This preview shows page 1. Sign up to view the full content.
Unformatted text preview: benius theorems, which
are useful in their own right and also provide the necessary results about [P ]n in general. 4.4 PerronFrobenius theory A real vector x (i.e., a vector with real components) is deﬁned to be positive, denoted x > 0,
if xi > 0 for each component i. A real matrix [A] is positive, denoted [A] > 0, if Aij > 0 for
each i, j . Similarly, x is nonnegative, denoted x ≥ 0, if xi ≥ 0 for all i. [A] is nonnegative,
denoted [A] ≥ 0, if Aij ≥ 0 for all i, j . Note that it is possible to have x ≥ 0 and x 6= 0
without having x > 0, since x > 0 means that all components of x are positive and x ≥ 0,
x 6= 0 means that at least one component of x is positive and all are nonnegative. Next,
x > y and y < x both mean x − y > 0. Similarly, x ≥ y and y ≤ x mean x − y ≥ 0. The
corresponding matrix inequalities have corresponding meanings.
We start by looking at the eigenvalues and eigenvectors of positive square matrices. In
what follows, when we assert that a matrix, vector, or number is positive or nonnegative,
we implicitly mean that it is real also. We will prove Perron’s theorem, which is the
critical result for dealing with positive matrices. We then generalize Perron’s theorem to
the Frobenius theorem, which treats a class of nonnegative matrices called irreducible
matrices. We ﬁnally specialize the results to stochastic matrices. 4.4. PERRONFROBENIUS THEORY 151 Perron’s theorem shows that a square positive matrix [A] always has a positive eigenvalue
∏ that exceeds the magnitude of all other eigenvalues. It also shows that this ∏ has a right
eigenvector ∫ that is positive and unique within a scale factor. It establishes these results
by relating ∏ to the following frequently useful optimization problem. For a given square
matrix [A] > 0, and for any nonzero vector6 x ≥ 0, let g (x ) be the largest real number a
for which ax ≤ [A]x . Let ∏ be deﬁned by
∏= sup g (x ). (4.20) x 6=0 ,x ≥0 P
We can express g (x ) explicitly by rewriting ax ≤ Ax as axi ≤ j Aij xj for all i. Thus,
the largest a for which this is satisﬁed is
P
j Aij xj
g (x ) = min gi (x )
where gi (x ) =
.
(4.21)
i
xi
P
Since [A] > 0, x ≥ 0 and x 6= 0 , it follows that the numerator i Aij xj is positive for all i.
Thus gi (x ) is positive for xi > 0 and inﬁnite for xi = 0, so g (x ) > 0. It is shown in Exercise
4.10 that g (x ) is a continuous function of x over x 6= 0 , x ≥ 0 and that the supremum in
(4.20) is actually achieved as a maximum.
Theorem 4.5 (Perron). Let [A] > 0 be a M by M matrix, let ∏ > 0 be given by (4.20)
and (4.21), and let ∫ be a vector x that maximizes (4.20). Then
1. ∏∫ = [A]∫ and ∫ > 0.
2. For any other eigenvalue µ of [A], µ < ∏.
3. If x satisﬁes ∏x = [A]x, then x = β∫ for some (possibly complex) number β .
Discussion: Property (1) asserts not only that the solution ∏ of the optimization problem
is an eigenvalue of [A], but also that the optimizing vector ∫ is an eigenvector and is
strictly positive. Property (2) says that ∏ is strictly greater than the magnitude of any
other eigenvalue, and thus we refer to it in what follows as the largest eigenvalue of [A].
Property (3) asserts that the eigenvector ∫ is unique (within a scale factor), not only among
positive vectors but among all (possibly complex) vectors.
Proof* Property 1: We are given that
∏ = g (∫ ) ≥ g (x ) for each x ≥ 0 , x 6= 0 .
(4.22)
P
We must show that ∏∫ = [A]∫ , i.e., that ∏∫i = j Aij ∫j for each i, or equivalently that
P
j Aij ∫j
∏ = g (∫ ) = gi (∫ ) =
for each i.
(4.23)
∫i
Thus we want to show that the minimum in (4.21) is achieved by each i, 1≤i≤M. To
show this, we assume the contrary and demonstrate a contradiction. Thus, suppose that
6 Note that the set of nonzero vectors x for which x ≥ 0 is diﬀerent from the set {x > 0} in that the
former allows some xi to be zero, whereas the latter requires all xi to be zero. 152 CHAPTER 4. FINITESTATE MARKOV CHAINS g (∫ ) < gk (∫ ) for some k. Let e k be the kth unit vector and let ≤ be a small positive number.
The contradiction will be to show that g (∫ + ≤e k ) > g (∫ ) for small enough ≤, thus violating
(4.22). For i 6= k,
P
P
j Aij ∫j + ≤Aik
j Aij ∫j
gi (∫ + ≤e k ) =
>
.
(4.24)
∫i
∫i
gk (∫ + ≤e k ), on the other hand, is continuous in ≤ > 0 as ≤ increases from 0 and thus remains
greater than g (∫ ) for small enough ≤. This shows that g (∫ + ≤e k ) > g (∫ ), completing the
contradiction. This also shows that ∫k must be greater than 0 for each k.
Property 2: Let µ be any eigenvalue of [A]. Let x 6= 0 be a right eigenvector (perhaps
complex) for µ. Taking the magnitude of each side of µx = [A]x , we get the following for
each component i
X
X
µxi  = 
Aij xj  ≤
Aij xj .
(4.25)
j j Let u = (x1 , x2 , . . . , xM ), so (4.25) becomes µu ≤ [A]u . Since u ≥ 0,...
View
Full
Document
This note was uploaded on 09/27/2010 for the course EE 229 taught by Professor R.srikant during the Spring '09 term at University of Illinois, Urbana Champaign.
 Spring '09
 R.Srikant

Click to edit the document details