Answer:
a) Entries of A are non negative, that is ai,j ≥ 0 for all 1 ≤ i ≤ n and all 1 ≤ j ≤ n.
c) yes MS is stochastic
Step-by-step explanation:
a) A stochastic matrix is a square matrix whose columns are probability vectors. A probability vector is a numerical vector whose entries are real numbers between 0 and 1 whose sum is 1.
B)
a) Now, suppose Sx=λx for some λ>1. Since the rows of S are nonnegative and sum to 1, each element of vector Sx is a convex combination of the components of x, which can be no greater than maximum of x the largest component of x. On the other hand, at least one element of λx is greater than maximum of x, which proves that λ>1 is impossible and Hence λ = 1.
b) Then [tex]S^{2}[/tex] =[tex]P^{2}_{ij}[/tex] is also stochastic; it is the two-step transition matrix for the chain {Xn, n = 0,1,…}. To every stochastic matrix S, there corresponds a Markov chain {Xn} for which S is the unit-step transition matrix.
However, not every stochastic matrix is the two-step transition matrix of a Markov chain.
c) Let A and B be two row-stochastic matrices and suppose we know the product of column stochastic matrices is column-stochastic. Observe that,
MS = [tex]((MS)^{T})^{T} = (S^{T}M^{T})^{T}[/tex]
by properties of transpose of a matrix. Let us consider [tex]S^{T} M^{T}[/tex]. It is easy to see that the transpose of a row-stochastic matrix is column-stochastic by definition (and vice versa). Thus, [tex]S^{T}[/tex]and [tex]M^{T}[/tex] are column stochastic and by our assumption, it must then be the case that [tex]S^{T} M^{T}[/tex] is column-stochastic. Since [tex]S^{T} M^{T}[/tex] is column-stochastic, then it's transpose [tex](S^{T} M^{T})^{T}[/tex]=MS is row stochastic.
