Showing $\boldsymbol{Q}$ is stochastic given a transition matrix on $k$ states It is known that a stochastic matrix is a square matrix $\boldsymbol{T}$ that satisfies


*

*$\boldsymbol{T_{ij}} \geq 0$ for all $i,j$.

*$\sum_{j}^{ }\boldsymbol{T_{ij}}=1.$ 


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? 
 A: 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 
$$
((1-\alpha)I_n + \alpha P)e = (1-\alpha)e +\alpha Pe = (1-\alpha)e +\alpha e = e.
$$
A: 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.
