Chapter 2 Linear maps and ODEs - Universiteit Utrechtkouzn101/NLDV/Lect2_3.pdf42 CHAPTER 2. LINEAR...

Chapter 2

Linear maps and ODEs

The aim of this chapter is to study linear dynamical systems and in particular

– to obtain insight in the qualitative features of their phase portrait and howthese relate to eigenvalues of the defining matrices;

– to derive estimates that settle the stability issue for the zero steady state;

– to discuss the more subtle issue of topological equivalence (conjugacy) of thesesystems.

2.1 Linear maps in Rn

Consider a linear mapx 7→ Ax, x ∈ R

n, (2.1)

where A = (aij)i,j=1,...,n is an n× n real matrix and Ax is the product of the matrixA and the vector x = (x1, x2, . . . , xn)

T :

(Ax)i =

n∑

j=1

aijxj , i = 1, . . . , n.

The map (2.1) will be denoted by A as well. Note however, that if another basis inRn is selected, the same map will be represented by a different matrix. We extend(2.1) to complex vectors by the rule

A(u+ iv) = Au+ iAv, u+ iv ∈ Cn, u, v ∈ R

n.

Now consider a discrete-time dynamical system {N,Rn, Ak} or {Z,Rn, Ak} definedby iteration of the map A (and – if defined – its inverse).

2.1.1 Dynamics in eigenspaces

Recall that an eigenvalue of A is a number λ ∈ C satisfying

Av = λv,

37

38 CHAPTER 2. LINEAR MAPS AND ODES

for a nonzero vector v ∈ Cn, which is called a corresponding eigenvector. Theeigenvalues of a real matrix A can be complex but then occur in complex-conjugatepairs. The set of all eigenvalues of A is denoted by σ(A) and is called the spectrum

of A. It should be known to the reader that λ ∈ C is an eigenvalue of A if and onlyif it is a root of the characteristic polynomial

h(λ) = det(λI − A), (2.2)

where I is the unit n × n matrix. The polynomial h and hence the characteristic

equation

h(λ) = 0 (2.3)

are independent of the choice of the basis, so one can speak about the eigenvaluesof a linear map A. Note that h(A) = 0 (Hamilton-Cayley Theorem).

First consider algebraically simple eigenvalues of A, i.e. simple roots of thecharacteristic polynomial.

(Simple real eigenvalue) Suppose that λ is a simple real eigenvalue of A with theeigenvector v ∈ Rn, i.e. Av = λv. The line

Xλ = {x ∈ Rn : x = sv, s ∈ R}

is invariant under A, since x ∈ Xλ implies Ax ∈ Xλ. If x = sv, then

Ax = A sv = sAv = λsv.

This means that on the line Xλ the dynamics is simply the multiplication by thenumber λ in every time step: if x = s0v then Akx = skv where sk = λsk−1, so that

sk = λks0.

If |λ| < 1, |sk| → 0 as k → +∞, while |sk| → +∞ as k → −∞ if s0 6= 0. Incontrast, if |λ| > 1 and s0 6= 0 then |sk| → +∞ as k → +∞, while |sk| → 0 ask → −∞. Notice that if λ > 0, then the time-series is monotone, while if λ < 0, thetime-series exhibits oscillation. Also notice that for λ = 0 the decay of sk to zero isinstantaneous. Finally, note that we did not use the assumption that λ is simple.The assumption is neverthelass made, because for simple eigenvalues the formulatedresults provide the whole story, while for multiple eigenvalues they capture only partof it, see below.

(Simple complex-conjugate pair of eigenvalues) Suppose λ, λ is a pair of simplecomplex eigenvalues with eigenvectors v, v ∈ Cn, i.e.

Av = λv, Av = λv.

(Again, the observations below are correct even if λ and λ are not simple.) Sinceλ 6= λ the vectors v and v are complex-linearly-independent (see Exercise 2.5.2) andthe plane in Rn

Xλ,λ = {x ∈ Rn : x = zv + zv, z ∈ C}

2.1. LINEAR MAPS IN RN 39

is invariant under A. Indeed, if x = zv + zv then

Ax = A(zv + zv) = zAv + zAv = λzv + λzv.

This means that A acts asz 7→ z′ = λz,

i.e. as a multiplication by λ ∈ C. If λ = ρeiψ and z = reiϕ, then

z′ = (ρr)ei(ψ+ϕ).

The motion in Xλ,λ is therefore a superposition of the rotation through the angleψ = argλ and the radial expansion (ρ > 1) or contraction (ρ < 1) with factorρ = |λ|.

Alternatively (but equivalently) we can work with real coordinates (a, b) andwrite

Xλ,λ = {x ∈ Rn : x = a Re v − b Im v, a, b ∈ R}.

The relationship between the coordinates is given by

z =1

2(a+ ib).

Writing Av = λv as

A(Re v + i Im v) = (Re λ+ i Im λ)(Re v + i Im v)

we obtain

A(Re v) = Re λ Re v − Im λ Im v,

A(Im v) = Im λ Re v + Re λ Im v.

Therefore, in the coordinates (a, b) the map A restricted to Xλ,λ is represented bythe matrix

(

Re λ −Im λIm λ Re λ

)

= ρ

(

cosψ − sinψsinψ cosψ

)

.

Whenever the characteristic polynomial (2.2) has n distinct (and therefore nec-essarily simple) roots λi, i = 1, . . . , n, we can combine our understanding of thedynamics in each of the invariant subspaces Xλi

(for real λi) or Xλi,λi(for complex

λi) to obtain an overall description of the dynamics. The key point is that Rn canbe decomposed into these subspaces.

Definition 2.1 Let X1, . . . , Xr be linear subspaces of a linear space X. We say that

X is the direct sum of X1, . . . , Xr, and write

X = X1 ⊕X2 ⊕ · · · ⊕Xr,

whenever every x ∈ X can be written in a unique way as

x = x1 + x2 + · · · + xr

with xi ∈ Xi for i = 1, 2, . . . , r.


As our building blocks Xλiand Xλi,λi

are linearly independent and together havedimension n, they form a direct sum decomposition of Rn. So an arbitrary x ∈ Rn

can be uniquely written as a linear combination of elements of Xλiand Xλi,λi

.The action of A on x can then be simply reconstructed from that on the invariantsubspaces by superposition. Since a generic matrix A has only simple eigenvalues,the above consideration completely describes the dynamics generated by such a map.If there are no eigenvalues with |λ| = 1, we get exponential contraction/expansionon the linear eigenspaces, combined with rotation in the case of nonreal eigenvalues.

What happens when the characteristic equation (2.3) has multiple roots?

(Real multiple eigenvalue) First assume that the multiple (i.e., not simple) eigen-value λ is real. The preceding analysis still applies when there are n linearly-independent eigenvectors. But this is an exceptional case. Generically, there arem ≥ 1 Jordan chains of linearly-independent vectors v(j,k) ∈ Rn such that

Av(j,1) = λv(j,1),Av(j,2) = λv(j,2) + v(j,1),

· · ·Av(j,nj) = λv(j,nj) + v(j,nj−1),

(2.4)

for j = 1, 2, . . . , m. The integer nj is called the length of the Jordan chain. Thenumber of Jordan chains with length l can be computed by the formula

N(l, λ) = Rl+1 − 2Rl +Rl−1,

where R0 = n and Rk = rank (A− λI)k for k ≥ 1. The vectors v(j,1) are the eigen-vectors of A corresponding to the eigenvalue λ. The vectors v(j,2), v(j,3), . . . , v(j,nj)

are called generalized eigenvectors corresponding to the eigenvalue λ. It follows from(2.4) that these vectors are null-vectors of the matrix (A− λI)nj .

We now have an nj-dimensional subspace of Rn defined by

X(j)λ = {x ∈ R

n : x =

nj∑

k=1

c(j)k v(j,k), c

(j)k ∈ R} (2.5)

that is invariant under (2.1). A coordinate vector c = (c1, c2, . . . , cnj)T ∈ Rnj is

mapped toc′ = Jc,

where the nj × nj matrix J is given by the Jordan block

J =

λ 1 0λ 1

λ 10 λ

. (2.6)

Therefore, to understand how orbits of (2.1) behave in X(j)λ , we must consider Jk.


The low-dimensional examples (see Exercise 2.5.8 for the general case)(

λ 10 λ

)k

=

(

λk kλk−1

0 λk

)

and

λ 1 00 λ 10 0 λ

k

=

λk kλk−1 k(k−1)2

λk−2

0 λk kλk−1

0 0 λk

clearly indicate that all elements of Jk tend to zero as k → +∞ when |λ| < 1,whereas some elements of Jk diverge as k → +∞ when |λ| > 1. Moreover, in thecritical case |λ| = 1 only higher multiplicity can and, for suitable initial vectors, willlead to divergent orbits.

The direct sum of all linear subspaces X(j)λ ,

Xλ = X(1)λ ⊕X

(2)λ ⊕ · · · ⊕X

(m)λ ,

is an invariant linear subspace of Rn called the generalized eigenspace of A corre-sponding to the eigenvalue λ. Its dimension is (n1 + · · ·+ nm). Note that Xλ is thenull-space of (A− λI)l for l = max1≤j≤m nj .

