Download Let [R denote the set of real numbers and C the set of complex

4
actions result.
and vice versa.
L-U'LUHULLI.HU
facilitates a stochastic inter-
of
For the historical survey in this chapter we
used the MacTutor
Mathematics Archive from the
of St. Andrews in Scotland (2003) and the
Furthermore, the paper by
outline of the history of game theory by Walker
Breitner (2002) was used for a reconstmction of the early days of dynamic game theory.
This chapter reviews some basic linear algebra including material which in some
instances goes beyond the introductory level. A more detailed treatment of the basics
of linear algebra can for example, be found in
(2003), whereas Lancaster and
Tismenetsky (1985) provide excellent work for those who are interested in more details at
an advanced level. We will outline the most important basic concepts in linear algebra
together with some theorems we need later on for the development of the theory. A
detailed treatment of these subjects and the proofs of most of these theorems is omitted,
since they can be found in almost any textbook that provides an introduction to linear
algebra. The second part of this chapter deals with some subjects which are, usually, not
dealt with in an introduction to linear algebra. This part provides more proofs because
either they cannot easily be found in the standard linear algebra textbooks or they give an
insight into the understanding of problems which will be encountered later on in this
book.
Let [R denote the set of real numbers and C the set of complex numbers. For those who
are not familiar with the set of complex numbers, a short introduction to this set is given
where each
is
in section 2.3.
let [Rll be the set of vectors with n
an element of [R. Now let Xj, ... ,Xk E [Rll. Then an element of the form Cl:jXj + ... + Cl:kXk
with Cl:i E [R is a linear combination of XI ,
,Xk. The set of all linear combinations of
XI, X2, ... ,Xk E [Rll, called the
of XI ,X2,
,Xk, constitutes a linear
of [Rll.
That is, with any two elements in this set the sum and any scalar multiple of an element
also belong to this set. We denote this set by Span {Xl,X2,'" ,Xk}.
A set of vectors Xl, X2, ... ,Xk E [Rll are called
if there exists
Cl:], ... ,Cl:k E [R, not all zero, such that Cl:]Xl + ... + Cl:kXk = 0; otherwise they are said to
be liin'O>OJIrhT nl(le:pe]l1dc~nt.
Upt'imi:Gatil'Jn and Differential Games
Sons, Ltd
© 2005 John Wiley &
J. Engwerda
7
16
Let S be a
of [H1l, then a set of vectors {b I , b 2 ...
this set of vectors are linearly
and S
Consider the vectors
el :=
[b]
/ ]} are a basis for
and e2:=
[n
, bk}
is called a basis for S if
... ,bd·
Given a basis
VI
= [~~] then, nsing the theorem of Pythagoras, the length of x is II x
Using induction it is easy to verify that the length of a vector x
112 =
=
E [H1l is
II x
J
aT + ... + a~. Introducing the superscript T for
[al ... all], we can rewrite this result in shorthand as
Two vectors x, y E
[H1l
are perpendicular if and only if x T y
II x
= O.
D
Based on this result we next introduce the concept of orthogonality. A set of vectors
{XI, ... ,xll } are mutually
if Xj = 0 for all i i=- j and orthonormal if
X.i = bu. Here bU is the Kronecker delta function with bU = 1 for i = j and bU = 0
for i i=- j. More generally, a collection of subspaces SI, ... ,Sk are mutually orthogonal if
x T y = 0 whenever x E Si and y E Sj, for i i=- j.
The
of a subspace S is defined by
xf
xf
S1- :=
{y E [HnlyTx =
0 for all
bf
._ b
b~VI
b~V2
V,n ·- m --T-VI - - T - V2 _ ...
VI VI
v 2 V2
Then
{II~: I '
b~Vm-1
T
Vm_ 1Vm-I
,... , II~;;;: II} is an orthonormal basis for S.
Vm-l'
..,,-,,"~
I-;·nc..th."'......
Span{vI, V2,"" vd = Span{h, b 2,···, bd·
In the above sketched case, with
Ui
D
=
S1- = Span{uk+l, ... ,un}'
of a vector, i.e.
112 = ~. Now, two
vectors x and yare perpendicular if they enclose an angle of 90°. Using Pythagoras
theorem again we conclude that two vectors x and yare perpendicular if and only if the
length of the hypotenuse II x - y 11 2 = II X 11 2 + II Y 11 2 . Or, rephrased in our previous
terminology: (x - y)T (x - y) = x TX + yTy. Using the elementary vector calculation
rules and the fact that the transpose of a scalar is the same scalar (i.e. x Ty = yT x)
straightforward calculation shows that the following theorem holds.
112 =
=
:=h
bf
all
xT
define
VI
V2
V3 := b 3 - -T-VI - -T-- V2
VI VI
v 2 V2
.jai + a~.
[~I]
[H1l,
V2 := b 2 - y-VI
VI VI
. The set {e I, e2} is called the standard basis for
So, a basis for a subspace S is not unique. However, all bases for S have the same number
of elements. This number is called the dimension of S and is denoted by dimeS). In the
above example the dimension of [H2 is 2.
Next we consider the problem under which conditions are two vectors perpendicular.
of a vector x is introduced which will be denoted by II x 112.
First the
for a sut)space S of
b~vI
in 1hl2 Then, both {el, e2} and
D
If x
{b I , b 2 , ... , b m}
Since {u I, ... , Un} form a basis for [Hn, {Uk+ I, ... , Un} is called an orthonormal
cmr:npletion of {u I , U2, ... , Uk}.
An ordered array of mn elements au E [H, i = 1, ... ,n; j = 1, ... ,m, written in the
form
al2
A=
aim]
[ all
a21
a22
a2m
a:71
a ,J2
al~1II
is said to be an n x m-matrix with entries in lH. The set of all n x m-matrices will be
denoted by [HnxIII. If m = n, A is called a square matrix. With matrix A E [HIlXIII one can
associate the linear map x ---7 Ax from [Hm ---7 [H1l. The kernel or null space of A is defined
by
ker A = N(A) := {x E
and the
[Hili
= O},
or range of A is
1m A = R(A) := {y E [Hnl y = Ax,X E [Hili}.
XES}.
A set of vectors {u I, U2, ... , Uk} is called an orthonormal basis for a subspace S c [H1l if
they form a basis of S and are orthonormal. Using the orthogonalization procedure of
Gram-Schmidt it is always possible to extend such a basis to a full orthonormal basis
{UI' U2, ... ,ull } for [H1l. This procedure is given in the following theorem.
IGram was a Danish mathematician who lived from 1850-1916 and worked in the insurance
business. S~hn:idt was a German mathematician who lived from 1876-1959. Schmidt ~eproved the
orthogonahzatlOn procedure from Gram in a more general context in 1906. However, Gram was not
the first to use this procedure. The procedure seems to be a result of Laplace and it was essentially
used by Cauchy in 1836.
is a
Let A E
of
of [R\m and
fundamental result.
[R\1l.
2. A is invertible if and
[Rllxm.
+
5. det([~ ;])
= in.
D
= dim((kerA)l-).
2.
Let ai, i = 1, ...
,in,
denote the columns of matrix A E
[RIlX'Il,
then
1m A = Span{a1,"" am}.
The rank of a matrix A is defined by rank(A) = dim(ImA), and thus the rank of a matrix
T
is just the number of independent columns in A. One can show that rank(A) = rank(A ).
Consequently, the rank of a matrix also coincides with the number of independent rows in
A. A matrix A E [RIlX'1l is said to have full row rank if n ::; in and rank(A) = n. Equally, it
is said to have full column rank if in ::; nand rank (A ) in. A full rank square matrix is
called a
or invertible
otherwise it is called singular. The following
result is well-known.
if
:f: 0;
D
=
Next we present Cramer's rule to calculate the inverse of a matrix. This way of
calculating the inverse is sometimes helpful in theoretical calculations, as we will see
later on.
For any n x n matrix A and any b E [R\'\ let Ai (b) be the matrix obtained from A
replacing column i by the vector b
Let A be an invertible n x n matrix. For any b E
entries given
[R'\
the
solution x of Ax
xi=-d-e-tA-' i= 1,2, ... ,n.
Let A E
[Rllxm
and b E
[R1l.
Then, Ax = b has
b has
D
Cramer's rule leads easily to a
formula for the inverse of an n x n matrix A. To
see this, notice that the jth column of
is a vector x that satisfies
1. at most one solution x if A has full column rank;
2. at least one solution x if A has full row rank;
3. a solution if and only if rank([Alb]) = rank(A);
a unique solution if A is invertible.
where ej is jth column of the identity matrix.
Cramer's rule,
D
xij
If a matrix A is invertible one can show that the matlix equation AX = 1 has a unique
matrix with entries eij := Dij, i, j =
solution X E [RIlXIl. Here 1 is the n x n
1, ... , n, and Dij is the Kronecker delta. Moreover, this matrix X also satisfies the matrix
equation XA = 1. Matrix X is called the inverse of matrix A and the notation A-I is used
:= [(i,j) - entry of
_ detAi(ej)
detA
(2.1.1)
Let
denote the submatrix of A formed by deleting row j and column i from matrix A.
An expansion of the determinant down column i of A i ( ej) shows that
to denote this inverse.
A notion that is useful to see whether or not a square n x n matrix A is singular is the
determinant of
denoted by det(A). The next theorem lists some properties of
determinants.
Let
Then,
B E [RIlXIl; C E [RIlXIIl; D E [RlIlxm; and 0 E [R'IlXIl be the matrix with all entries zero.
