The Frobenius/Perron theorem [Takayama, 1984, p367-376]

Let A be a nonnegative irreducible nxn-matrix. Then

1. A has an eigenvalue λ^*>0 such that
2. An eigenvector x^*>>0 can be associated with λ^*.
3. The eigenvector x^* is unique up to a scalar multiple; that is, if y^* is an eigenvector associated with λ^*, then y^*=θx^* for some scalar θ<>0.
4. If Ax=λx for some λ>=0 and x>0, then λ=λ^*.
5. If ω is any eigenvalue of A, then |ω|<=λ^*.
6. The eigenvalue λ^* increases when any element of A increases; that is, if A₁>A₂>=0 with A₁ irreducible, then λ^*(A₁)>λ^*(A₂), where λ^*(A₁) and λ^*(A₂) respectively, denote the λ^* associated with A₁ and A₂.
7. The eigenvalue λ^* is a simple root and is called the dominant eigenvalue.

Let A be a nonnegative nxn-matrix. Then

1. The matrix A has an eigenvalue λ^*>=0 such that
2. An eigenvector x^*>0 can be associated with λ^*.
3. If Ax>=μx for some real number μ and x>0, then λ^*>=μ.
4. If ω is any eigenvalue of A, then λ^* >=|ω|.
5. If A₁>=A₂>=0, then λ^*(A₁)>=λ^*(A₂).

1. The root λ^* can be zero.
2. Some, but not all, elements of x^* can be zero.
3. Both x^* and y^*, with y^*<>θx^* and θ element of R can be eigenvectors associated with λ^*.
4. The root λ^* is not necessarily a simple root.

Let A be an irreducible nxn-matrix. Let B be defined as ρI-A, with ρ element of R. Let λ^* be the dominant eigenvalue of A, and I the identity matrix. Then the matrix B is nonsingular and

B^-1>>if and only if ρ>λ^*.

The subinvariance theorem [Seneta, 1981, p20-21]

Suppose T is a nonnegative irreducible matrix, s a positive number, and x>0 is a vector which satisfies Tx<=sx. This implies:

1. x>>0
2. s>=λ^*, with λ^* the dominant eigenvalue of T.
3. s=λ^* if and only if Tx=λ^*x.

Now, since T is irreducible, for this i and any j, there exists a k such that T_ij^(k)>0; and since x>0 for some j, it follows that x_i>0, which is a contradiction. Thus x>>0. Now, premultiplying the relation Ty<=sy by q', a positive left eigenvector of T corresponding to λ^*, we get

sq'y>= q'Tx=λ^*q'y  <=>  s>= λ^*

Now suppose Tx<=λ^*x with strict inequality in at least one place; then the preceding argument, on account of the strict positivity of Tx and λ^*x, yields λ^*<λ^*, which is impossible. The implication s=λ^* follows from Tx=λ^*x similarly.
 
 

An eigenvectortheorem (special case) [Dietzenbacher, 1991, p216]

Let A be reducible, and partitioned as follows

with A₁ and A₂ square, nonnegative and irreducible. Suppose B>0. Suppose furthermore that v'=(v₁,v₂)' is the left dominant eigenvector of A, and w=(w₁,w₂)' is the right dominant eigenvector of A.
 

if λ1=λ* and λ2<λ*	then v1>>0, v2=0, w1>>0 and w2>>0.
if λ1<λ* and λ2=λ*	then v1>>0, v2>>0, w1=0 and w2>>0.
if λ1=λ* and λ2=λ*	then v1>>0, v2=0, w1=0 and w2>>0.

with the eigenvectors all unique up to a scalar multiple

MAIN