(Complex multiple eigenvalue) A similar construction applies when λ is a multiplecomplex eigenvalue. In this case, v(j,k), j = 1, 2, . . . , m, k = 1, 2, . . . , nj, arecomplex vectors and the Jordan chains corresponding to the eigenvalue λ can becomposed of complex-conjugate vectors v(j,k). Then

X(j)

λ,λ= {x ∈ R

n : x =

nj∑

k=1

(

c(j)k v(j,k) + c

(j)k v(j,k)

)

, c(j)k ∈ C} (2.7)

is a real 2nj-dimensional invariant subspace of Rn parameterized by c ∈ Cnj . Alsoin this case, ‖Jkc‖ → 0 as k → +∞ when |λ| < 1, and ‖Jkc‖ → ∞ as k → +∞when |λ| > 1 and c 6= 0. The direct sum of all linear subspaces X

(j)

λ,λ,

Xλ,λ = X(1)

λ,λ⊕X

(2)

λ,λ⊕ · · · ⊕X

(m)

λ,λ,

is an invariant linear subspace of Rn called the generalized eigenspace of A corre-sponding to the eigenvalues λ, λ. Its dimension is 2(n1 + · · ·+ nm).

It is also possible to define

X(j)λ = {x ∈ C

n : x =

nj∑

k=1

c(j)k v(j,k), c

(j)k ∈ C}

and consider the generalized eigenspace corresponding to λ

Xλ = X(1)λ ⊕X

(2)λ ⊕ · · · ⊕X

(m)λ

as a complex-linear subspace of Cn. Then Xλ,λ = (Xλ ⊕Xλ) ∩ Rn.

Since Rn can be decomposed into the direct sum of all generalized eigenspacesXλ for real eigenvalues λ and Xλ,λ for complex-conjugated eigenvalue pairs λ, λ, wemay summarize the main results of this section as follows.


Theorem 2.2 If all eigenvalues of A are located strictly inside the unit circle in the

complex plane, then for any x ∈ Rn we have that

‖Akx‖ → 0,

as k → +∞.

If there is an eigenvalue outside the unit circle, or at least one generalized eigen-

vector corresponding to some eigenvalue on the unit circle, then there exists x ∈ Rn

such that

‖Akx‖ → ∞,

as k → +∞. 2

In the next section we shall prove a somewhat stronger result using more generaltechniques. Moreover, each statement of the above theorem has a suitable converse,see Exercise 2.5.10.

Example 2.3 (Dynamics generated by planar linear maps)

When n = 2, equation (2.1) takes the form

(

x1

x2

)

7→(

a11 a12

a21 a22

)(

x1

x2

)

and defines a linear dynamical system {N,R2, Ak}, where

A =

(

a11 a12

a21 a22

)

(2.8)

Let τ = Tr(A) = a11 + a22, ∆ = det(A) = a11a22 − a21a12. Then the characteristic

∆

0

+1

−1

−1 +1 τ

Figure 2.1: Stability triangle for planar linear maps.

equation (2.3) can be written as

λ2 − τλ+ ∆ = 0


and has two (possibly complex) solutions

λ1,2 =τ

2±√

τ 2

4− ∆.

It can easily be shown that both eigenvalues λj satisfy |λ| < 1 if and only if thepoint (τ,∆) falls inside the triangle

{(τ,∆) : ∆ < 1, ∆ > −τ − 1, ∆ > τ − 1}

depicted in Figure 2.1. In this case, in accordance with Theorem 2.2, ‖Akx‖ → 0for all x ∈ R2.

The reader is also invited to verify that A has

– an eigenvalue λ1 = 1 along the boundary where ∆ = τ − 1;

– an eigenvalue λ1 = −1 along the boundary where ∆ = −τ − 1;

– eigenvalues λ1,2 = e±iθ with cos θ = 12τ along the boundary where ∆ = 1. 3

2.1.2 Growth estimates and the spectral radius

For any norm in Rn we define the associated operator norm for linear maps:

‖A‖ = supx 6=0

‖Ax‖‖x‖ = sup

‖x‖=1

‖Ax‖ . (2.9)

The number ‖A‖ describes a rate of expansion/contraction of the map A that de-pends on the choice of the norm in Rn. A norm independent measure is given bythe spectral radius.

Definition 2.4 The spectral radius of a linear map A is defined by

r(A) = supλ∈σ(A)

|λ| .

The relation between these quantities is specified by the following lemma, whichonce again shows that the eigenvalues of A yield information about the growth ordecay of the time-series obtained by iterating A.

Lemma 2.5 (Gelfand’s formula)

r(A) = limk→∞

‖Ak‖1/k = infk≥1

‖Ak‖1/k. (2.10)

Proof: Definer = inf

k≥1‖Ak‖1/k.

Then for all ε > 0 there exists m = m(ε) such that ‖Am‖1/m ≤ r + ε. For arbitraryk, put k = pm+ q with 0 ≤ q ≤ m− 1. Then

‖Ak‖1/k ≤ ‖Am‖p/k‖A‖q/k ≤ (r + ε)pm/k‖A‖q/k.


As k → ∞, necessarily q/k → 0 and pm/k → 1, hence

lim supk→∞

‖Ak‖1/k ≤ r + ε.

By letting ε ↓ 0 we deduce that

lim supk→∞

‖Ak‖1/k ≤ r = infk≥1

‖Ak‖1/k,

from which we conclude that the limit of ‖Ak‖1/k for k → ∞ exists and that thislimit equals the infimum.

For |λ| > r the series

λ−1

∞∑

k=0

λ−kAk (2.11)

converges absolutely. If we multiply (2.11), either from the left or from the right,by (λI − A), we obtain

∞∑

k=0

λ−kAk −∞∑

k=0

λ−(k+1)Ak+1 = λ0A0 = I.

Thus the series (2.11) defines the inverse (λI −A)−1 and, accordingly, λ /∈ σ(A). Itfollows that r(A) ≤ r.

It remains to exclude the possibility that r(A) < r. The matrix-valued function

f(z) = (I − zA)−1

is analytic in {z ∈ C : |z| < 1/r(A)}. Since f (k)(0) = k!Ak, the Cauchy integral

formula yields the identity

Ak =1

2πi

∫

γ

f(z)

zk+1dz ,

where γ is a circle centered at the origin with radius (r(A)+ε)−1 for arbitrary ε > 0.It follows that

‖Ak‖ ≤ (r(A) + ε)k supz∈γ

‖f(z)‖

and hencer = lim

k→∞‖Ak‖1/k ≤ r(A) + ε .

By letting ε ↓ 0, we obtain that r ≤ r(A). 2

Remark: The matrix-valued function λ 7→ (λI − A)−1 is called the resolvent of A.The series representation above is the Taylor series at λ = ∞. By working withz = λ−1 we avoid having to deal with the point at infinity. 3

Theorem 2.6 For every ρ > r(A) there exists M ≥ 1 such that

‖Ak‖ ≤ Mρk

for all k ≥ 1.


Proof: If ρ > r(A) then Lemma 2.5 implies ‖Ak‖ ≤ ρk for large k and consequently

M = supk≥0

ρ−k‖Ak‖ <∞ .

Thus for all k ≥ 0 the inequality

‖Ak‖ ≤Mρk

holds. For k = 0 this gives M ≥ 1, since A0 = I and ‖I‖ = 1. 2

Using (2.9), we find that

‖Akx‖ ≤Mρk‖x‖, x ∈ Rn, k ∈ N. (2.12)

This estimate establishes the global asymptotic stability of the origin when r(A) < 1.Indeed, in this case we can choose ρ such that r(A) < ρ < 1. Then for any pointx ∈ R

n with ‖x‖ ≤ ε/M the estimate ‖Akx‖ ≤ ρkε holds, which implies the (global)asymptotic stability of x = 0. In Exercise 2.5.12 the reader is asked to explore thegeometric manifestation of the fact that, possibly, Mρ > 1.

We now show that one can introduce an equivalent norm1, tailor-made for thelinear map A, such that M reduces to one (the advantage being that the estimatefor k = 1 carries all information, in the sense that the estimate for arbitrary k > 1is obtained by iteration of the one for k = 1).

Theorem 2.7 Let ρ > r(A). There exists an equivalent norm ‖ ·‖1 on Rn such that

‖A‖1 ≤ ρ.

Proof: Define ‖ · ‖1 for x ∈ Rn by the formula:

‖x‖1 =

∞∑

k=0

ρ−k‖Akx‖.

Formula (2.10) implies that this series converges. Indeed, for k sufficiently large andsome q < 1,

ρ−1‖Ak‖1/k ≤ q

and henceρ−k‖Akx‖ ≤ ρ−k‖Ak‖ ‖x‖ ≤ ‖x‖qk .

Clearly, ‖x‖1 ≥ 0 for all x ∈ Rn and ‖x‖1 = 0 if and only if x = 0. Likewise theproperties ‖αx‖1 = |α|‖x‖1 and ‖x + y‖1 ≤ ‖x‖1 + ‖y‖1 for x, y ∈ Rn and α ∈ R

follow straightaway from the corresponding properties of ‖ · ‖. So ‖ · ‖1 is a norm onRn. Moreover, we have

‖x‖ ≤ ‖x‖1 ≤(

∞∑

k=0

(

ρ−1‖Ak‖1/k)k

)

‖x‖,

1Recall that two norms on a linear space E are called equivalent, if there are two constants,C2 ≥ C1 > 0, such that for all x ∈ E: C1‖x‖ ≤ ‖x‖1 ≤ C2‖x‖.


i.e. ‖ · ‖1 is equivalent to ‖ · ‖ on Rn. Now, for x ∈ Rn,

‖Ax‖1 =

∞∑

k=0

ρ−k‖Ak+1x‖ = ρ(‖x‖1 − ‖x‖)

so that‖Ax‖1 ≤ ρ‖x‖1, x ∈ R

n. 2

Note that by induction the estimate ‖A‖1 ≤ ρ implies that

‖Akx‖1 ≤ ρk‖x‖1, x ∈ Rn,

for all k ≥ 0.