2Cramer was a well-known Swiss mathematician who lived from 1704-1752. He showed this
rule in an appendix of his book (Cramer, 1750). However, he was not the first one to give this rule.
The Japanese mathematician Takakazll and the German mathematician Leibniz had considered this
idea already in 1683 and 1693, respectively, long before a separate theory of matrices was
developed.
20
If the entries of
Thus
divided
is
the
Qii.
are
next
Qij,
i ,j
= 1, ... ,n, the
of
is defined as
are well-known.
ell
=
C.12
_1_
det
:
[
\fA E
= trace
The entries
are the so-called cofactors of matrix A. Notice that the subscripts on
are the reverse of the entry number (i,j) in the matrix. The matrix of cofactors on the
light-hand side of (2.1.2) is called the
of
denoted
adj A. The next theorem
.2).
simply restates
Let A be an invertible n x n matrix. Then
1
adj (A).
(2.1.3)
D
If
A
=
1
[2 -0 A 2-A
-1o J,
1
-1
= trace
3.
~IlXIl
and a E
~;
+ trace
\fA E
~Ilxm,
B E
~mxll.
D
Let A E ~IlXIl, then A E ~ is called an
of A if there exists a vector x E ~1Z,
different from zero, such that Ax = Ax. If such a scalar A and conesponding vector x
the vector x is called an
If A has an eigenvalue A it follows that there
exists a nonzero vector x such that (A = O. Stated differently, matrix A - AI is
singular.
according to Theorem 2.5, A is an eigenvalue of matrix A if and
if
= O. All vectors in the null space of A AI are then the eigenvectors
corresponding to '\. As a consequence we have that the set of eigenvectors conesponding
with an eigenvalue A forming a subspace. This subspace is called the
of ,\ and
we denote this subset by EA' So, to find the eigenvalues of a matrix A we have to find
= O. Since p(A) := det(A is a polynomial of
those values ,\ for which det(A degree n, p(A) is called the characteristic
of A. The set of roots of this
polynomial is called the
of A and is denoted by a(A).
An important property of eigenvectors corresponding to different eigenvalues is that
they are always independent.
3-A
then
[2 - A 30A] -det[ 1 -1 ] det[ 1 ~1 ]
2-A
-1 3-A
-1
1] -det[2 A 01]
-det[~ 3~ A] det [2-A
0
13,\
A] -det [2 -1 A ~1 ] det [2 - A 2~ A]
det[~ 20
-1
[A 5A + 6 ,\2 ,\-2
-H2
o J.
5,\ + 7
0
Let A E ~IlXIl and AI, A2 be two different eigenvalues of A with conesponding eigenvectors Xl and X2, respectively. Then {Xl, X2} are linearly independent.
det
adj(A)
=
j1XI, for some nonzero scalar j1 E ~. Then 0 =
/-LX]) =
j1AIX] = A2j1Xj - j1AjXj = j1('\2 - )1]
I- 0, due to the stated assumptions. So this yields a contradiction and therefore our assumption that X2 is a multiple
Assume X2
= A7X7
x]
D
must be inconect.
2
-
=
A-2
-,\ + 3
A2 - 4A +4
Notice that all entries of the adjoint matrix are polynomials with a degree that does not
exceed 2. Furthermore, det(A) = -A 3 + 7 A2 - 17 A + 14, which is a polynomial of
degree 3.
D
1. Consider matrix
__ [-21
-3]
4 . The characteristic polynomial of
p(A) = det(A j -'\I) = (,\ - 1)('\ - 2).
is
22
) = {I,
and
i].
2. Consider matrix
3. Consider matrix
=
=
=[
={a[i],aE~}
+
[~ ~ ] . The characteristic polynomial of
= {3}. Furthermore,
Consider matrix
is (A +
The characteristic
={-2}.
=
3I)
for [Rn
with columns b 1 , ..• ,bn •
=
is (A - 3
f
where ltc denotes the k x k
[R2.
Then
~2 : ] . The characteristic polynomial of A 4 is (A 2 -
4A + 5).
has no real eigenvalues.
D
The above
UHfJ~V illustrates a number of properties that hold in the general
(Lancaster and
1985).
too
This polynomial has no real roots. So, matrix
""A •.
, we have A = S [0
this relation by
det(A -
det
[
D
0
([
= det( [(AJ
polynomial p(A) can be factorized as the product of different linear and quadratic
terms, i.e.
a square (n - k) x (n - k) matrix.
matrix and
. Therefore
]S-l _
D
0
D]
]
-
D
0
])
= (AI - A) kdet(En-k
D
It turns out that
for some scalars c, Ai, hi and Ci·
quadratic terms do not have real roots.
for i f j, ,\, f A, and
ni
[~:]
f
[~;]
and the
+ 2 L:;~k-I-l ni = n.
D
The power index ni appearing in the factorization with the factor A - Ai is called the
~I<Il'4J>h.1r'Qli ... nlUltlJlIUcny of the eigenvalue Ai. Closely related to this number is the soof the eigenvalue Ai, which is the dimension of the
cOlTesponding eigenspace
In Example 2.3 we see that for every eigenvalue the
geometric multiplicity is smaller than its algebraic multiplicity. For instance, for
both
multiplicities are 1 for both eigenvalues, whereas for
the geometric multiplicity of the
eigenvalue -2 is 1 and its algebraic multiplicity is 2. This property holds in general.
Let Ai be an eigenvalue of A. Then its geometric multiplicity is always smaller than
to) its algebraic multiplicity.
for an eigenvalue AI, there holds a strict inequality between its
geometric and algebraic multiplicity, there is a natural way to extend the eigenspace
towards a larger subspace whose dimension has the cOlTesponding algebraic multiplicity.
of the eigenvalue A1 and is
This larger subspace is called the
by
. In fact, there exists a minimal index p ::; n for which
follows from the property that
C
and the fact that whenever
(see Exercises).
In the previous subsection we saw that the characteristic polynomial of an n x n
involves a polynomial of degree n that can be factorized as the product of different linear
and quadratic terms (see Theorem 2.10). Furthermore, it is not possible to factorize any of
these quadratic terms as the product of two linear terms. Without loss of generality such a
quadratic term can be written as
(2.3.1)
Next introduce the
Assume Al is an eigenvalue of A and {b I , ... ,bk } is a set of independent eigenvectors
that span the cOlTesponding eigenspace. Extend this basis of the eigenspace to a basis
i to denote the square root of
i :=
1. So, by definition
25
24
o has two
this notation the eqllarJlon
\. _ 2a- ±- : - - - - - ' - - - - ' -
A] -
L>VJ.IUU\.JUL>,
=a±
(2.3.1) has, with this notation, the two square roots
Al = a + hi and
OPE~ratlOn
also induces a mIL1tl1Pw::atlon rule of two vectors in
as
= a ± hi, j = 1,2.
2
stated differently, p(A) in
Note that this
I.e.
=a -
(2.3.2)
hi.
An expression z of the form
We will not,
elaborate this point.
From equation (2.3.2) it is clear that closely related to the complex number z = x + yi
is the complex number x - yi. The complex number x yi is called the
of z
and denote it by z (read as 'z bar').
the conjugate of a complex number z is obtained
by reversing the sign of the imaginary part.
z = x + yi
where x and yare real numbers and i is the formal symbol satisfying the relation i2 = -1
is called a
number. x is called the real
of z and y the
of z.
Since any complex number x + iy is uniquely determined by the numbers x and y one can
visualize the set of all complex numbers as the set of all points (x,y) in the plane [R2, as in
Figure 2.1. The horizontal axis is called the real axis because the points (x,O) on it
= -2 + 4i.
The conjugate of z = -2 - 4i is z = -2 + 4i, that is
Geometrically, is the minor image of z in the real axis (see Figure 2.2).
z
Imaginary
axis
Imaginary
axis
y
z
x + iy
)'
.........•
x + iy
i
x
Real axis
-y
2.2
2.1
D
The complex plane C
conespond to the real numbers. The vertical axis is the
axis because the points
(O,y) on it conespond to the pure
numbers of the form 0 + yi, or simply yi.
Given this representation of the set of complex numbers it seems reasonable to introduce
the addition of two complex numbers just like the addition of two vectors in [R2, i.e.
x- iy
The conjugate of a complex number z
+ yi is the length of the
in ~2 That is, the absolnte value of z is the real number Izi defined
The absolute value or modulus of a complex number z = x
associated vector [; ]
by
Izi =
This number
Note that this rule reduces to ordinary addition of real numbers when Yl and Y2 are zero.
Furthermore, by our definition of i 2 = -1, we have implicitly also introduced the
of two complex numbers. This operation is defined by
operation of
:z
Real axis
Izi = ViZ,
Izi coincides with the square root of the product of z with its conjugate z, i.e.
We now turn to the division of complex numbers. The objective is to define devision as
the inverse of multiplication. Thus, if z =I=- 0, then the definition of ~ is the complex
number w that satisfies
wz = 1.
(2.3.3)
27
26
number Z #- 0 there
relatlon~shlp. The next theorem states that this
c.V'~"'"'1i- n~presemt(lt1(m of this number.
it is not a
number w
number w
exists and
vVJ.HjJ''-'''
ro""Y>-nIc.~
Just as vectors
IR Il and matrices in IR llxm are
can define vectors in CIl and
llxm
as vectors and matrices whose entries are now
numbers. The
matrices in C
VIJ'~H-<UVJ'hJ of addition and
are defined in the same way. FurtherZ of Z is defined as
more for a matrix Z with elements Zij from C the
ro"'.","". ....
the matrix obtained from Z
all its entries to their
other
the entries of Z are
VV'.HlJl.vA
If Z
#- 0,
(2.3.3) has a
then
L)Vl.UUVH.
which is
Let Z be any complex numtler. z\ =
4 + 2i
[ -2 - i
Let Z = a + bi and w
ax-by
l+(bx+ay)i=O.
