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the equation eX = A. This is not a very suitable de¬nition, but it shows
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)

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.

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

• 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

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

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

(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,
7. (a) Expanding tk (z) =
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:
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
2πi 2πi
X ’1 (ue2πi ) b(u) du
x(ze ) = X(ze ) c+c+
X ’1 (u) b(u) du ,
= X(z) e (c + c) +


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