Definition 2.8 The map A is called a (strict) linear contraction if for some ρwith 0 < ρ < 1 we have

‖Ax‖1 ≤ ρ‖x‖1, x ∈ Rn.

Since a linear contraction obviously has the origin as its asymptotically stable fixedpoint, we have proved the following criterion, which we have already mentionedbefore.

Theorem 2.9 If all eigenvalues of a matrix A are located strictly inside the unit

circle, then A defines a linear contraction with respect to an appropriate norm and

the origin is globally asymptotically stable. 2

As with Theorem 2.2 a certain converse of the statements above holds (see Exercise2.5.10).

Since σ(A−1) = {λ−1 : λ ∈ σ(A)} whenever 0 /∈ σ(A), we can obtain similarbounds for A−1 in terms of infλ∈σ(A) |λ|.

Theorem 2.10 Assume that |λ| > ρ > 0 for all λ ∈ σ(A). Then there exists M ≥ 1such that

‖A−k‖ ≤Mρ−k, k ≥ 0.

Moreover, there exists an equivalent norm ‖ · ‖1 on Rn such that

‖A−1‖1 ≤ ρ−1

and hence ‖A−k‖1 ≤ ρ−k for all k ≥ 0.

2.1.3 Hyperbolic linear maps and spectral projectors

In this section we focus on the situation when

σ(A) = σs(A) ∪ σu(A),

where σs(A) is located strictly inside the unit circle but does not contain zero, whileσu(A) is located strictly outside the unit circle.


Definition 2.11 A linear map x 7→ Ax in Rn is called hyperbolic if A is invertible

and has no eigenvalue λ with |λ| = 1.

The direct sum of all generalized eigenspaces of A corresponding to σs(A) forman invariant subspace T s of A, called the stable eigenspace, while the direct sum ofall generalized eigenspaces of A corresponding to σu form the unstable eigenspace T u

of A. Moreover, we have T s ⊕ T u = Rn, meaning that any x ∈ Rn can be uniquelydecomposed as

x = xs + xu

with xs ∈ T s and xu ∈ T u. Clearly, σ(A|T s) = σs(A) and σ(A|Tu) = σu(A).There are explicit methods to find xs (and hence xu) for a given x, using the

so called spectral projectors. We begin by assuming that σs(A) consists of just onesimple eigenvalue λ ∈ R with |λ| < 1. Let v ∈ Rn be the corresponding eigenvector,so that

Av = λv.

Let AT denote as usual the transposed matrix corresponding to A. It also has asimple eigenvalue λ with some eigenvector w ∈ Rn,

ATw = λw.

Since the eigenvalue λ is simple, the Fredholm decomposition for linear maps (seeLemma 6.14 in Chapter 6) tells us that 〈w, v〉 6= 0 and therefore w can be selectedto satisfy the normalization condition

〈w, v〉 = 1.

Now the mapx 7→ Px = 〈w, x〉v

assigns to each x ∈ Rn a vector xs = Px ∈ T s ⊂ Rn. The map P is linear and hasobviously the property P 2 = P . We say that P projects Rn onto T s. Note that

T s = {x ∈ Rn : Px = x} = {sv : s ∈ R}.

The linear map Q = I − P or, in more detail,

x 7→ Qx = x− 〈w, x〉v

projects Rn onto T u and

T u = {x ∈ Rn : Px = 0} = {x ∈ R

n : 〈w, x〉 = 0} = w⊥.

If σs is composed of ns eigenvalues λ1, λ2, . . . , λns, all of which are simple and

satisfy |λi| < 1, the corresponding projectors can be defined by

P =ns∑

i=1

Pi, Q = I − P,


wherePix = 〈w(i), x〉v(i), (2.13)

is the projection of Rn onto the one-dimensional invariant eigenspace Xλicor-

responding to the eigenvalue λi. Here Av(i) = λiv(i) and ATw(i) = λiw

(i) fori = 1, 2, . . . , ns, and

〈w(i), v(j)〉 =

{

1, i = j,0, i 6= j.

(Verify!) The projectors defined by (2.13) have the following properties:

P 2i = Pi, PiPj = 0, for j 6= i. (2.14)

Notice that the eigenvalues (and the corresponding eigenvectors) can be complex,provided that 〈·, ·〉 is interpreted as the standard scalar product in Cn.

There is an alternative method to construct a projector Pi onto the generalizedeigenspace corresponding to an eigenvalue λi ∈ C, which is applicable for bothsimple and multiple eigenvalues and does not require computing any eigenvectoror generalized eigenvector. In this method, the projection matrix Pi is constructedas a polynomial function of the matrix A. We describe this method briefly below.Actually, we explain how to compute Pi corresponding to any eigenvalue λi of A.

Suppose that the characteristic polynomial h(λ) = det(λI − A) has d distinctroots: λ1, λ2, . . . , λd ∈ C. Then, it can be written as

h(λ) =d∏

i=1

(λ− λi)mi ,

where mi are the algebraic multiplicities of the eigenvalues and hence m1+· · ·+md =n. This implies that the following partial fraction decomposition holds:

1

h(λ)=

d∑

i=1

gi(λ)

(λ− λi)mi, (2.15)

where gi(λ) are polynomials of degree mi − 1 which can be found, for example, bythe method of unknown coefficients.

Multiplying both sides of (2.15) by h(λ) leads to the equation

1 =d∑

i=1

pi(λ), (2.16)

where

pi(λ) =h(λ)gi(λ)

(λ− λi)mi= gi(λ)

∏

j 6=i

(λ− λj)mj . (2.17)

Each pi is a polynomial of degree less than n in λ. Moreover, all the polynomialsp2i − pi and pipj with j 6= i are divisible by the characteristic polynomial h. Indeed,

pi(λ)pj(λ) =gi(λ)gj(λ)h2(λ)

(λ− λi)mi(λ− λj)mj= h(λ)

(

gi(λ)gj(λ)∏

k 6=i,k 6=j

(λ− λk)mk

)

.


Moreover,

p2i (λ) − pi(λ) = pi(λ)(pi(λ) − 1) = −pi(λ)

∑

j 6=i

pj(λ) = −∑

j 6=i

pi(λ)pj(λ),

with each term in the sum divisible by h due to the previous formula.Define now

Pi = pi(A), i = 1, 2, . . . , d. (2.18)

Since h(A) = 0 by the Hamilton-Cayley Theorem, we see that the linear maps Pithus constructed satisfy (2.14), i.e., are projectors. Because each Pi is a polynomialfunction of A, it is linear and commutes with A. This implies that each range Pi(C

n)is an invariant subspace of A. The formula (2.17) implies that Pi(A)x = 0 if Ax = λjwith j 6= i. Therefore, since

∑di=1 Pi = I, Pi must project Cn onto the generalized

eigenspace of A corresponding to the eigenvalue λi:

Xλi= Pi(C

n).

If there are ds distinct eigenvalues λ1, λ2, . . . , λdssatisfying |λi| < 1, then the pro-

jectors onto T s and T u are given by

P =

ds∑

i=1

Pi and Q = I − P,

repectively.

Example 2.12 (Practical computation of spectral projectors)

Consider

A =

(

−1 −332

72

)

.

Then

λI − A =

(

λ+ 1 3−3

2λ− 7

2

)

so that the characteristic polynomial is

h(λ) = det(λI − A) = λ2 − 52λ+ 1 = (λ− 1

2)(λ− 2)

and the eigenvalues of A are λ1 = 12

and λ2 = 2. Hence A is hyperbolic.The vector

v =

(

2−1

)

is an eigenvector corresponding to the eigenvalue λ1 = 12, while the vector

u =

(

−11

)


is an eigenvector corresponding to the eigenvalue λ2 = 2. The transposed matrix

AT =

(

−1 32

−3 72

)

has

w =

(

11

)

as an eigenvector corresponding to its eigenvalue λ1 = 12. Notice that

〈w, v〉 = 1, 〈w, u〉 = 0.

For any vector

x =

(

x1

x2

)

∈ R2,

we have 〈w, x〉 = x1 + x2 and, therefore,

Px = 〈w, x〉v =

(

2x1 + 2x2

−x1 − x2

)

.

This gives

P =

(

2 2−1 −1

)

(2.19)

which is indeed a projection matrix:

P 2 =

(

2 2−1 −1

)(

2 2−1 −1

)

=

(

2 2−1 −1

)

= P.

The map x 7→ Px projects the space R2 onto the eigenline T s spanned by theeigenvector v. The matrix

Q = I − P =

(

−1 −21 2

)

defines a projection x 7→ Qx onto the eigenline T u spanned by the eigenvector u.Clearly, PQ = QP = 0.

To find the spectral projectors P and Q using the partial fraction decomposition(2.15), write

1

h(λ)=

1

(λ− 12)(λ− 2)

=−2

3

λ− 12

+23

λ− 2.

This gives 1 = p1(λ) + p2(λ), where

p1(λ) = −23(λ− 2), p2(λ) = 2

3(λ− 1

2).