[Z(2
Therefore, the equation (2.3.3) has a unique solution if and only if the next set of two
equations has a unique solution x, y
ax - by
has a unique solution. Note that det ([
~ ~b]) =
unique. It is easily verified that x = a2~b2 and y
the stated result follows straightforwardly.
a2
+
b oF O. So, the solution [~] is
2
-4z
=
l],z[ 2-4i.] _ [
3
- 3 + 2z
-4i)],
+
[(2 ~13~32+~)3 :42i 6i] = [~~ ~:]
(1+i)(4+2i)+i(-2-i)
[ (23i)(4+
+2(-2-i)
l
(l+i)(l-i)+i = [3+4i
(2-3i)(l-i)+2J
10-lOi
2+
i].
1 - 5i
vector Z E CIl can be written as Z = x + yi, where x, y E 1R1l •
any matrix
Z E C llxm can be written as Z = A + Bi, where B E IR IlX11l •
The eigenvalue-eigenvector theory already developed for matrices in IR IlXIl applies
well to matrices with complex entries. That is, a
scalar A is called a
IlXIl
,",'U"lUV"'"'"'" ej~~envalille of a complex matrix Z E C
if there is a nonzero
vector Z
such that
ZZ:::: AZ.
can be written in the
Before we elaborate this
we first
the notion of determinant to
complex matrices. The definition coincides with the definition of the determinant of a real
Z = [Zij] is
matrix. That is, the determinant of
o
detZ
z1 and Z2
+ 22
2. Z1Z2 = 2122 (and consequently
o
= Zll detZIl - Z12detZ12 + ... + (-1
r+ Zlll detZ
1
lll,
where
denotes the submatrix of Z formed
deleting row i and column j and the
determinant of a complex number Z is z. One can now copy the theory used in the real
in the
case, to derive the results of Theorem 2.5, and show that these properties also
case. In particular, one can also show that in the complex case a square matrix Z
if its determinant differs from zero. This result will be used to
is nonsingular if and
analyze the eigenvalue problem in more detail. To be self-contained, this result is now
shown. Its proof
the basic facts that, like in the real case, adding a multiple of
one row to another row of matrix Z, or adding a multiple of one column to another
column of Z does not change the determinant of matrix Z. Taking these and the fact that
'-'V'ClllJL'-'''
ZI.
[5+~i
0
Theorem 2.13 lists some useful properties of the complex conjugate. The proofs are
elementary and left as an exercise to the reader.
For any cOlTInlex numbers
+
~;;)~:J,
satisfy the equation, from which
= 3 + 4i, then ~ =:is (3 - 4i). The complex number Z =
standard form z =:is (2 i)(3 - 4i) =:is (2 lIi) =:fs - *i.
If Z
=
1 + ..]
[ 2 - ~i ~ and
o
stated differently, the equation
3. 21
!3-+4~i l
0
bx+ay = O.
Z2 = 21
ZF [
1 -. i]
1 .
ZI+ Z2= [3-3i.]
-1 - z
or
ZI
[21+Ll
= x + yi. Then equation (2.3.3) can be written as
(x + yi)(a + bi) = 1,
1.
0'0
29
28
det( 0
;]) =
the next fundamental prop-
for
elty on the existence of solutions for the set of linear eOlLlatlOflS Zz
0 is
"..O"Hror>rr.... C'
=0.
oon,esI:lonc11l112: to A.
[ ~2
also Example 2.3 part 4) Let
Let Z E C IlXIl • Then the set of
- 4A + 5. The
A2
= A + iB and z = x + iy, with
B E IR llxll and x,y E IR II .
Zz = Ax - By + i(Bx +
Therefore Zz = 0 has a unique solution if and only if the next set of equations are
uniquely solvable for some vectors x,y E 1R1l :
Ax
By = 0 and Bx +
= O.
Since this is a set of equations with only real entries, Theorem 2.4 and 2.5 can be used to
conclude that this set of equations has a solution different from zero if and only if
det([
~
-:]) =
o.
Since adding multiples of one row to another row and adding multiples of one column to
another column does not change the determinant of a matrix
that the above mentioned addition operations can indeed be represented in this
way.) Spelling out the right-hand side of this equation yields
det ( [ A
B
So, det (
[~
CII (i- 0)
D
I]. Its characteristic
is
=2-
VV<UV'VA
are AI
roots of this
i. The eigenvectors conesponding to AJ are
= 2 + i and
- (2+
z i- 0 if and only if det Z = O.
a E C}. The
Let Z
zE
'-''-I'UUllVlh)
Zz = 0
has a COJmPleX solution
all
- : ]) = det(A
-AB ] ) = det ( [A +
B iB
+ iB)det(A
0] ).
eH!(~nv(=ctr)r~
conesponding to
are
D
a E C}.
From Example 2.7 we see that with AI = 2 + i being an eigenvalue of
its
This property is, of course, something one would
2 - i is also an eigenvalue
given the facts that the characteristic polynomial of a matrix with real entries can be
factorized as a product of linear and quadratic terms, and equation (2.3.2).
Let A E IRllxll • If A E C is an eigenvalue of A and z a corresponding eigenvector, then ,\ is
also an eigenvalue of A and z a conesponding eigenvector.
definition x and A satisfy Az = AZ. Taking the conjugate on both sides of this equation
so that its
gives
Using Theorem 2.13 and the fact that A is a real
by definition, z is an
conjugate is matrix A again, yields Az = ,\z.
D
conesponding with the eigenvalue '\.
eigenvalue, then
Theorem 2.17 below, shows that whenever A E IRllxll has a
has a so-called two-dimensional invariant subspace
Section 2.5 for a formal
introduction to this notion) a property that will be used in the next section.
Let A E IR llxll . If A = a + bi (a, b E IR, b i- 0) is a complex eigenvalue of A and
z = x + iy, with x, y E 1R1l , a conesponding eigenvector, then A has a two-dimensional
y]. In particular:
invariant subspace S =
A - iB
AS =
s[
a
-b
b].
a
iB). Therefore, Zz = 0 has a solution differ-
ent from zero if and only if det(A + iB)det(A - iB) = O. Since w := det(A + iB) is a
complex number and det(A iB) = w (see Exercises) it follows that det(A + iB)
2
det(A iB) = 0 if and only if ww Iwl = 0, i.e. w = 0, which proves the claim. D
- (2-
Theorem 2.16 shows that both
Az = AZ and Az = ,\z.
30
out both
=ax-
Ax+
+
+
=axand
"~<~'UU.vLU.'';;;
~1
2
arnF'T1Anc
both eqllatJlOnS,
+
the next two
n~SDI~ctJlve.Lv
"''-!
Sion~
~'~'UVH"
= ax - by
= bx+ay.
VH'-'LLuv~'-".<'JU'-
one real root Al =
are
= 2 + 3i and
roots of this
+
The
inv::ant
{Q
=
two COJl11nlex
2 - 3i. The plCrpnUpf,tfY"C
[o~iL:,:[I ~J
indeed with a = 2 and h = 3, AS = S
c()rr{~sDon(1-
has a two-dimenVerification shows
[~h ~].
D
y] =
So what is left to be shown is that
y} are linearly independent. To show this, assume
that y = /-LX for some real /-L =J O. Then from equations (2.3.4) and (2.3.5) we get
Ax = (a - b/-L)x
Ax
1
= -(b + a/-L)x.
(2.3.7)
~t
\cc:onjmlll to equation
x is a real eigenvector corresponding to the real
Illp
a - b/-L; whereas according to equation (2.3.7) x is a real eigenvector
to the
real eigenvalue 1 (b + ap). However, according to Theorem 2.9 eigenvectors correspondfJ
ing to different eigenvalues are always linearly
the
a bp
and 1 (b + ap) must coincide, but this implies p2 = 1, which is not possible.
D
A theorem that is often used in linear
is the
theorem. the
section this theorem will be used to derive the Jordan canonical form.
To introduce and prove this
theorem let p ( A) be the characteristic
of
that is
1"1 Opn1J'A
/1.
1. (see also Example 2.7) Let A
(A 2
-
i ~ J.
[~2
The characteristic
this eigenvector is x:= [
m
I - (2 + i)l) = {Q [ I
~l J and
~i
l
Q E iC}. The real
.-
we
obtain the
+ bi
1] =s[ -b
a
a str;al,ll.httopW'a:rd
side of the next
'HJVUiUI;;:.
+ ... +
+
of
+
'-''-I,.LUL'VU,
+ ... +
can
+
+
A has a two-dimensional invariant subspace
2
with
is
the imaginary part of this
with 2 + i =: a
+···+an ·
of A is
A has one real root Al = 1 and two complex roots. The
roots of this equation are
= 2 + i and A3 = 2 - i. The
are
=Xl+al
1)
4A + 5)(A -
to
y:=
=
p(A)
VV'LL"',JUL·<F
of
if p ( A) = Xl +
+ ... + an, let
defined as the matrix formed
each power of A in p(A)
the '-'VJl.l'-"~I.J'"'HULH"" power of A
'-'IJLuvUil;;:.
+ ... +
we have
Theorem 2.17)
Comparing the right-hand sides of equations (2.4.1) and (2.4.3) it follows that the next
holds
b].
a
- AI)adj(A -
AI)Q(A) +
(2.4.5)
33
32
Next, assume that the characteristic
Theorem 2.10. That is
where we used the shorthand notation
Q(A)
+
:=
+ ... +
+
Using equation (2.4.2),
~r""n"H"'"
+ Q(A)} = p(A).
