Showing $\boldsymbol{Q}$ is stochastic given a transition matrix on $k$ states

Question

It is known that a stochastic matrix is a square matrix $\boldsymbol{T}$ that satisfies

1. $\boldsymbol{T_{ij}} \geq 0$ for all $i,j$.
2. $\sum_{j}^{ }\boldsymbol{T_{ij}}=1.$

Assume $\boldsymbol{P}$ is the transition matrix of a Markov chain on $k$ states. $\boldsymbol{I}$ is the $k \times k$ identity matrix.

Consider the matrix $\boldsymbol{Q}=(1-p)\boldsymbol{I}+p\boldsymbol{P}, 0<p<1$.

How would one go about showing that $\boldsymbol{Q}$ is a stochastic matrix?

Answer 1

If $Q= [q_{ij}]$ and $P=[p_{ij}]$, then the $(i,j)$-entry of $(1-\alpha)I_n + \alpha P$, $\alpha \in (0,1)$, is given by $$ [(1-\alpha)I_n + \alpha P]_{ij} = \begin{cases} 1-\alpha+\alpha p_{ii}, & i=j \\ \alpha p_{ij}, & i \ne j. \end{cases} $$ Since the sum of nonnegative matrices is nonnegative and since a scalar multiple of a nonnegative matrix is nonnegative, it follows that this matrix must be nonnegative. Thus, it suffices to show that the matrix is row stochastic.

If $P$ is row stochastic, then for $i \in \{1,\dots, n\}$, notice that the $i^\text{th}$ row-sum is $$ 1 - \alpha + \alpha \left(\sum_{j=1}^n p_{ij} \right) =1 -\alpha + \alpha = 1. $$

A similar argument applies to column stochastic matrices. Thus, the set of all (doubly) stochastic matrices is star-convex with star-center at the identity.

Answer 2

Here's another proof, which is far more elegant.

Exercise: If $A$ is a square complex matrix, then every row of $A$ sums to one if and only if $Ae = e,$ i.e., $(1,e)$ is a right eigenpair of $A$ ($e$ denotes the all-ones vector).

Thus, if $P$ is row stochastic, then $Pe = e$ and $$ ((1-\alpha)I_n + \alpha P)e = (1-\alpha)e +\alpha Pe = (1-\alpha)e +\alpha e = e. $$