# Let $P$ be an n th order matrix whose row sums equal $1$. Then for any positive integer m the row sums of the matrix $P^m$ equal $1$.

Let $P$ be an n th order matrix whose row sums equal $1$. Then for any positive integer m the row sums of the matrix $P^m$ equal $1$.

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Another approach: Let $\,A=(a_{ij})\,\,,\,B=(b_{ij})\,$ be two $\,n\times n\,$ matrices s.t. their rows sums is $\,1\,$ ,i.e.

$$\forall\,i\,\,,\,\sum_{j=1}^na_{ij}=\sum_{j=1}^nb_{ij}=1$$

Lema: Then also $\,AB\,$ has rows sum equal to $\,1\,$

Proof: By definition we have:

$$AB=\left(\sum_{k=1}^na_{ik}b_{kj}\right)\Longrightarrow\,\, \text{For any}\,\,\,1\leq i\leq n:\,$$

$$\sum_{j=1}^n\sum_{k=1}^na_{ik}b_{kj}=\sum_{k=1}^na_{ik}\sum_{j=1}b_{kj}=\sum_{k=1}^na_{ik}\cdot 1=1\;\;\;\;\;\;\;\;\;\;\square$$

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Let $\mathbf{v} = \begin{pmatrix}1 & 1 & \cdots & 1\end{pmatrix}^\rm{T}$ be the vector of $n$ ones. The condition that your matrix $P$ has row sums to one is precisely the condition that $$P\mathbf{v} = \mathbf{v}$$ This means that $\mathbf{v}$ is an eigenvector with eigenvalue $1$. Now re-apply $P$ as you wish.

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