(2.4.6)
+ Q(A) =
AI)
adj(A
p(A)
(2.4.7)
p(A).
Using the definition of the adjoint matrix, it is easily verified that every entry of the
adjoint of A - Al is a polynomial with a degree that does not exceed n - 1 (see also
Example 2.2). Therefore, the left-hand side of equation (2.4.7) is a polynomial matrix
function. Furthermore, since the degree of the characteristic polynomial p(A) of A is n,
p(A)
withpi(A) = (A - Aiyzi, i = 1, ... ,k, andpi(A) = (A 2 + biA + CiYZi, i = k + 1, ... , r. The
next lemma shows that the null spaces of Pi(A) do not have any points in common
for the zero vector). Its proof is provided in the Appendix to this chapter
(2.4.6) can be rewritten as
adj(A - AI)
adj(A - AI)
is factorized as
+ ... +
+
rewrite equation (2.4.5) as
- AI){adj
p(A) of
1
= Po p(A) +
-A+ ... +
p(A)
p(A)
Let Pi (A) be as described above. Then
= {O}, if i =/-J.
The next lemma is rather elementary but, nevertheless, gives a useful result.
Let
B E [f:RIlXIl. If AB
= 0 then
dim(kerA)
is a strict rational function of A. Therefore, the left-hand side and right-hand side of
equation (2.4.7) coincide only if both sides are zero. That is,
adj (A
p(A)
AI) ( ) = 0
PA
.
D
+ dim (ker B)
;:::: n.
;:::: dim(ker B)
+ ~UU\ ~UL
(2.4.8)
Note that 1mB C kerA. Conse;oulentlv
In particular this equality holds for A = .\, where .\ is an arbitrary number that is not an
eigenvalue of matrix A. But then, adj - .\I) is inver1ible. So, from equation (2.4.8) it
follows that p(A) = O. This proves the Cayley-Hamilton theorem.
Therefore, using Theorem 2.3,
dim(ker B)
Let A E
[f:RIlXIl
+ dim (ker A)
D
=n.
and let p(A) be the characteristic polynomial of A. Then
p(A) = O.
D
Reconsider the matrices
i = 1, ... ,4 from Example 2.3. Then straightforward
2I) = 0;
+ 21/ = 0; (A 3 - 3I)2 = 0 and
calculations show that (AI - I)
- 4A 4 + 51 = 0, respectively.
D
We are now able to prove the next theorem which is essential to constmct the Jordan
Canonical form of a square matrix A in the next section.
Let the characteristic polynomial of matrix A be as described in equation
that the set {bjl , ... , bjmJ forms a basis for kerpj(A), with pj(A) as before, j
Then,
the set of vectors {b u , ... , b Irlll , ... , brl , ... , bnn,.} forms a basis for
3Cayley was an English lawyer/mathematician who lived from 1821-1895. His lecturer was the
famous Irish mathematician Hamilton who lived from 1805-1865 (see also Chapter 1).
Assume
=
1 ... , r.
[f:R1l;
2. lni=ni, i= l, ... ,k, and lni=2ni, i=k+l, ... ,r. That is, the algebraic multiplicities of the real eigenvalues coincide with the dimension of the cOlTesponding
34
and the
twice their
'-"PlAJlU''--
E
that any
result in linear
UHUVU,HVH
~IlXIl
a
HH.UUl-/uvA.L'''''.
First construct for the
of each
a basis {bjl , ... , bjmj }. From
Lemma 2.19 it is clear that all the vectors in the collection {b lI ,···, bIml"'" b rl , · · · ,
b./Ill,. } are
Since all vectors are an element of ~Il this
that
the number of vectors that
to this set is smaller than or
to the dimension
of ~Il, i.e. n.
On the other hand it follows "llu~f"ro;Lht·.j..L·~,·nF, ...rlh, by induction from Lelmna 2.20 that,
= 0, the sum of the
since
theorem PI
combining both
dimensions of the nullspaces
should be at least n.
we conclude that this sum should be exactly 11.
Since the dimension of the nullspace of PI (A) is m I, the dimension of the nullspace of
is at most mi. So A has at most In I 1l1(1el:lendelllt
with the
AI. A reasoning similar to Theorem 2.11 then shows that the
characteristic polynomial p(A) of A can be factorized as (A - Adlllh(A), where h(A)
Il1h
of degree n-ml' Since by
p(A) = (A-Ad 2(A),
17 2(A) does not contain the factor A AI, it follows that
mi :S ni, i = I, ... , k. In a similar way one can show that mi:S
1=
k + 1, ... ,r. Since according to Theorem 2.10
To grasp the idea of the Jordan
first consider the case that A has n different
Ai, i = 1, ... ,n. Let
i = 1, ... ,n, be the
From Theorem 2.9 it then follows
that
, X2, ... ,x,z} are
im1er,enaeJllt and in fact constitute a basis for ~Il. Now let matrix S := [XI X2
Since this matrix is full rank its inverse exists. Then AS =
AI
]. So if matrix A has n
where 11 =
[
different eigenvalues it can be factorized as
. Notice that this same DHKedm"e
can be used to factorize matrix A as long as matrix has n
The
fact that the eigenvalues all differed is not crucial for this construction.
to the
Next consider the case that A has an eigenvector Xl
the
and that the generalized eigenvectors X2 and X3 are obtained
A -
=0
=Xl
= X2·
From the second
= X2
can
hold if Ini
=
ni, i
= 1, ... ,k, and
mi
o
we see that
With S:=
X2 X3]
= Xl +
we
and from the third eOllatlon
therefore have AS =
I~
l
~.
if E
1
J. In
0
0 Al
and {Xl, X2, X3} are linearly independent, we conclude that S is inveliible and therefore
can be factorized as A = ShS- 1 •
For the
case, assume that matrix A has an o-"",-o""'ro1',,," Xl COJlTe~mClndll1g with
X2,' .. ,Xk are obtained
the eigenvalue Al and that the gel.1eraWwd
the following eql.latJlons:
Xl
it is clear that this
i = k + ,... , r.
~~"n~"~~
+
+
x,
+ )\jx,] =
with
Jz =
IJLUU ... LUUl.
LH.L'V""'::.H
Let A be an eigenvalue of A E ~IlXIl, and x be a corresponding eigenvector. Then Ax = Ax
and
A( ax) for any a E ~. Clearly, the eigenvector x defines a one-dimensional
that is invariant with respect to pre-multiplication
A since
VIc In
general, a subspace S C ~Il is called A-invariant if Ax E S for every xES. In other
words, S is A-invariant means that the image of S under A is contained in S, i.e. 1m
AS C S. Examples of A-invariant subspaces are the trivial subspace {O}, ~n, kerA and
=0
=XI
0UlJ0!.JUv,--
ImA.
A-invariant subspaces play an
role in calculating solutions of the so-caned
algebraic Riccati equations. These solutions constitute the basis for determining various
equilibria as we will see later on. A-invariant subspaces are intimately related to the
generalized eigenspaces of matrix A. A complete picture of all A-invariant subspaces is
in fact provided by considering the so-called Jordan canonical form of matrix A. It is a
=Xk-I·
let S be an A-invariant subspace that contains X2. Then Xl should also be in S.
since from the second equality above
= AIX2 + Xl, and, as both
and AI X2 belong
to S, Xl has to belong to this subspace too. Similarly it follows that if X3 belongs to some
A-invariant subspace S, X2 and, consequently, Xl have also to belong to this subspace S.
So, in general, we observe that if some Xi belongs to an A-invariant subspace S all its
'predecessors' have to be in it too.
36
37
= O. That
Furthermore note that
1)
An A-invariant subspace S is called a stable
of A constrained to Shave
'.d.""'-'UY <''''''''''-'u
Consider a set of vectors {Xl, ... ,xp } which are
= 0;
obtained as follows:
= Xi,
i
= 1, ... ,p
1.
(2.5.2)
Consider the
{b l , ••• ,bd constitute a basis for the COlTe~;oo'nC11l12: elg~ensp(lce
the Jordan chain generated
the basis vector bi.
basis for
with S :=
), ... ,
and
constitutes a
holds.
Then, with S :=
AS=
where A E !RIPxp is the matrix
(2.5.3)
S has full column
From Lemma 2.22 it is obvious that for each bi , J (b i ) are a set of 111"~Clr", 1I1C1c~penC1lent
vectors. To show that, for example, the set of vectors {J(bd,
inctepenlctellt too, assume that some vector Yk in the Jordan chain
is in the span
of).
) = {b I ,X2,""
and
= {b 2 ,Y2,'" ,Ys}, then there exist
ILi, not all zero, such that
rank.
From the reasoning
the first part of the lemma is obvious. All that is left to be
shown is that the set of vectors {Xl, ... ,Xp } are linearly independent. To prove this we
similarly.
assume that
consider the case p = 3. The general case can be
(2.5.4 )
the definition of the Jordan chain we conclude that if Ie > r all vectors on the
contradicts our
hand side of this equation become zero. So b 2 = 0, which
that b 2 is a basis vector. On the other
if Ie :::; r, the
reduces to
(2.5.5)
From this we see,
again, that
U""'UULIJLlVH
Then also
both sides of this
~'-l'~~"'~H
on the left
0=
consequently,
0=
(2.5.6)
Since Xi i=- 0 it is clear from equations (2.5.4) and (2.5.5) that necessarily Pi = 0,
i = 1,2, 3. That is {XI, X2, X3} are linearly independent.
