Linear Algebra, Theory And Applications, 2012a

11.1. REGULAR MARKOV MATRICES 277
One of the entries of N s is nonzero by the definition of s. Let this entry be n s ij . Then this
implies that one of the entries of (I + N) n is of the form ( n
s)
n
s
ij . This entry dominates the
ij th entries of ( n
k)
N k for all k 0, not just ≥ 0, then
the eigenvalue condition of the above theorem is valid.
Lemma 11.1.4 Suppose A =(a ij ) is a stochastic matrix. Then λ =1is an eigenvalue. If
a ij > 0 for all i, j, then if μ is an eigenvalue of A, either |μ| < 1 or μ =1. In addition to
this, if Av = v for a nonzero vector, v ∈ R n , then v j v i ≥ 0 for all i, j so the components of
v have the same sign.
Proof: Suppose the matrix satisfies
∑
a ij =1.
j
Then if v = ( 1 ··· 1 ) T
, it is obvious that Av = v. Therefore, this matrix has λ =1
as an eigenvalue. Suppose then that μ is an eigenvalue. Is |μ| < 1orμ =1?Letv be an
eigenvector and let |v i | be the largest of the |v j | .
μv i = ∑ j
a ij v j
and now multiply both sides by μv i to obtain
|μ| 2 |v i | 2 = ∑ j
a ij v j v i μ = ∑ j
a ij Re (v j v i μ)
≤
∑ j
a ij |μ||v i | 2 = |μ||v i | 2
Therefore, |μ| ≤1. If |μ| =1, then equality must hold in the above, and so v j v i μ must be
real and nonnegative for each j. In particular, this holds for j = 1 which shows μ and hence
μ are real. Thus, in this case, μ =1. The only other case is where |μ| < 1.
If instead, ∑ i a ij =1, consider A T . Both A and A T have the same characteristic polynomial
and so their eigenvalues are exactly the same.