Therefore

P = p1(A) = −23(A− 2 I) = −2

3

(

−3 −3

32

32

)

=

(

2 2−1 −1

)

,


which (re-assuringly) is the same projector (2.19) as before. For the projector Q,we have

Q = p2(A) = 23(A− 1

2I) = 2

3

(

−32

−3

32

3

)

=

(

−1 −21 2

)

.

Note that P +Q = I, so that the computation of Q as p2(A) is redundant: It canbe found by merely evaluating I − P . 3

Remark:There is another alternative formula for the projector Pi, namely:

Pi =1

2πi

∫

γi

(λI −A)−1 dλ. (2.20)

Here the integral is interpreted (for each matrix element) as the contour integralin the complex plane along the small circle γi around λi. It can be computed bythe method of residues. Notice that the formula (2.20) is valid in case of multipleeigenvalues as well. Moreover, by a suitable choice of contour γ it works for anyspectral set, i.e. any subset of σ(A). For example, γ = S1, where S1 is the unitcircle, gives the projector onto T s of a hyperbolic map A. The formula (2.20) canalso be generalized to an operator A in a Banach space. 3

Theorem 2.13 For any hyperbolic map A, there is an equivalent norm ‖ · ‖1 in

which A|T s and A−1|Tu become linear contractions, i.e. for some ρ with 0 < ρ < 1we have

‖Ax‖1 ≤ ρ‖x‖1, x ∈ T s, (2.21)

and

‖A−1x‖1 ≤ ρ‖x‖1, x ∈ T u. (2.22)

Proof:Take some ρ such that 0 ≤ r(A|T s) < ρ < 1. According to Theorem 2.7, there

exists an equivalent norm ‖ · ‖1 on T s such that (2.21) holds.Next note that A−1|Tu is defined and that σ(A−1|Tu) is contained strictly in-

side the unit circle. Therefore, it follows from Theorem 2.10 that there exists anequivalent norm ‖ · ‖1 on T u such that

‖A−1x‖1 ≤ ρ‖x‖1, x ∈ T u,

for some ρ < 1. To obtain the estimates (2.21) and (2.22) with the same ρ, we cansimply take ρ := max{ρ, ρ}.

Finally, extend the norm to all of Rn by setting

‖x‖1 = ‖xs‖1 + ‖xu‖1 for x = xs + xu ∈ T s ⊕ T u. 2

Remarks:


(1) Note that, by induction,

‖Akx‖1 ≤ ρk‖x‖1, x ∈ T s,

and‖A−kx‖1 ≤ ρk‖x‖1, x ∈ T u.

(2) Note that we did not use in the proof the invertibility of A, i.e. the absenceof eigenvalue 0. All that matters for the estimates is that there are no eigenvalueson the unit circle. Indeed, we only used A−1|Tu and we can replace this by (A|Tu)−1

since 0 /∈ σ(A|Tu). 3

Due to Theorem 2.13 we can give an alternative definition of a hyperbolic linearmap, which does not use eigenvalues.

Definition 2.14 An invertible linear continuous map A : X → X on a Banach

space is called hyperbolic if X = T s ⊕ T u, where both T s and T u are invariant

under A, and with respect to an equivalent norm ‖ · ‖1 we have

‖A|T s‖1 ≤ ρ, ‖A−1|Tu‖1 ≤ ρ,

for some 0 < ρ < 1.

Remark:Not all authors include “invertibility” in the definition of “hyperbolicity”. We

do this to facilitate the formulation of a topological equivalence result ahead (seeTheorem 2.30). For linear maps, the sign of an eigenvalue has no impact on growthor decay of the image length, but it does have an impact on orientation. In par-ticular, the border between orientation preserving and orientation reversing linearcontractions corresponds to zero eigenvalue. 3

2.2 Linear autonomous systems of ODEs

Consider an autonomous system of linear ODEs

x = Ax, x ∈ Rn. (2.23)

The associated flow is given by ϕt(x) = etAx where

etA =∞∑

k=0

tk

k!Ak.

(Note in particular the group property e(t+s)A = etAesA.) Now consider the linearcontinuous-time dynamical system {R,Rn, ϕt}.

For the spectra we have

σ(etA) = etσ(A) = {µ ∈ C : µ = etλ, λ ∈ σ(A)},for any t ∈ R, so in particular zero is never an eigenvalue of etA (indeed, etA hasinverse e−tA).

Note that z 7→ etz maps the imaginary axis onto the unit circle and the left halfplane inside the unit circle. Accordingly, the condition |λ| < 1 translates in thecondition Re λ < 0.

2.2. LINEAR AUTONOMOUS SYSTEMS OF ODES 53

2.2.1 Dynamics in the eigenspaces

First consider the dynamics in one- and two-dimensional invariant eigenspaces as-sociated to, respectively, a simple real eigenvalue and a simple complex pair ofeigenvalues.

(Simple real eigenvalue) Let λ be a simple real eigenvalue of A with eigenvectorv ∈ R

n, i.e. Av = λv. The line

Xλ = {x ∈ Rn : x = sv, s ∈ R}

is invariant with respect to the flow, since x ∈ Xλ implies x ∈ Xλ. The functionx(t) = s(t)v ∈ Xλ satisfies the equation (2.23) if and only if

s = λs, (2.24)

sincesv = x = Ax = Asv = sAv = λsv.

Thus we find thats(t) = s0e

λt

and that s(t) → 0 as t→ +∞ if λ < 0, while s(t) → 0 as t→ −∞ if λ > 0.

(Simple complex-conjugate pair of eigenvalues) Let λ, λ be simple nonreal eigen-values with the eigenvectors v, v ∈ Cn, i.e.

Av = λv, Av = λv.

The planeXλ,λ = {x ∈ R

n : x = zv + zv, z ∈ C} ⊂ Rn

is invariant with respect to the flow generated by (2.23). Moreover, x(t) = z(t)v +z(t)v is a solution if and only if

z = λz.

This impliesz(t) = z0e

λt.

If λ = α + iω and z0 = r0eiϕ0 , then

z(t) = (r0eαt)ei(ϕ0+ωt).

If α 6= 0, the motion in Xλ,λ is a superposition of rotation (with angular speed ω)and radial divergence (if α > 0) or convergence (if α < 0).

If α = Re λ = 0 then the invariant subspace Xλ,λ is filled with periodic orbitsparametrized by r0 > 0. The corresponding periodic solutions are given by

x(t) = r0(

ei(ϕ0+ωt)v + e−i(ϕ0+ωt)v)

,

where the phase ϕ0 determines the position of x(0) on the closed orbit. Of course,the period of the orbit is p = 2π

ω.


Since a generic n×n matrix A has n different simple eigenvalues and therefore nlinearly independent eigenvectors, the above consideration completely describes thedynamics of a generic linear system (2.23). If there are no eigenvalues with Re λ = 0,we get exponential contraction/expansion on the linear eigenspaces, combined withrotations in the case of nonreal eigenvalues.

There is a possibility, however, that the matrix A has multiple eigenvalues.

(Real multiple eigenvalue) Assume first that the eigenvalue λ ∈ R. If there isa basis of eigenvectors in the corresponding invariant subspace, then the solutionsbehave as in the simple real case. Otherwise, there are Jordan chains (2.4) ofvectors v(j,k) ∈ Rn, j = 1, 2, . . . , m, k = 1, 2, . . . , nj . The nj-dimensional subspace

X(j)λ defined by (2.5) is invariant and c = (c1, c2, . . . , cnj

)T ∈ Rnj satisfies

c = Jc,

where J is again given by (2.6). Therefore, any solution component is the sum ofscalar multiples of tkeλt with k < nj (see Exercise 2.5.8 for an explicit expressionfor etJ).

(Complex multiple eigenvalue) A similar construction applies when λ ∈ C is amultiple eigenvalue. In this case, v(j,k) are complex vectors and the Jordan chainscorresponding to the eigenvalue λ can be composed of complex-conjugate vectorsv(j,k). Then Xλ,λ defined by (2.7) is a real 2nj-dimensional invariant subspace pa-rameterized by c ∈ Cnj . All components of Re c(t) and Im c(t) are the sums ofscalar multiples of tkeαt sin(ωt), tkeαt cos(ωt), where λ = α + iω, k < nj .

Thus, the following theorem holds.

Theorem 2.15 If all eigenvalues λ1, λ2, . . . , λn of the matrix A satisfy Re λi < 0,then for any x ∈ Rn

‖etAx‖ → 0,

as t→ ∞.

If at least one eigenvalue of A satisfies Re λ > 0 or there is at least one gener-

alized eigenvector with corresponding eigenvalue Re λ = 0, then there exists x ∈ Rn

such that

‖etAx‖ → ∞,

as t→ ∞. 2

Remark:Notice that when ‖x(t)‖ → 0 as t → ∞, the decay need not to be monotone,

see Exercise 2.5.13. Moreover, as in the discrete time case the statements in thistheorem can be reversed, see Exercise 2.5.11. 3

Example 2.16 (Phase portraits of planar linear autonomous ODEs)

When n = 2, equation (2.23) takes the form(

x1

x2

)

=

(

a11 a12

a21 a22

)(

x1

x2

)


or{

x1 = a11x1 + a12x2,x2 = a21x1 + a22x2.

(2.25)

Let

τ = Tr(A) = a11 + a22, ∆ = det(A) = a11a22 − a21a12.