0
Since in !RIll one can find at most 17 linearly independent vectors, it follows from Lemma
2.22 that the set of vectors defined in equation (2.5.2) has at most 17 elements. A sequence
+ ... +
= Pk+l (A
= Pk+lbI + ... + PrXr-k·
according Lemma 2.22 the vectors in the Jordan chain J(h) are linearly
independent, so Pk+I = ... = Pr = 0; but this implies according to equation (2.5.7) that
b2 = pkbI, which violates our assumption that {h, b 2 } are linearly independent. This
shows that in general {J(bd, ... ,J(bd} are a set of linearly independent vectors.
Furthermore we have from equation (2.5.1) and its preceding discussion that all vectors
in J(b i ) belong to Ef. So, what is left to be shown is that the number of these independent
38
the dimension of
that purpose we show that every vector
chosen Jordan chain for some vector Xl E
Let X E
=0
n 1 is the
of
or,
vectors
C'111TClr\H,
Below we
=0.
to denote the
matrices
for square, but
r\Clrt1t1AT,=rI
= 0, where
= Xl·
= Xl ..
introduce X2 :=
see that {Xl, .. ' ,XIl1 -1,X}
L".Vl-'v,-,-ULL.I:;,
'-''-'iva,;;.u
0
For any square matrix A
E [RIIlXIl
there exists a
rran~';lrr:g'\J.l~u
matrix
such that
this process we
where I =
extended
i=1, ... ,nl·
From this it is
to a Jordan chain for some Xl E
Which cOJnplet(~s the
... ,
and Ii either has one of the
COJmpleX (C) or
forms
a matrix A has the
real
verified
o
i)
'n~~~nn
[~
A=
=Xi-l,
....
matrix
-IX. Then,
Let Xl :=
X
use the notation
ii)
form
A=
iii)
Then it is easy to
iv)
that
= Span{x1};
S3 = Span{x3};
= Span{x4};
Sl
Here { Ai, i
o
are all A-invariant subspaces.
i
= k + ,... , r, and is the 2 x 2
hV"Vi'U.HL~VU
",-v\JU<'vUJlv
+
=
A1(a1 x l
hi
ai
= , ... , r} are the distinct real
eigenvalue Al and that its
Assume that matrix A has a
multiplicity is 2. Then A has infinitely many invariant subspaces.
in the case of Al E [RI, let {Xl,
be a basis for the nullspace of A Then any
linear combination of these basis vectors is an A-invariant subspace, because
+
ai
-hi
of
c, =
[
matrix.
corresDl)n(11l12: to the
definition
AI, which has an
Vi"'.'-'UL>U'-'-VV
i
?:. O.
+
Similarly, if A = a + bi (with b =I- 0) is a
with geometric
there exist four linearly independent vectors Xi,Yi, i = 1,2 such that
2,
=0.
4
•
u,
Jordan was a famous French engineer/mathematician who lived from 1838-1922.
40
From which we conclude that
Let A =
c
Consequently, 1111rocLUcmg
some matrix
such that
=SI
for
So, assuming the characteristic polynomial of A is factorized as in Theorem 2.10, we
i = 1, ... ,r, such that
conclude that there exist matrices Si and
. . .Sr] = [SI ... Sr]diag{ )q/ +
, where I is the Jordan canonical form COITeSp4Jnl1mg to A. Then
since Tis
ImST = ImS.
ImAS = ImSJ.
This
that the columns of matrix S are either
eigenvectors of A.
ImAST = ImSTJ.
, ... ,
+ Vr } .
This proves the first part of the theorem. What is left to be shown is that for each
generalized eigenspace we can find a basis such that with respect to this basis
=SJi,
where Ii has either one of the four representations
or
this was
basically shown in Proposition 2.23. In the case where the eigenvalue )\1 is real and the
In the case
geometric multiplicity of )\1 coincides with its algebraic multiplicity, Ii =
where its geometric multiplicity is one and its algebraic multiplicity is larger than one,
Ii =
If the geometric multiplicity of Al is larger than one but differs from its
and
occurs. The exact form of this
algebraic multiplicity, a mixture of both forms
mixture depends on the length of the Jordan chains of the basis vectors chosen in
- A1I). The corresponding results Ii =
and Ii =
in the case where Al = ai +
bii, bi =I=- 0, is a complex root, follow similarly using the result of Theorem 2.17.
D
i
. the boxes Ci = [a
In the above theorem the numbers ai and bi 111
-bi
2. From part 1 it follows that all A-invariant subspaces can be determined from the Jordan
canonical form of A. In Example 2.11 we already showed that if there is an eigenvalue
which has a geometric multiplicity larger than one, there exist infinitely many invariant
subspaces. So what remains to be shown is that if all real eigenvalues have exactly one
there correcorresponding eigenvector and with each pair of conjugate
sponds exactly one two-dimensional invariant subspace, then there will only exist a
finite number of A-invariant subspaces.
However, under these assumptions it is obvious that the invariant subspaces corresponding with the eigenvalues are uniquely determined. So, there are only a finite
number of such invariant subspaces. This implies that there are also only a finite
number of combinations possible for these subspaces, all yielding additional A-invariant
subspaces. Therefore, all together there will be only a finite number of invariant
D
subspaces.
From the above corollary it is clear that with each A-invariant subspace V one can
associate a part of the spectrum of matrix A. We will denote this part of the ",,,,c'0h"'H~~
CT(Alv)'
generically, a polynomial of degree 11 will have 11 distinct (possibly complex)
roots. Therefore if one considers an arbitrary matrix A E IR. IlXIl its Jordan form is most of
the times a combination of the first,
and third,
Jordan form.
bi]
ai come f rom the
complex roots ai + bJ, i = k + 1, ... ,r, of matrix A. Note that the characteristic
polynomial of Ci is A2 + 2aiA + aT + bT·
An immediate consequence of this theorem is the following corollary.
If A E IR.
IlXIl
has
11
distinct (possibly complex) eigenvalues then its Jordan form is
AI
Let A E IR. IlXIl •
All A-invariant subspaces can be constructed from the Jordan canonical form.
2. Matrix A has a finite number of invariant subspaces if and
multiplicities of the (possibly complex) eigenvalues are one.
if all geometric
I=
ak+I
-bk+I
bk+I
ak+l
ar
1. Let 1m S be an arbitrarily chosen k-dimensional A-invariant subspace. Then there
exists a matrix A such that
ImAS
= ImSA.
br
ar
where all numbers appearing in this matrix differ. If A has only real roots the numbers
ai, bi disappear and k = 11.
D
42
-4A+
2+i
J =
l~1 ~ :l
Therefore it has one real
and two
the Jordan canonical form of A is
2 - i.
o
start this section with the formal introduction of the tra.nsDositiion
The
a matrix E ~Ilxm, denoted
, is obtained
H!lVL~,uau~"LH~therows
if
and columns of matrix A. In more
VIJ'-'H.HJ.\JH.
tnlm~p()Se of
[all
a2l
al~z1
an
a22
al"
a2n ]
a 1l1 2
al~lll
then
all
a21
an
an
a m2
al n
a21l
al~lll
ami]
If A
J=
then there exists an orthonormal matrix U and a
such that
~HLIL;;;'VHLH
matrix
A=
The ith-column of U is an
o
Ai.
correSDl)nl1m£ to the
transpose, i.e. A = , the matrix is called
One
matrices is that a
matrix has no
to different
are per-
A
Let A E
~nxll
>0
(~
0) if and
if all
,001cr~'nu"
be SVlnnletnc.
if A is an
then A is a real
i = 1,2, are two different
= O.
then
Let X be an ,001.r,oor,u,oo0tr,r "r.'"1·,ooC'·nr.-nrl-,,,
eigenvalue of
too and x is a
xT,\x =
On the other
clude that ,\
of A and
Xi,
i = ,2,
A
from Theorem 2.27. Choose
Consider the
= Ai and the result is obvious.
the ith-column of U. Then
The converse statement is left as an exercise for the reader.
X
=
Ui,
where
Ui
is
0
~~r,_r!H'~ to
Theorem 2.16 ,\ is then an
Therefore
is a scalar and A is
E ~, different from zero, we con-
= A. So A E R
p=
On the other
p=
on the one
= AIXf X2. SO comparing both results we conclude that
it follows that
= O.
0
Since Al f
In this subsection we
the so-called
or ARE for short.
AREs have an impressive range of applications, such as linear quadratic optimal control,
SCount Jacopa Francesco Riccati (1676-1754) studied the differential equation x(t)+
t- n x 2 (t) ntlll +n - 1 = 0, where 111 and n are constants (Riccati, 1724). Since then, these kind of
equations have been extensively studied in literature. See Bittanti (1991) and Bittanti, Laub and
Willems (1991) for an historic overview of the main issues evolving around the Riccati equation.
44
45
and stochastic
of linear
networks and robust
control.
this book we will see that
also
a
central role in the determination of 'A..
in the
of linear
differential
games.
Q and R be real n x n matrices with Q and R symmetric. Then an algebraic
Let
Riccati eOllatJlOn in the n x n matrix X is the following quadratic matrix equation:
I'.HH.UULU
a solution to the Riccati ~~"~ ...;~,~
the solution is
of
+
of the basis of V.
Since V is an H-invariant
+
+
'-''-«J'-'I_'U'v'v,
there is a matrix A E
~IlXIl
such that
(2.7.1)
+Q=O.
