of

the initial value problem.

(d) If x(z) is any other solution of the initial value problem, then

˜

for every ρ < ρ there exist K = K(˜) and M = M (˜) such that

˜ ρ ρ

x(z) ’ x(z) ¤ M K |z ’ z0 | /k!, |z ’ z0 | ¤ ρ, k ≥ 0; thus for

k k

˜ ˜

k ’ ∞ we conclude x(z) ≡ x(z).

˜

3. Proceed as above, estimating on compact sets that are star-shaped

with respect to z0 .

244 Solutions

4. For G = C , the proof is completed. For every other G, there is a

function φ, holomorphic in the unit disc D and mapping D bijectively

˜

into G. For x(z) = x(φ(z)), A(z) = φ (z) A(φ(z)), verify that (1.1) is

˜

˜

equivalent to x = A(z) x, z ∈ D.

˜ ˜

6. (a) a1 (z) = 2z (1 ’ z 2 )’1 , a2 (z) = ’µ(1 ’ z 2 )’1 .

(b) (n + 1)(n + 2) xn+2 = [n(n + 1) ’ µ] xn , n ≥ 0. Since all solutions

are holomorphic in the unit disc, the power series converges for

at least |z| < 1. The quotient test, applied to the odd and even

part of a solution, shows that the radius of convergence equals

1, except when we have a polynomial solution.

(c) For m even (odd), let all xn with odd (even) n be zero. Then for

x0 = 1 (x1 = 1), we obtain a unique polynomial of degree m.

(d) Follows from the recursion formula.

∞ (k)

0 tn (z ’ z0 ) , for z ∈ D(z0 , ρ) ‚ G,

n

7. (a) Expanding tk (z) =

(k)

show by induction with respect to k that tn depends linearly

(0) (0) (0)

upon t0 , . . . , tn+k , but is independent of the remaining tm ,

(k) (0)

m ≥ n+k+1. In particular, verify tn = (n+k)! tn+k /n!+t, with˜

(0) (0)

˜

t depending only upon t0 , . . . , tn+k’1 . Next, show by induction

(0) (0)

that one can choose the vectors t0 , . . . , tk so that the matrix

(0) (k)

Tk with rows t0 , . . . , t0 has maximal rank.

(b) Show T (z) + T (z) A(z) = B(z) T (z) on D. Also, use Theorem 1

(p. 4) to conclude that x(z) = T ’1 (z) x(z), which is de¬ned and

˜

holomorphic on D, can be holomorphically continued into all of

the region G.

Section 1.2:

2. Compare Exercise 5 on p. 239 for how to di¬erentiate the exponential

of a matrix.

3. As above.

5. Set A(z) = X (z) X ’1 (z).

6. Di¬erentiate the identity X(z) X ’1 (z) ≡ I to obtain [X ’1 (z)] =

’X ’1 (z) X (z) X ’1 (z), z ∈ G.

˜ ˜

7. For a fundamental solution X(z) of (1.1), conclude X(z) = X(z) C,

with a constant matrix of suitable size. Use this and elementary re-

sults on the rank to see that X(z) has the same rank as C.

Solutions 245

8. Take xµ+1 (z), . . . , xν (z) constant so that the said determinant is

nonzero at some selected point z0 ∈ G, and then conclude by holo-

morphy that it is nonzero on some disc around z0 . The rest should

be straightforward.

Section 1.3:

1. Compare Section C.1 for how to di¬erentiate the matrix z M .

2. Take M = »I, hence z M = z » I, and distinguish the cases » ∈ Z,

» ∈ Q \ Z, » ∈ R \ Q, » ∈ C \ R.

4. Consider X(z) = (z’z1 )M1 ·. . .·(z’zµ )Mµ and use the commutativity

of the Mk .

5. Compare Exercise 5 on p. 7.

6. The proof of the statement concerning monodromy factors is straight-

forward. Concerning the monodromy matrices, conclude that for some

constant invertible matrix C we must have that z M1 C z ’M2 is single-

valued. Then, show the same for each Mk replaced by its Jordan

canonical form Jk , and conclude that this implies e2πiJ1 C = Ce2πiJ2 .

Finally, relate the eigenvalues of e2πiJk and Jk .

7. Use properties of logarithms of matrices, as stated in Section C.2.

8. Show existence of a fundamental solution X(z) with a monodromy

matrix in (lower triangular) Jordan form, and consider the last col-

umn of X(z).

9. Use the previous exercise, together with Exercise 5 on p. 4.

Section 1.4:

n

1. (n + 1)xn+1 = bn + m=0 An’m xm , n ≥ 0; thus, x0 can be chosen

arbitrarily, while the other coe¬cients are determined.

d d

n n

2. For b(z) = n=0 bn z , x(z) = n=0 bn z , we have (1.10) if and

only if (n + 1)xn+1 = bn + A xn , 0 ¤ n ¤ d ’ 1, and 0 = bd + A xd .

In order that the last equation can be satis¬ed for arbitrary bd , we

have to assume that A is invertible. If this is so, then the previous

equations can all be solved for xn . Hence, the polynomial solution

even is unique!

246 Solutions

4. For x(z) as in (1.11), let x(z exp[2πi]) denote its holomorphic contin-

uation about the origin in the positive sense. Show

z

2πi 2πi

X ’1 (ue2πi ) b(u) du

x(ze ) = X(ze ) c+c+

˜

z0

z

2πiM

X ’1 (u) b(u) du ,

= X(z) e (c + c) +

˜

z0