Then the characteristic equation (2.3) can be written as

λ2 − τλ+ ∆ = 0

and has two (possibly complex) solutions

λ1,2 =τ

2±√

τ 2

4− ∆.

If follows right away that both eigenvalues λj satisfy Re λ < 0 if and only if

τ < 0 and ∆ > 0.

Under this condition, Theorem 2.15 implies that ‖etAx‖ → 0 as t → ∞ for anyx ∈ R2.

Traditionally, three generic cases are distinguished, characterized by inequalitiesin terms of τ and ∆:

(b)(a)

s2

s1 x1

x2

v(2)

v(1)

Figure 2.2: Canonical (a) and original (b) saddles.

(Saddle: ∆ < 0) The matrix A has one positive and one negative eigenvalue:λ2 < 0 < λ1. Let v(1) and v(2) be the corresponding eigenvectors:

Av(j) = λjv(j), j = 1, 2.

Let us use these eigenvectors as new basis vectors in R2, i.e. write any x ∈ R2 inthe form

x = s1v(1) + s2v

(2),


with some s1,2 ∈ R. Notice that s1,2 can be considered as new coordinates in R2. Inthese new coordinates, the system (2.25) takes a particularly simple form, namely

{

s1 = λ1s1,s2 = λ2s2.

(2.26)

So each of the equations can be solved independently. This gives

s1(t) = s1(0)eλ1t, s2(t) = s2(0)eλ2t,

yielding a phase portrait in the (s1, s2)-plane presented in Figure 2.2(a). The phaseportrait in the (x1, x2)-plane (see Figure 2.2(b)) appears then by applying a lineartransformation to the phase portrait in the (s1, s2)-plane. The equilibrium at theorigin is a saddle.

(Node: τ 2 > 4∆ > 0) The matrix A has either two positive eigenvalues λ1 >λ2 > 0 (when τ > 0) or two negative eigenvalues λ2 < λ1 < 0 (when τ < 0). Theanalysis here is very similar to that in the saddle case. Using the eigenvectors v(j) as

(a) (b)

s1

s2

x1

x2

v(1)

v(2)

Figure 2.3: Canonical (a) and original (b) stable nodes.

new basis vectors in R2, we obtain once again system (2.26) and solve it as above.When τ < 0, this leads to a phase portrait shown in Figure 2.3(a), since both s1(t)and s2(t) tend to zero exponentially fast as t → ∞. Notice, however, that due tothe difference in the convergence rates, a generic orbit of (2.26) tends to the originhorizontally. The phase portrait in the (x1, x2)-plane (see Figure 2.3(b)) is obtainedby a linear transformation. The equilibrium in the origin is a stable node. The caseτ > 0 gives an unstable node. The corresponding phase portrait can be obtainedfrom that of the stable node by reversing time.

(Focus: 0 < τ 2 < 4∆) The matrix A has now two complex-conjugate eigenvaluesλ1,2 = α± iω with

α =τ

2, ω =

1

2

√4∆ − τ 2.

The corresponding eigenvectors v(1) and v(2) are also complex but can be selectedto be complex-conjugate:

v(1) = v(2) = w ∈ C2.


The vector w can be used to introduce a new coordinate z ∈ C which specifies apoint x ∈ R2 according to the formula

x = zw + zw.

As we have already seen, this coordinate satisfies

z = λz

with λ = α + iω, so thatz(t) = r0e

αtei(ϕ0+ωt),

where r0 and ϕ0 are specified by the initial point at t = 0. For s1 = Re z ands2 = Im z, we get oscillatory behaviour superimposed with growth (α > 0) or decay(α < 0):

{

s1(t) = r0eαt cos(ϕ0 + ωt),

s2(t) = r0eαt sin(ϕ0 + ωt).

(2.27)

When τ < 0, we have α < 0 and the phase portrait in the (s1, s2)-plane is as shown

(b)(a)

s2 x2

s1 x1

Figure 2.4: Canonical (a) and original (b) stable foci.

in Figure 2.4(a). All orbits spiral to the origin. A linear transformation producesthe phase portrait in the (x1, x2)-plane (see Figure 2.4(b)). The equilibrium at theorigin is a stable focus. When τ > 0, one obtains an unstable focus. As we havealready noted (see Example 1.17 and Exercises 1.5.13 and 1.5.14 in Chapter 1), anode and a focus of the same stability type are topologically equivalent. We returnto this issue in Section 2.3.1 ahead.

We leave it to the reader to carry out the analysis of all critical cases (definedby at least one equality relation involving τ and/or ∆). Here we only mention theso-called center, which occurs when τ = 0 but ∆ > 0. In this case both eigenvaluesof A are purely imaginary: λ1,2 = ±iω with ω =

√∆ > 0. From (2.27) it follows

that all solutions of (2.25) are p0-periodic, where

p0 =2π

ω

and, therefore, all nonequilibrium orbits of (2.25) are closed (see Figure 2.5). 3


(b)(a)

x2

x1s1

s2

Figure 2.5: Canonical (a) and original (b) centers.

2.2.2 Growth estimates and the spectral bound

Motivated by the observations so far, we can provide the following definition.

Definition 2.17 The spectral bound of a linear map A is defined by

s(A) = supλ∈σ(A)

Re(λ).

Theorem 2.18 For every ω > s(A) there exists M ≥ 1 such that

‖etA‖ ≤ Meωt

for t ≥ 0.

Proof: Define B = −ωI + A. Since σ(eB) = e−ω+σ(A) and |eλ| = eRe(λ), we knowthat all eigenvalues of eB are inside the unit circle. Let ρ be such that r(eB) < ρ < 1.By Theorem 2.7 we know that

‖x‖1 =∞∑

k=0

ρ−k‖ekBx‖

defines an equivalent norm such that ‖eB‖1 ≤ ρ.We claim that, for some M ≥ 1, the inequality

‖etB‖1 ≤ ρtM

holds for all t ≥ 0. Indeed, write t = m− ν with m ∈ N and ν ∈ (0, 1] and define

M = sup0<s≤1

‖e−sB‖1.

Then 1 ≤ M <∞ and, moreover,

‖etB‖1 = ‖e(m−ν)B‖1 ≤ ‖emB‖1‖e−νB‖1 ≤ ρmM ≤ ρm−νM = ρtM.


It follows that‖etA‖1 ≤ e(ω+ln ρ)tM ≤ eωtM

(here we use that ln ρ < 0). Finally, the equivalence of ‖ · ‖1 and ‖ · ‖ yields theestimate for ‖etA‖ with an adjusted constant M . 2

Like in the discrete time case, we can eliminate the constant M by modifyingthe norm on R

n such that it is tailor-made for A.

Theorem 2.19 Let ω > s(A). There exists an equivalent norm ‖ · ‖2 on Rn such

that

‖etA‖2 ≤ eωt

for all t ≥ 0.

Proof: As in the proof of Theorem 2.18, let

‖x‖1 =

∞∑

k=0

ρ−k‖ekBx‖

where B = −ωI + A. Now define

‖x‖2 =

∫ ∞

0

‖eτBx‖1dτ.

The integral converges and

‖x‖2 =

∫ ∞

0

‖eτBx‖1dτ ≤∫ ∞

0

‖eτB‖1‖x‖1dτ ≤ M

(∫ ∞

0

ρτdτ

)

‖x‖1 = − M

ln(ρ)‖x‖1 ,

i.e.

‖x‖2 ≤ − M

ln(ρ)‖x‖1.

Moreover, since x = e−tBetBx, we have

‖x‖1 ≤ ‖e−tB‖1‖etBx‖1 ≤ et‖B‖1‖etBx‖1

and hence‖etBx‖1 ≥ e−t‖B‖1‖x‖1.

It follows that

‖x‖2 ≥(∫ ∞

0

e−τ‖B‖1dτ

)

‖x‖1 =1

‖B‖1‖x‖1.

Combining the two estimates above, we obtain

1

‖B‖1‖x‖1 ≤ ‖x‖2 ≤ − M

ln(ρ)‖x‖1,

meaning that ‖ · ‖2 is equivalent to ‖ · ‖1.


Finally, we compute

‖etAx‖2 =

∫ ∞

0

‖eτBetAx‖1dτ =

∫ ∞

0

‖e−ωτe(t+τ)Ax‖1dτ

=

∫ ∞

t

‖e−ω(θ−t)eθAx‖1dθ = eωt∫ ∞

t

‖e−ωθeθAx‖1dθ

≤ eωt∫ ∞

0

‖e−ωθeθAx‖1dθ ≤ eωt∫ ∞

0

‖eθBx‖1dθ = eωt‖x‖2,

from which it follows that ‖etA‖2 ≤ eωt for all t ≥ 0. 2

Definition 2.20 A norm ‖ · ‖2 in Rn is called a Lyapunov norm for a linear

system x = Ax if

‖etAx‖2 ≤ e−αt‖x‖2, t ≥ 0,

for some α > 0.

Theorem 2.21 If all eigenvalues of a matrix A have negative real part, then there

is an equivalent Lyapunov norm ‖ · ‖2 on Rn for the linear system x = Ax and the

origin is a globally asymptotically stable equilibrium of this system.

Proof: When all eigenvalues of A have negative real part, we have by definitions(A) < 0. Thus, there exists ω < 0 such that s(A) < ω < 0. Theorem 2.19 impliesthat ‖·‖2 is the Lyapunov norm with α = −ω > 0. The global asymptotical stabilityof x = 0 in this case is obvious. 2

As in the discrete time case one can show that stability can be fully characterizedin terms of eigenvalue conditions, see Exercise 2.5.11 for details.