The above equation can be rewritten as
Post-multiplying the above equation
Q
[I
From this we infer that the image of matrix [I
Q
R] [X]·
orthoQ:o:nal complement of the image of matrix [I
ARE has a solution if and
(2.7.3)
is orthogonal to the image of
stated differently, the image of
of the image of matrix [I
we get
[~ R] [~]
belongs to the
Now pre-multiply equation (2.7.3)
I] to get
It is easily veIified that the
is given by the image of [ -:].
if there exists a matrix A E ~IlXIl such that
Rewriting both sides of this equality
-XA-
Premultiplication of both sides from the above equality with the matrix
then
-XRX
Q= 0,
which shows that X is indeed a solution of equation (2.7.1). Some
(2.7.3) also gives
[~I ~]
""""W,t"~,,
of eqllatJlon
A+RX=
stated
the symmetric solutions X of ARE can be obtained by considering
the invariant subspaces of matrix
= 0-( A). However, by definition, A is a matrix rer)re~~entatlOn of the
= o-(Hlv)' Next notice that any other basis
V can be
therefore,
map Hlv' so
represented as
(2.7.2)
H:= [ A
for some nonsingular P. The final conclusion follows then from the fact that
= X2X 11 .
Theorem 2.29 gives a precise formulation of this observation.
D
The converse of Theorem 2.29 also holds.
Let V C ~2n be an n-dimensional invariant subspace of H, and let
real matrices such that
E
~Ilxn
be two
If X E ~IlXIl is a solution to the Riccati equation (2.7.1), then there exist matrices
,
E
~"n"
with
invertible, such that X = X2Xj 1 and the columns of
basis of an n-dimensional invariant subspace of H.
[~~] form a
46
are ,2, - ,
The
Define A := A
+ RX.
l\/h'<lb •..,I""-. ...
this
and
",nl,,,,1"",nn
(2.7.1)
+
Write these two relations as
All solutions of the Riccati
follows.
the columns of [ ~] span an n-dimensional invariant snbspace of
:= I,
and
:=
and defining
"'-Lll.LLlLJlVU
l
1. Consider Span{ VI, V2} =:
yields
X:=
Matrix H is called a Hamiltonian matrix. It has a number of nice properties. One of
them is that whenever A E
then also - A E
That is, the spectrum of a
Hamiltonian matrix is symmetric with respect to the
axis. This fact is
easily established by noting that with J:=
+
[~ ~ll
H
det( -JH JAI) =
+ =p(-A).
det(
_HT -
From Theorems 2.29 and 2.30 it will be clear
the Jordan canonical form is so
important in this context. As we saw in the previous section, the Jordan canonical form of
a matIix H can be used to construct all invariant subspaces of H. So, all solutions of
equation (2.7.1) can be obtained by
all n-dimensional invariant subspaces
~nxn, that have the additional
is invertible. A subspace V that satisfies this property is called a
it can be 'visualized' as the graph of the map: x -7
X:=
=
= det
D
E
=
2. Consider Span{vI, v-d =:
which
ptA) =
=
1
V = 1m [~~] of equation (2.7.2), with
that
=
X:=
4. Let Span{ V2,
yields
V-I}
=:
= [ 1 -3]
4 '
2
R = [2
0
0]
1 ' and Q = [00
0]
0 .
X'-
2
H=
[
o
o
-3
4
o
o
2
0
1
3
6. Let Span{ V-I, V-2} =:
yields
X:=
l
l
=
Then
-1
[~ ~ ] , which
=
[~ ~]
[~~].
and
Then
Then
~I [~
[~:l Then
1 [ 66
- 67 -114
and
+
= {1,
-33]
=
[~2
72]
and
+
[ ~I
4
108
Then
~2 [i
= { ,2}.
+
Then
~I [~ ~]
-I [48
51 72
=
5. Let Span{ V2, V-2} =:
yields
(see also Example 2.3) Let
l
3. Let Span{vI, V-2} =:
yields
X:=
A
Then
D
X completes the proof.
T
combinations of these vectors as
are obtained
:] and
=
~]
[~I
5
1}.
[~
' and
24]
36 ' which
{1, -2}.
13] '
+
and
n
[~
=
which
={2,-1}.
-33]
5
'
and a-(A +
and
=
[~
24]
. h
36 ' WhIC
[~
24]
.
36 ' whIch
= {2, -2}.
[-2 -33]
5
-114]
-270 and a-(A +
' and
, -2}.
D
48
49
natural
that arises is whether one can make any statements on the number of
solutions of the algebraic Riccati
(2.7.1). As we already noted, there is a one-toone relationship between the number of solutions and the number of graph subspaces of
matrix H. So, this number can be estimated by the number of invariant subspaces of
matrix H. From the Jordan canonical form, Theorem 2.24 and more in particular
Corollary
we see that if all eigenvalues of matrix H have a geometric multiplicity
of one then H has only a finite number of invariant subspaces. So, in those cases the
algebraic Riccati equation (2.7.1) will have either no, or at the most a finite number, of
solutions. That there indeed exist cases where the equation has an infinite number of
solutions is illustrated in the next example.
Let
A=R
[b ~]
and Q =
[~
n
3.
[
-2
o
0]
-2
+
= {-1,-
D
to this
we have said nothing about the structure of the solutions
Theorems 2.29 and 2.30. In Chapters 5 and 6 we will see that
solutions that are
C C- (the so-called
will
and for which a(A +
interest us. From the above example we see that there is only one stabilizing solution and
that this solution is symmetric. This is not a coincidence as the next theorem shows. In
fact the propeliy that there will be at most one stabilizing solution is already indicated
if matrix H has n different eigenvalues in C-, then
our Note following Theorem 2.30.
it also has n different eigenvalues in C+. So, there can exist at most one appropriate
context,
invariant subspace of H in that case. To prove this observation in a more
we use another well-known lemma.
Consider the Sylvester equation
Then
AX+XB=C
(2.7.4)
1
o
-1
o
The eigenvalues of Hare {I, 1,
1, -I} and the corresponding eigenvectors are
The set of all one-dimensional H-invariant subspaces, from which all other H-invariant
subspaces can be determined, is given by
where A E [RIlXIl, BE IR lIlxlIl and C E [R/lXIIl are given matrices. Let {/\, i = 1, ... ,n} be
the eigenvalues (possibly complex) of A and {Pj, } = 1, ... ,m} the eigenvalues (possibly
complex) of B. There exists a unique solution X E [RIlXIIl if and only if Ai(A)+
!-L.i(B) =1= 0, Vi = 1, ... ,n and} = 1, ... ,m.
First note that equation (2.7.4) is a linear matrix equation. Therefore, by rewriting it as a
set of linear equations one can use the theory of linear equations to obtain the conclusion.
For readers familiar with the Kronecker product and its corresponding notation we will
provide a complete proof. Readers not familiar with this material are referred to the
literature (see, for example, Zhou, Doyle and Glover (1996)). Using the Kronecker
product, equation (2.7.4) can be rewritten as
= vec(C).
All solutions of the Riccati equation are obtained by combinations of these vectors. This
yields the next solutions X of equation (2.7.1)
1.
2.
[~ ~]
An immediate consequence of this lemma is the following corollary.
yielding cr(A + RX) = {I, I}.
[~2 ~l[ ~2 ~l[~ ~2l[~ ~J
a(A +RX) = {-I, I}.
This is a linear equation and therefore it has a unique solution if and only if matrix
E9 A
is nonsingular; or put anotherway, matrix B T E9 A has no zero eigenvalues. Since the eigenvalues of BT E9 A are Ai(A) + pj(B T ) = Ai(A) + /lj(B), the conclusion follows.
D
and
OIl]
[ =~ 1J
yielding
6English mathematician/actuary/lawyer/poet who lived from 1814-1897. He did important work
on matrix theory and was a friend of Cayley.
50
First, we prove the uU'''''I'_'.'-'11'-'OO n,"r,np,"tu To that end
~~lnT,/.n" of A""QT'"''
Consider the so-called
AX+
=c
+
where
C E IR
are
matrices. Let {Ai, i = 1, ... ,11} be the eigenvalues
(2.7.5) has a
solution E IR llxll if and
(possibly COlnplex of A. Then
+ .\(A) i- 0, 'Vi,) = 1, ... ,n.
0
only if
that
+ Q = 0, i =
,2.
llxll
After SUl)tf(lctmg the
+
the
C'U"""'Y'Ah"u
Consider the matrix equation
of matrix
+
+XA=
we find
'"''-I''''U •.''JUu,
+
=0.
this equation can be rewritten as
{(X1 -
+
+
=0.
The above equation is a Sylvester
Since
both A +
A+
have all their eigenvalues in ([:-, A +
and
+
have no nA'-";tQ I,"PC'
in common.
it follows from Lemma 2.31 that the above eqllatllOn (2.7.6) has a
= satisfies the
So
unique solution
Obviously,
which proves the
result.
then this solution
Next, we show that if equation (2.7.1) has a stabilizing solution
with the definition of 1 and H as
will be symmetric. To that end we first note
the matrix lH defined
p,
where a(A) C ([:-. This matrix equation has a unique solution for every choice of matrix
be in ([:- and
C. This is because the sum of any two eigenvalues of matrix A will
0
thus differ from zero.
°
Consider the matrix equation
+
with
[-;1
as in
~3] , and
C,
2.3, and C an arbitrarily chosen 2 x 2
[=:
~].
that is:
According to Example 2.3 the eigenvalues of
are
is a SYlnnletnc matrix.
let X solve the Riccati
where
""r",,,+""'"
{ I , 2} and
has only one eigenvalue {- 2}.
according to Lemma 2.31, there
exist matrices C for which the above equation has no solution and also matrices C for
which the equation has an infinite number of solutions. There exists no matrix C for
0
which the equation has exactly one solution.
with
This
lH
and
invertible.