By time reversal, we obtain the following theorem.

Theorem 2.22 Assume that Re(λ) > ω for all λ ∈ σ(A). Then there exists M ≥ 1such that

‖e−tA‖ ≤Me−ωt, t ≥ 0,

and there exists an equivalent norm ‖ · ‖2 on Rn such that

‖e−tA‖2 ≤ e−ωt, t ≥ 0.

2.2.3 Hyperbolic linear ODEs

Definition 2.23 A linear ODE x = Ax is called hyperbolic if A has no eigenvalue

λ with Re λ = 0.

The stable and unstable invariant subspaces T s,u of a linear hyperbolic ODE aredefined like those of a linear hyperbolic map. Namely, we can write as before

σ(A) = σs(A) ∪ σu(A),

where now

σs(A) = {λ ∈ σ(A) : Re λ < 0} and σu(A) = {λ ∈ σ(A) : Re λ > 0}.

2.3. TOPOLOGICAL CLASSIFICATION OF LINEAR SYSTEMS 61

Then T s and T u will be spanned by all eigenvectors and generalized eigenvectors ofA corresponding to σs(A) and σu(A), respectively. Spectral projectors can also becomputed as before, with γi in the formula (2.20) replaced by any contour encirclingσs(A).

The dynamics on T s is described by Theorem 2.21. There is no need to studythe dynamics on T u separately, since the results can be obtained from those aboutT s by reversal of time, i.e. the substitution t 7→ −t. Thus, we arrive at the followingtheorem (we leave it to the reader to provide the details of the proof).

Theorem 2.24 For any hyperbolic system x = Ax, there is an equivalent norm

‖ · ‖2 on Rn such that for some α > 0 we have for all t ≥ 0

‖etAx‖2 ≤ e−αt‖x‖2, x ∈ T s,

and

‖e−tAx‖2 ≤ e−αt‖x‖2, x ∈ T u. 2

2.3 Topological classification of linear systems

2.3.1 Topological equivalence of hyperbolic linear ODEs

We consider first the problem of topological classification of linear ODEs. We advisethe reader to make Exercises 1.5.13 and 1.5.14 before proceeding.

Theorem 2.25 The linear system

x = Ax, x ∈ Rn, (2.28)

where all eigenvalues of A have negative real part, is topologically conjugate to the

system

y = −y, y ∈ Rn. (2.29)

Proof: Let ‖ · ‖2 be a Lyapunov norm: ‖etAx‖2 ≤ e−αt‖x‖2, α > 0. Any nontrivialorbit of (2.28) crosses

Σn−1 = {x ∈ Rn : ‖x‖2 = 1}

exactly once. Denote by H : Σn−1 → Sn−1 the map that assigns to the point ofintersection of the ray {sv : s > 0} with Σn−1 the point of intersection of the sameray with Sn−1 (see Figure 2.6), i.e.

H(x) =x

‖x‖

where the standard (Euclidian) norm ‖ · ‖ in Rn is used.Define now the map h : Rn → Rn as follows. For any x 6= 0, let τ(x) 6= 0 be the

(positive or negative) time such that

x = eτ(x)Ax with x ∈ Σn−1.


Σn−1

vv

y

x

x

Sn−1

y

H

Figure 2.6: Construction of the homeomorphism x 7→ y = h(x).

We have x = e−τ(x)Ax. Map x 7→ y = H(x) and then set

y = h(x) = e−τ(x)y = e−τ(x)H(e−τ(x)Ax).

Let h(0) = 0. The map h : Rn → Rn thus constructed has the following properties:(i) h is a homeomorphism on Rn (a detailed check of this is left to the reader);(ii) h maps orbits of (2.23) onto orbits of (2.29), preserving time parametrization.

For the proof of the last statement use the fact that τ(esAx) = τ(x) + s. 2

Applying Theorem 2.25 in the stable eigenspace directly and after time reversalin the unstable eigenspace, we arrive at the following result.

Theorem 2.26 Two hyperbolic linear systems in Rn, x = Ax and y = By, are

topologically conjugate if the matrices A and B have the same number of eigenvalues

with negative real part.

Remark:Actually, the inverse result is also valid, i.e. one can write “if and only if” in

Theorem 2.26. However, this fact is only of theoretical interest, since it is rarelyknown a priori that two linear ODE systems are topologically conjugate. 3

2.3.2 Topological equivalence of hyperbolic linear maps

When are two hyperbolic linear maps topologically equivalent ? We begin with anexample that illustrates that that we can use essentially the same idea as used in thecontinuous time case, but, because now we deal with “jumps” rather than with acontinuous orbit, we need to replace the boundary of a ball by a set with non-emptyinterior.

Example 2.27 (Topologically equivalent scalar linear maps)

Consider two linear scalar maps:

f : x 7→ 1

2x and g : y 7→ 1

3y.


13

yh(x)x

y′x′

H

0

x′ = 12x

01

y′ = 13y

y′ = yx′ = x

x12

1

Figure 2.7: The maps f : x 7→ 12x and g : y 7→ 1

3y are topologically conjugate.

These maps are not smoothly conjugate, since they have different eigenvalues, i.e.12

and 13

(see Exercise 1.5.12). However, they are topologically conjugate. Indeed,the following construction (see Figure 2.7) gives a conjugating homeomorphism h :R → R.

Take any, for example linear, homeomorphism

H : [12, 1] → [1

3, 1],

satisfying H(12) = 1

3, H(1) = 1. The intervals are called the fundamental domains.

For x ∈ (0, 12) take

h(x) = gk(x)(H(f−k(x)(x))),

where k(x) is the minimal2 positive integer k, such that

f−k(x) ∈ [12, 1].

For x > 1 takeh(x) = g−l(x)(H(f l(x)(x))),

where l(x) is the minimal positive integer l, such that

f l(x) ∈ [12, 1].

Make the analogous construction for x < 0. Set h(0) = 0. The map h constructedin this manner has the following properties:

(i) h : R → R is a homeomorphism; if H is linear, h is piecewise-linear.(ii) h sends orbits of f to orbits of g, i.e. f = h−1 ◦ g ◦ h.

This means that f ∼ g.Now consider another map g defined by

g : y 7→ −1

3y.

2The adjective “minimal” is added to be specific but it is, in fact, superfluous. Ambiguity canonly arise when x = 2−n but then both choices k(x) = n − 1 and k(x) = n yield h(x) = 3−n so,after all, the ambiguity is harmless.


This map is not topologically equivalent to f . Indeed, suppose that there exists acontinuous map h : R → R sending orbits of f to orbits of g. In particular, it shouldmap the orbit of f

. . . , 1,1

2,1

4, . . .

to the orbit of g

. . . , a,−a3,a

9, . . . ,

where a 6= 0. Therefore, we must have

h(1) = a, h

(

1

2

)

= −a3, h

(

1

4

)

=a

9,

showing that h is not monotone. Therefore, h is not invertible. 3

Lemma 2.28 Let

A : x 7→ Ax, x ∈ Rn, (2.30)

be an invertible linear map which is a contraction with respect to a norm ‖ · ‖ in Rn.

Then any linear map B,

B : y 7→ By, y ∈ Rn, (2.31)

with ‖B −A‖ sufficiently small, is topologically conjugate to A.

Proof: We have‖Ax‖ ≤ ρ‖x‖, x ∈ R

n,

for some ρ < 1. Take the unit sphere

Sn−1 = {x ∈ Rn : ‖x‖ = 1}

and consider its image ASn−1 ⊂ IntSn−1. Define a fundamental domain for A:

ASn−1

DA

Sn−1

v

w

A−1v

A−1v‖A−1v‖

Figure 2.8: Fundamental domain for the contraction A.

DA = Sn−1 ∪ (Int Sn−1 \ Int ASn−1).


ASn−1 BSn−1

DA

DB

H(w)

w

A−1

B

H

Sn−1Sn−1

id

Figure 2.9: Constraints on the map H : H|Sn−1 = id, H|ASn−1 = BA−1.

This is a closed annulus between Sn−1 and ASn−1 (see Figure 2.8).If ‖B − A‖ ≤ ε with sufficiently small ε, we obtain by the triangle inequality

‖B‖ = ‖B − A+ A‖ ≤ ‖B − A‖ + ‖A‖ ≤ ε+ ρ =: ρ1 < 1

and therefore‖By‖ ≤ ρ1‖y‖, y ∈ R

n.

Define the fundamental domain for B by

DB = Sn−1 ∪ (Int Sn−1 \ Int BSn−1).

Suppose that we can construct a homeomorphism

H : DA → DB,

which maps the corresponding boundaries of the fundamental domains DA and DB

into each other and which “agrees” with the maps A and B on them, in the sensethat

H|Sn−1 = id. (2.32)

andH|ASn−1 = BA−1 (2.33)

(see Figure 2.9). Then we can define a map

h : Rn → R

n

by setting first h|DA= H and then defining for x ∈ IntASn−1, x 6= 0,

h(x) = Bk(x)H(A−k(x)x),

where k(x) is the minimal integer k > 0, such that A−kx ∈ DA. Similarly defineh(x) for x ∈ ExtSn−1, and finally set h(0) = 0. A careful reasoning using (2.32)and (2.33) shows that the resulting map h : Rn → Rn is a homeomorphism with theproperty

Ax = h−1(Bh(x)), x ∈ Rn,


meaning that B is topologically conjugate to A. Therefore, the lemma will be provedif we construct a homeomorphism H : DA → DB, satisfying (2.32) and (2.33). Therest of the proof is devoted to this construction.

First, we notice that any ray

{x ∈ Rn : x = sv, s ∈ (0,∞)}

for some given v ∈ Rn with ‖v‖ = 1, intersects ASn−1 at a unique point

w = A

(

A−1v

‖A−1v‖

)

=v

‖A−1v‖(see Figure 2.8). This allows us to introduce new coordinates (v, t) in DA. Namely,identify x ∈ DA with a point (v, t) ∈ Sn−1 × [0, 1] as follows. For each x ∈ DA, take

v =x

‖x‖ ∈ Sn−1,

compute

a =1

‖A−1v‖ =‖x‖

‖A−1x‖ ∈ R,

so that w = av, and then set

t =a− ‖x‖a− 1

∈ [0, 1].

(In Exercise 2.5.16 below you are asked to compute the inverse of the map x 7→(v, t).) Make a similar construction for DB, identifying each point y ∈ DB with(η, τ) ∈ Sn−1 × [0, 1].

The boundary constraints (2.32) and (2.33) motivate us to define two homeo-morphisms

δ1,0 : Sn−1 → Sn−1,

by δ1 = id and

δ0(v) =BA−1v

‖BA−1v‖ .

From the fact that B is close to A and, thus, BA−1 is close to the unit matrix, weinfer that δ0 is close to the identity map. The map

δt(v) =(1 − t)δ0(v) + tδ1(v)

‖(1 − t)δ0(v) + tδ1(v)‖,

is a homeomorphism on Sn−1 for all t ∈ [0, 1] (Prove this!). Therefore, the map∆ : Sn−1 × [0, 1] → Sn−1 × [0, 1] defined by the formula

∆(v, t) = (δt(v), t)

is a homeomorphism. Finally, the homeomorphism ∆ induces, via the above con-structed coordinates, a homeomorphism

H : DA → DB


with the required properties (2.32) and (2.33). Indeed, the property (2.32) is obvious,while the property (2.33) can be verified as follows. Take v ∈ Sn−1 and consider thepoint

w =v

‖A−1v‖ ∈ ASn−1

that has coordinates (v, 0) ∈ Sn−1 × [0, 1] in DA. Note that any point in ASn−1 canbe uniquely represented in this way. Then

BA−1w = BA−1

(

v

‖A−1v‖

)

=BA−1v

‖A−1v‖ ∈ BSn−1.

On the other hand,

δ0(v) =BA−1v

‖BA−1v‖ ∈ Sn,

which implies

η = δ0(v) = δ0(v) =BA−1v

‖BA−1v‖ ∈ Sn.

The point with coordinates (η, 0) ∈ Sn−1 × [0, 1] corresponds to the point

H(w) =η

‖B−1η‖ =BA−1v

‖A−1v‖ ∈ BSn−1,

from which we conclude that H(w) = BA−1w for w ∈ ASn−1. 2

Theorem 2.29 Two linear invertible contractions in Rn that either both preserve

or both reverse the orientation of Rn are topologically conjugate.

Proof: Let x 7→ Ax and x 7→ Bx be two such maps. By the assumption, thedeterminants of the corresponding matrices A and B are both positive or negative.Therefore, there exists a continuous family of matrices M(t), t ∈ [0, 1], such thatM(0) = A,M(1) = B, and

detM(t) 6= 0, r(M(t)) < 1

for all t ∈ [0, 1] (see Exercise 2.5.17). By Lemma 2.28, any two linear maps M(t′)and M(t′′) are topologically conjugate if |t′′− t′| is sufficiently small. Since [0, 1] canbe covered with finitely many such intervals, A is topologically conjugate to B. 2

Suppose that

f1 : Rn1 → R

n1 , g1 : Rn1 → R

n1 ,

are two topologically equivalent maps, i.e. f1 = h−11 ◦g1◦h1 for some homeomorphism

h1 : Rn1 → Rn1. Suppose also that

f2 : Rn2 → R

n2 , g2 : Rn2 → R

n2 ,


are two other topologically equivalent maps, i.e. f2 = h−12 ◦ g2 ◦ h2 for some

homeomorphism h2 : Rn2 → Rn2 . Then obviously the direct product maps f, g :Rn1 × Rn2 → Rn1 × Rn2 defined coordinate-wise,

f :

(

x1

x2

)

7→(

f1(x1)f2(x2)

)

, g :

(

x1

x2

)

7→(

g1(x1)g2(x2)

)

,

are topologically equivalent. Indeed,

f = h−1 ◦ g ◦ h,

where h : Rn1 × R

n2 → Rn1 × R

n2 is a homeomorphism defined by

h :

(

x1

x2

)

7→(

h1(x1)h2(x2)

)

.

The topology in the product space can be induced, for example, by the norm

‖x‖ = ‖x1‖ + ‖x2‖.

Let ns(M) and nu(M) denote the number of eigenvalues of the matrix M insideand outside the unit circle, respectively (taking multiplicities into account). Further,let ps(M) and pu(M) denote the products of the eigenvalues inside and outsidethe unit circle, respectively. Recall that topological conjugacy of two (in our caselinear) invertible maps is equivalent to the conjugacy of their inverses (with thesame conjugating homeomorphism). Applying Theorem 2.29 to the restriction of ahyperbolic linear map to its stable eigenspace, and to the inverse of the restrictionto the unstable eigenspace, we obtain the following theorem.

Theorem 2.30 Two linear hyperbolic maps in Rn, x 7→ Ax, y 7→ By, are topolog-

ically conjugate if the following properties hold simultaneously:

ns(A) = ns(B), ps(A)ps(B) > 0, pu(A)pu(B) > 0. 2

Remark:As in the continuous time case, the inverse for Theorem 2.30 is also valid but

again of minor interest. 3

2.4 References

There are many good books, where linear autonomous ODEs are treated. The stan-dard references here are [Arnol’d 1973, Hirsch & Smale 1974]. The analysis of linearmaps can be done using similar techniques, see, for example [Kato 1980, ChapterI]. The construction of spectral projectors using the partial fraction decompositionis inspired by unpublished notes of Joop Kolk (Utrecht University).

The fundamental domains are used in [Katok & Hasselblatt 1995] to prove thelocal topological equivalence of a map with a stable hyperbolic linear part to itslinearization.

2.5. EXERCISES 69

2.5 Exercises

E 2.5.1 (Exceptional cases of real simple eigenvalue)

Describe the dynamics generated by the linear map A on the line Xλ when

(i) λ = 1;

(ii) λ = −1;

(iii) λ = 0.

E 2.5.2 (Linear algebra refresher)

(a) Let λ be a nonreal eigenvalue of A with eigenvector v. Show that:

(i) v and v are linearly independent in Cn;

(ii) Re v and Im v are linearly independent in Rn.

(b) Let λ and µ be eigenvalues of A with eigenvectors v and w, respectively. Show that λ 6= µ

implies that these vectors are linearly independent.

(c) Let Av(i) = λiv(i) with v(i) 6= 0 and λi 6= λj for i 6= j, where i, j = 1, 2, . . . , n. Prove that the

vectors v(i) are linearly independent.

E 2.5.3 (Spectral projection)

Consider

A =

(

112 −36 − 7

2

)

as a generator of a discrete-time dynamical system in R2. Compute the projectors correspondingto the decomposition R

2 = T s ⊕ T u.

E 2.5.4 (Stability conditions for linear maps in R2 and R3)

(a) Prove all statements formulated in Example 2.3.

(b) Consider a three-dimensional linear map

x 7→ Ax, x ∈ R3

and write the characteristic equation det(λI −A) = 0 in the form

λ3 + a0λ2 + a1λ+ a2 = 0.

Show that all its roots lie strictly inside the unit circle if and only if the following inequalities holdsimultaneously:

1 + a0 + a1 + a2 > 0, 1 − a0 + a1 − a2 > 0, |a2| < 1, and 1 − a22 > a1 − a0a2.

Also verify that λ1 = 1 is a root if 1+a0+a1+a2 = 0, that λ1 = −1 is a root if 1−a0+a1−a2 = 0,and that there are two roots λ1,2 with λ1λ2 = 1 if

1 − a22 = a1 − a0a2.

Show that in the latter case λ1,2 = e±iθ with 0 < θ < π where cos θ = 12 (a2 − a0), provided that

|a2 − a0| < 2.

E 2.5.5 (Operator norm of matrices)


Let ‖x‖ be the standard Euclidian norm of x ∈ Rn. Consider a n× n matrix A with real elementsand define its operator norm ‖A‖ by the formula (2.9) with Ax understood as the matrix-vectorproduct.

(a) Prove that ‖A‖ =√µmax, where µmax is the largest eigenvalue of the symmetric matrix

ATA. (Hint: ‖Ax‖2 = 〈Ax,Ax〉 = 〈ATAx, x〉, where 〈x, y〉 = xT y, the standard scalar product ofx, y ∈ R

n. Recall that a symmetric matrix has an orthonormal basis consisting of eigenvectors.)(b) Using (a), show that for a (2 × 2)-matrix

A =

(

a b

c d

)

, a, b, c, d ∈ R,

the operator norm satisfies

‖A‖2 =1

2

[

a2 + b2 + c2 + d2 +√

[(a− d)2 + (b+ c)2][(a+ d)2 + (b − c)2]]

.