]=
Since the left-hand side of this eqllatlon is syrnnletlnC, we conclude that the flQ'm·'l1a.nu
side of this equation has to
stated
7Russian mathematician who lived from 1857-1918. Schoolfriend of Markov and student of
Chebyshev. He did important work on differential equations, potential theory, stability of systems
and probability theory.
c ([:-
Theorem
that
The next theorem states the important result that if the algebraic Riccati equation (2.7.1)
has a stabilizing solution then it is
moreover, symmetric.
The algebraic Riccati equation (2.7.1) has at most one stabilizing solution. This solution
is symmetric.
"",-,,,·rl,,.., ....
rl,-t'+"U'A,,+lu
+
=0.
53
52
This is a
obvious from
2.32 that this
o satisfies the
'-'\.I'.HtLLVH
has a
of
solution
. But, then
c
C - it is
Riccati
of equation (2.7.1), then X ::;
has a
U","',,",Ul.UH.
~~""TH~n
Since both
and X
.... '-, .....
~,u'Vu
"tUU.LUlLoHlj;:.
and
solution
is another
(2.7.1)
+
XA+
D
of the algebraic Riccati
it
Apart from the fact that the stabilizing solution,
exists) is characterized
its uniqueness, there is another characteristic property. It is the
"""0.",."...... 1 solution of equation (2.7.1). That is, every other solution X of equation (2.7.1)
Notice that maximal (and minimal) solutions are unique if
exist.
satisfies <
To prove this property we first show a lemma whose proof uses the concept of the
like the scalar
exponential of a matrix. The reader not familiar with this, can think of
exponential function eat. A formal treatment of this notion is given in section 3.1.
If Q ::; 0 and A is stable, the Lyapunov equation
AX+
=Q
Since A is stable eAt converges to zero if t becomes arbitrarily large (see section 3.1
again). Consequently, since e A .O = I, and the operation of integration and differentiation
'cancel out' we obtain on the one hand that
= 0-
(2.7.9)
...,,"'U.l.JL.l.L.d.H6
Let X be the stabilizing solution of equation (2.7.
equation (2.7.1), yield
Combining equations (2.7.9) and (2.7.10) yields then that X 2: O.
Simple manipulations, using
R
+
As
I
[-X
0]-1
[I
I
= X
~],
we conclude from the above
(J(H) = (J(A
"i",nh1ru
1
that
+ RX) U (J( -(A +
Since X is a stabilizing solution, (J(A +
is contained in C- and (J( - (A
H does not have eigenvalues on the imaginary axis.
and on the other hand, using equation (2.7.8), that
V(t)dt =
D
This maximality property has also led to iterative procedures to
the
solution (see, for example, Zhou, Doyle and Glover (1996).
Next we provide a necessary condition on the spectrum of H from which one can
conclude that the algebraic Riccati equation has a stabilizing solution. Lancaster and
Rodman (1995) have shown that this condition together with a condition on the
associated so-called matrix sign function are both necessary and sufficient to conclude
that equation (2.7.1) has a real, symmetric, stabilizing solution.
(2.7.8)
+
- X 2: O.
The algebraic Riccati equation (2.7.1) has a stabilizing solution only if H has no
eigenvalues on the imaginary axis.
Since A is stable we immediately infer from Corollary 2.32 that equation (2.7.7) has a
unique solution X. To show that X 2: 0, consider V(t) := ~ (eATtXeAt). Using the product
eAtA (see section 3.1), we have
rule of differentiation and the fact that
V(t)dt
is stable Lemma 2.34 gives that
Since, by assumption, A -
(2.7.7)
has a unique semi-positive definite solution X.
V(t) =
+
+
+
(2.7.10)
D
+
in C+.
D
The next example illustrates that the above mentioned condition on the spectrum of
matrix H in general is not enough to conclude that the algebraic Riccati equation will
have a solution.
54 linear
~ln<Olhl'~
55
solution of the
Riccati
This
rise to an iterative
been formalized
Kleinman
"l£lUU'"'-'U"'b
Let A
= [
01
~ ] , R = [~ ~]
and Q =
[~ ~ ] . Then it is readily verified that the
is based on
to calculate this solution
'-'-I,.lUljlVH
1I1'-"'U~~.
eigenvalues of Hare {- 1, I}. A basis for the conesponding eigenspaces are
[~]
,b4
=
[~],
1. Consider
respectively.
Consequently, H has no graph subspace associated with the eigenvalues {-1, -I}.
0
We conclude this section by providing a sufficient condition under which the algebraic
Riccati equation (2.7.1) has a stabilizing solution. The next theorem shows that the
converse of Theorem 2.36 also holds, under some additional assumptions on matrix R.
One of these assumptions is that the matrix pair (A, R) should be stabilizable. A more
detailed treatment of this notion is given in section 3.5. For the moment it is enough to
R) is called stabilizable if it is possible to steer any initial
bear in mind that the pair
state xo of the system
x(t) =
+ Ru(t),
x(O)
xo,
independent.
Show that {VI, V2} are
Show that {V 1, V2, V3} are linearly dependent.
(c) Does a set of vectors Vi exist such that they constitute a basis for a:;R3?
(d) Determine Span{V2, V3, V 4}. What is the dimension of the subspace sp,mrLea
these vectors?
Determine the length of vector V3.
(f) Determine all vectors in a:;R3 that are perpendicular to V4.
2. Consider
towards zero using an appropriate control function u(.). The proof of this theorem can be
found in the Appendix to this chapter.
R) is stabilizable and R is positive semi-definite. Then the algebraic
Assume that
Riccati equation (2.7.1) has a stabilizing solution if and only if H has no eigenvalues on
0
the imaginary axis.
Good references for a book with more details on linear algebra (and in particular section
2.5) is Lancaster and Tismenetsky (1985) and Horn and Johnson (1985).
For section 2.7 the book by Zhou, Doyle and Glover (1996) in patiicular chapters 2 and
13, has been consulted. For a general treatment (that is without the positive definiteness
assumption) of solutions of algebraic Riccati equations an appropriate reference is the
book by Lancaster and Rodman (1995). Matrix H in equation (2.7.2) has a special
structure which is known in literature as a Hamiltonian structure. Due to this structure, in
particular the eigenvalues and eigenvectors of this matrix have some nice prclpertle:s.
Details on this can be found, for example, in chapter 7 of Lancaster and Rodman (1995).
A geometric classification of all solutions can also be found in Kucera
The
eigenvector solution method for finding the solutions of the algebraic Riccati equation
was popularized in the optimal control literature by MacFarlane (1963) and Potter (1966).
Methods and references on how to evercome numerical difficulties with this approach can
be found, for example, in Laub (1991). Another numerical approach to calculate the
(a) Use the orthogonalization procedure of Gram-Schmidt to find an orthonormal
basis for S.
(b) Find a basis for Sl-.
3. Consider
A
=
[1 ~l
n
Determine Ker A and 1m A.
Determine rank(A), dim(Ker
and
Determine all b E a:;R4 for which Ax = b has a solution x.
If the equation Ax = b in item (c) has a solution x, what can you say about the
number of solutions?
4. Let A E a:;Rllxm and S be a linear subspace. Show that the following sets are linear
subspaces too.
(a)
(b)
(c)
(d)
(a) Sl-;
(b) Ker A;
(c) 1m A.
57
56
Show that for any cOJmvlex numbers ZI and
5. Consider the
i(t)
x(o) = Xo
y(t) = Cx(t),
(a) ZI
(b) ZIZ2
where x(.) E ~n and y(.) E ~m. Denote by y(t,xo) the value ofy at time t induced by
the initial state xo. Consider for a fixed tl the set
V(td
:=
{xoly(t,xo)
=
0, t E [0, td}·
Z2 = 2:1 + 2:2,
= 2:12:2 and Z]
Z2
= 2:1
z\ = ZI·
Show that for any complex vector
(c)
Z
E
en Izi =
Show that for any complex matrix Z
Show that V(td is a linear subspace (see also Theorem 3.2).
i
=
[Z:I] , where Zi
= 1, ... , n + 1,
=
[Zil, ... ,Zin],
Zn
6. Use Cramer's rule to determine the inverse of matrix
Z]
det
+z~Zn+] ]
..
[
.
[Z/;:] ]
= detZ + Adet
Zn
...
.
Zn
7. Consider
vi = [1,
2, 3, 0, -1, 4] and
vI = [1,
-1, 1, 3, 4, 1].
Let A:= VIVI(a) Determine trace(13 * A).
(b) Determine trace(2 * A + 3 * I).
(c) Determine trace(A * AT).
8. Determine for each of the following matrices
A
=
[~
n [7 ~];
B
=
C
=
[i il
D
=
[~
:
n
(a) the characteristic polynomial,
(b) the eigenvalues,
(c) the eigenspaces,
(d) for every eigenvalue both its geometric and algebraic multiplicity,
(e) for every eigenvalue its generalized eigenspace.
9. Assume that A = SBS- 1 • Show that matrix A and B have the same eigenvalues.
10. Let AE TR
Zl
I Zi II,
and
16. Starting from the assumption that the results shown in Exercise 14 also hold when we
consider an arbitrary row of matrix Z, show that
det([~
;])
det(A)det(D).
17. Determine for each of the following matrices
-1
1
llXU
(a) Show that for any p 2: 1, N(AP) C N(Ap+I).
(b) Show that whenever N(A k ) N(A k + 1) for some k 2: 1 also N(AP)
for every p 2: k.
(c) Show that N(AP) = N(A n ), for every p 2: n.
11. Determine for the following complex numbers
plane.