(c) Apply the formula derived in (b) to prove that

‖A−1‖ =1

| detA| ‖A‖

for any nonsingular 2 × 2 real matrix A. Is this result valid for n > 2 ?

E 2.5.6 (Fibonacci numbers)

Define a sequence of integers by the recurrence relation

ak = ak−1 + ak−2, k ≥ 2,

with initial condition a0 = a1 = 1. Find an explicit expression for ak.Hint: Write

(

ak

ak+1

)

= A

(

ak−1

ak

)

,

where

A =

(

0 11 1

)

,

and consider a discrete-time dynamical system generated by the linear map x 7→ Ax in R2 (cf.Exercise 1.5.6 in Chapter 1). Find eigenvalues λ and µ and corresponding eigenvectors v and w ofA and compute the associated projection operator(s).

Answer :

ak =1

2k+1√

5

[

(1 +√

5)k+1 − (1 −√

5)k+1]

.

E 2.5.7 (Quantum oscillations)

In Quantum Mechanics, a system with two observable states is characterized by a vector

ψ =

(

a1

a2

)

∈ C2,

where ak are complex numbers called amplitudes, satisfying the condition

|a1|2 + |a2|2 = 1.

The probability of finding the system in the kth state is equal to pk = |ak|2, k = 1, 2.The behaviour of the system between observations is governed by the Heisenberg equation:

i~dψ

dt= Hψ,

2.5. EXERCISES 71

where the real symmetric matrix

H =

(

E0 −A−A E0

)

, E0, A > 0,

is the Hamiltonian matrix of the system and ~ is Planck’s constant divided by 2π.(i) Write the Heisenberg equation as a system of two linear complex differential equations for

the amplitudes.(ii) Integrate the obtained system and show that the probability pk of finding the system in

the kth state oscillates periodically in time.(iii) How does p1 + p2 behave?

E 2.5.8 (Jordan blocks)

Write the Jordan block from (2.6) as J = λI +N with the matrix

N =

0 1 00 1

0 10 0

(nilpotent since Nnj−1 = 0) and use the binomial formula to find expressions for Jk and exp(tJ) =∑∞

k=0(tJ)k

k! .

E 2.5.9 (Jordan decomposition)

It is known3 that each n× n complex matrix A can be uniquely written as

A = S +N,

where S is diagonal in some basis (semisimple) and Nm = 0 for some m < n (nilpotent), and suchthat

SN = NS

(Jordan decomposition).Let Pi be the spectral projectors onto the generalized eigenspaces corresponding to the distinct

eigenvalues λ1, λ2, · · · , λd of A.(i) Prove that S =

∑di=1 λiPi is semisimple.

(ii) Define N = A− S and prove that it is nilpotent.(iii) Using the definition of exp(tJ) from Exercise 2.5.8, prove that

exp(tS) =

d∑

i=1

eλitPi, and exp(tN) =

m−1∑

j=1

tj

j!N j .

(iv) Show that

exp(tA) =

d∑

i=1

eλit

m−1∑

j=1

tj

j!N j

Pi.

(v) Combine the formula from step (iv) with the partial fraction decomposition method tocompute Pi and obtain an efficient algorithm for finding flows of linear autonomous systems.

E 2.5.10 (Stability and eigenvalues for maps)

3See, for example, Section 6.2 in Hirsch, M.W. and Smale, S. Differential Equations, Dynamical

Systems, and Linear Algebra. Academic Press, New York-London, 1974.


(i) Extend Theorem 2.9 by showing that asymptotic stability of the origin implies that all eigen-values of A lie strictly inside the unit circle.(ii) Prove that for a map x 7→ Ax the following are equivalent(a) The origin is stable;(b) Each eigenvalue either lies strictly inside the unit circle or lies on the unit circle but has nogeneralized eigenvector;(c) There exists a norm ‖ · ‖1 in Rn such that ‖A‖1 ≤ 1.

Hint: (a) ⇒ (b) follows from Theorem 2.2 and for (b)⇒ (c) prove the boundedness of Ak(k ≥ 0)and define ‖x‖1 = supk≥0 ‖Akx‖.

E 2.5.11 (Stability and eigenvalues for flows)

Make an analog of Exercise 2.5.10 for ODE’s x = Ax.(i) Asymptotic stability of the origin implies that all eigenvalues of A have negative real part.(ii)Lyapunov stability holds if and only if all eigenvalues either have negative real part or have zeroreal part but no generalized eigenvectors. Another equivalent condition says that there exists anequivalent norm ‖ · ‖1 in Rn such that ‖etA‖1 ≤ 1 for all t ≥ 0.

Hint: For the construction of a proper norm use ‖x‖1 = supt≥0 ‖etAx‖.

E 2.5.12 (Asymptotic stability and contraction)

Consider two maps in R2 generated by the matrices

A =

(

12 00 1

4

)

and B =

(

12 − 1

2

1 14

)

.

(i) Show that the origin is an asymptotically stable fixed points for both maps.(ii) Which of these maps is a linear contraction with respect to the standard norm ‖x‖ =

√

x21 + x2

2 in R2 ?Hint: Consider the images AU and BU of the unit disk

U = {x ∈ R2 : x2

1 + x22 ≤ 1}

under the maps A and B, respectively. Show that AU is located strictly inside U , while BU doesnot have this property (see Figure 2.10).

(iii) Construct several first elements of the sequences of the images

U,AU,A2U,A3U, . . .

andU,BU,B2U,B3U, . . .

and convince yourself that both sequences contract towards the origin.

E 2.5.13 (Asymptotic stability and monotonicity)

Consider a linear ODEx = Ax, x ∈ R

2, (2.34)

defined by the matrix

A =

(

− 25 2

− 12 − 2

5

)

.

(i) Prove that x = 0 is an asymptotically stable equilibrium of (2.34). What kind of equilibriumis it?

(ii) By numerical integration, show that the orbit starting at x0 = (1, 0)T crosses the unitcircle ‖x‖2 = x2

1 + x22 = 1 several times (see Figure 2.11 (a)). Prove this analytically.

(iii) Define ‖x‖22 = x2

1 + 4x22. Prove that ‖ · ‖2 is a norm in R

2 that is equivalent to ‖ · ‖.

2.5. EXERCISES 73

-2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2-2

-1.6

-1.2

-0.8

-0.4

0

0.4

0.8

1.2

1.6

2

x3

y3

-2

-1.6

-1.2

-0.8

-0.4

0

0.4

0.8

1.2

1.6

2

x3

y3

-2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2-2

-1.6

-1.2

-0.8

-0.4

0

0.4

0.8

1.2

1.6

2

x3

y3

-2

-1.6

-1.2

-0.8

-0.4

0

0.4

0.8

1.2

1.6

2

x3

y3

(a) (b)

x2

x1

x2

x1

U

BU

U

AU

Figure 2.10: Asymptotically stable fixed points: (a) a linear contraction; (b) not acontraction.

-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1.2

-0.9

-0.6

-0.3

0

0.3

0.6

0.9

1.2

1.5

x

y

-1.5

-1.2

-0.9

-0.6

-0.3

0

0.3

0.6

0.9

1.2

1.5

x

y

-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1.2

-0.9

-0.6

-0.3

0

0.3

0.6

0.9

1.2

1.5

x

y

-1.5

-1.2

-0.9

-0.6

-0.3

0

0.3

0.6

0.9

1.2

1.5

x

y

(b)(a)

x1

x2

x1

x2

Figure 2.11: Nonmonotone (a) and monotone (b) convergence.


(iv) Prove that any “circle” ‖x‖2 = R0 > 0 is crossed by any orbit of the system only once(see Figure 2.11 (b)).

Hint: Show thatd

dt‖x(t)‖2

2 = −4

5‖x(t)‖2

2

along any nonequilibrium solution of (2.34). Find R(t) = ‖x(t)‖2 by integrating this differentialequation and demonstrate that it converges to zero monotonously. Is ‖ · ‖2 a Lyapunov norm for(2.34)?

E 2.5.14 (Orientation and phase portraits)

In the two dimensional focus case derive a criterion for the matrix coefficients that characterizeswhether orbits are rotating clockwise or counterclockwise. Compare with the phase portraits fromthe previous exercise.

E 2.5.15 (Explicit conjugating homeomorphism)

Verify that map h : R → R defined by the formula

h(x) =

{

xν , x ≥ 0,−|x|ν , x < 0,

defines a homeomorphism conjugating the maps f and g from Example 2.27 for a suitable ν (whichone ?).

E 2.5.16 (Inverse map in Lemma 2.28) Compute the inverse of the map x 7→ (v, t) defined

in the proof of Lemma 2.28.

Answer:

(v, t) 7→(

t

(

1 − 1

‖A−1v‖

)

+1

‖A−1v‖

)

v.

E 2.5.17 (Connectedness of contractions)

Show that the set of contracting and positively oriented matrices M = {A : det(A) > 0, r(A) < 1}is connected, i.e. for any two A,B ∈M there is a continuous path M(t) ∈M, 0 ≤ t ≤ 1 such thatM(0) = A and M(1) = B.

Hint: It is sufficient to connect any A to the special matrix B = 12I ∈M . If A has no negative

eigenvalues, then take the line tB + (1 − t)A. If A has negative eigenvalues, they must occur inpairs. Isolate the invariant subspaces belonging to them and move the eigenvalues to the positivereal axis by a suitable rotation.

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