(b) Consider in (a) a matrix Z for which Z] = Z2. Show that det Z = 0.
(c) Show, using (a) and (b), that if matrix Z] is obtained from matrix Z by adding
some multiple of the second row of matrix Z to its first row, detZ] = detZ.
(d) Show, using (a), that if the first row of matrix Z is a zero row, detZ = 0.
15. Let w:= det(A + iB), where A, B E ~nx/l. Show that det(A - iB) = w. (Hint:
consider A + iB; the result follows then directly by a simple induction proof using
the definition of a determinant and Theorem 2.13.)
= 1 + 2i;
Z2
=2-
i; and Z3
= -1 -
1
= N(Ap+1),
2i,
±, respectively. Plot all these numbers on one graph in the complex
'-t
(a) the characteristic polynomial,
(b) the (complex) eigenvalues,
(c) the (complex) eigenspaces,
(d) for every eigenvalue both its geometric and algebraic multiplicity,
(e) for every eigenvalue its generalized eigenspace.
18. Show that kerA and lmA are A-invariant subspaces.
59
58
Consider
26.
nn,,,1"n.lp
definite and ,.. . """1"n'o
Sl~ml-Q(~nnlHe.
27. Determine all the solutions of the following algebraic Riccati eql1atlOns. Moreover,
+
determine for every solution
-A 3 +
Show that the characteristic .... "'h".."'.....-.F1 of A is p(A)
- 4A + 2/ = O.
Show that
+
Show that p(A) = (1 A)(A 2 - 2A + 2).
and
Determine
.Show that Nl n
= {O}.
(f) Show that any set of basis vectors for N] and
basis for
20. Consider the matrices
(b)
(c)
(d)
(e)
A
=
T]
[~ ~
and B
=
~ i] +X[~ ~ ]X+ [~ ~] = o.
;rX+X[~1 ;]+x[~ ~]x=o
ir X+
- 4A + 2.
2A+
X[
28. Verify whether the following matrix equations have a
respectively, form together a
[l ~
n
(a)
(b)
(c)
solution.
[~ ~3]X+X[~ ;] = [; 2]3 .
1
[~ ;]x +x[ 2 ~] = [;
[\ 1]1 X +X [-4-2 ;] = [;
n
n
29. Determine all solutions of the matrix equation
(a) Determine the (generalized) eigenvectors of matrix A and B.
(b) Determine the Jordan canonical form of matrix A and B.
21. Determine the Jordan canonical form of the following matrices
[~ ~]X+X[ 0 ~]
1
where C = [;
; ] and C =
[~ ~ ], respectively, Can you find a matrix
C for
which the above equation has a unique solution?
22. Determine the Jordan canonical form of the matrices
-I
j
[
-I
[~I
[~I n
0
A=
2
0
~
; B=
~1
0
1
0
~J c= [
2
0
1
1
0
1
~Il
-2
D=
2
2
23. Determine for each matrix in Exercise 22 all its invariant subspaces.
24. Factorize the matrices below as
diagonal matrix.
A =
[~ ~3];
B=
where U is an orthonormal matrix and D a
[~-1 0~ ~ll
0
25. Let A be a symmetric matrix. Show that A
Ai > 0 (2: 0).
c=
[;
1
We distinguish three cases:
i,j E {I, ... , k};
i E {I, ... , k}
E {k + 1, ... , r};
and
i,j E {k + 1, ... , r}. The proofs for all three cases are more or less the same (this
arithmetic
is not surprising, because if we could have allowed for more advanced
all the cases could have been dealt with as one case).
Assume without loss of
y E kerpl (A). Then there exists an index 1 ::;
111 such that y E
but ytj:.
Al1)I11-1
- A]// := / (see Exercises). Then for an arbitrary index i 2: 1
117 ::;
~ ~2l·
5
-2
> 0 (2: 0) if all eigenvalues Ai of A satisfy
= (A]
:f: O.
60
must differ from zero.
Using the same notation as in
+
, and thus in
that
]
now
+
+ (2A] + bk+d
- A]/yrl-] ((A -
was shown in the
]
1)
of Theorem 2.33, we have under these conditions that
(AT + bk+IAI + ck+dI)i y
= ((AT +
+ CHI
- AI/yrl-ly
i= 0,
from which in a similar way the conclusion results.
Again assume that 111 is such that y E ker(A 2 + bHIA + Ck+l/t and y~ ker(A 2+
bk+IA + Ck+lI)m-l. Moreover, assume y E ker(A 2 + bk+2A +
. Then
0=
m-]
i
+
+ Ck+l I)
+
+ cHII)m-]
+
+ cHd)m-I((bH2 - b H ]
+ bk+IA +
AW= WP.
+ b H2 A + cH2I) y
+ bk+IA + CHII + (h+2
= ((b k+2 - bk-1-dA + (CH2 - CHI
So, v:=
Then, according to Theorem 2.29, X:=
is the
stabilizing solution of
equation (2.7.1) provided
is invertible. So, what is left to be shown is that from our
assumptions on R it follows that
is invertible. To that end we first show that if matrix
W is a full column rank
such that 1mW :=
, then there exists a square
matrix P such that
+
bHdA + (Ck+2 - C/(+!
(2.10.3)
For that purpose consider the first n equations of equation (2.10.1). That
cHdI)i y
+
+ bk+IA + cHII)m-I y .
E ker((b k+2 - bk+]
+ (cH2 - cHdl(
Next assume that b H2 = bHI and thus Ck+2 i= CHI (otherwise PH] (A) =Pk-I-2(A)
holds). Then ker((h+2-bk+dA+(CH2-C/(+1
O. So in that case v=O and
y E ker(A 2 + bHIA + Ck_I_]I)m-1 which contradicts our assumption on y. So bk-I-2 i=
bk + l , and we conclude that v E ker(A + =~=
On the other hand we have
0= (A 2 + bk+!A + CHI/)'n-1
+ b H2 A + cH2I)i y
+ bH2 A + Ck+2I)i(A 2 + bk+1A + Ck+]I)m-l y .
(ii).
Therefore, also v E ker(A 2 + b H2 A +
However, according to (2) (i) and (ii) cannot both occur simultaneously, from which
the assertion now readily follows.
D
(2.10.4)
Pre- and post-multiplying this equation (2.10.4) by
yields
W+
From W = 0 and equation (2.10.2) it follows then that
semi-positive definite we conclude that
= O. Since R is
=0.
with W that
Using this it follows by post-multiplying equation (2.
stated differently, ImAW e . Obviously, this can be rephrased
(2.10.3).
let A be an eigenvalue of P and y a corresponding eigenvector.
equation (2.10.3),
=
From Theorem 2.36 we conclude the necessity of the condition. So what is left to be
shown is that if H has no eigenvalues on the imaginary axis, equation (2.7.1) has a
stabilizing solution.
From the Note following Theorem 2.30 it is clear that, since H has no eigenvalues on
the imaginary axis, H has an n-dimensional stable invariant subspace. That is, there exists
a matrix
with a(A) c C-, and a full column rank matrix
and
AWy.
from
(2.10.6)
That is, A is an eigenvalue of A and Wy a corresponding eigenvector. Since A is a stable
matrix, we conclude that A E C-. Next consider the second n-equations of
(2.10.1)
vLJ'U.UL.iVU
Post-multiplying this equation by
-QX 1
gives
62
and the fact that
W
= 0, we infer from this
+
Since
O.
. a f u11 column rank matrix, and
] IS
that
(2.
1
0
[XX,2 ]
] , it follows that
#- 0 if #- O. So, from equations (2.10.5) and (2.10.7) we conclude that if
is nonempty, then there exists a vector
#- 0 and a ,\ E C- such that
= 0 and
+
I
=0.
However, this implies that the equation
holds for some J-L E
and
#- O.
That is, the matrix
A linear
differential game studies situations
two or more decision
makers (individuals, organizations or
Decision makers are called the
players in the game. These players often have
interests and make
individual or collective decisions. In a linear differential game the basic
that all
can influence a number of variables which are crucial in
goals and that these variables change over time due to external
forces. These
variables are called the state variables of the system. It is assumed that the movement over
time of these state variables can be described by a set of linear differential
in
of the
actions is in an additive linear way.
which the direct
the extent to which the
succeed in
their goals
on the actions of
the other
if one
has information on the action that another
will
he can incorporate this information into the decision
about his
information plays a crucial role in the
of
actions
for the
to
linear
differential games one first
has to introduce systems of differential
criteria and information
sets.
This
the framework for this VHU.tJLV~,
avna]mllCal systems, linear systems are introduced. As
mentioned
these systems are assumed to describe the motion of the game over time. Some results on
the existence of
satisfying the associated set of differential
are
outlined.
The
behavior of the involved
will be formalized
of a
criterion. The associated optInlllz:atl,on . . .r,,.,.h.llan·H'
rise to the
of sets of nonlinear differential eqllatlons.
has a
solution.
this
result that not every set of differential
issue is dealt with in a separate section. In particular the consequences of this observation
for our
where players are free to
the state
are discussed.
The analysis of sets of nonlinear differential equations is a delicate matter.
one
starts this analysis
a study of the local behavior of trajectories near so-caned
ULhHUHI-/UVU
does not have a full row rank for some J-L E C+. But this implies, according to
Theorem 3.20, that the matrix pair
R) is not stabilizable, which contradicts our
0
assumption. So, our assumption that kerX 1 #- 0 must be wrong.
VI-/I.HHa!
V!-,UU.HL,UU.VU
LQ Dynamic Optimization and DUferential Games
2005 John Wiley & Sons, Ltd
J. Engwerda