I have the following $(n+1)\times (n+1)$ matrix


$$P = \begin{bmatrix}
f(0) & g(0) & 0 & 0 & 0 & \dots & 0\\
f(1) & 0 & g(1) & 0 & 0 & \dots & 0\\
f(2) & 0 & 0 & g(2) & 0 & \dots & 0\\
f(3) & 0 & 0 & 0 & g(3) & \dots & 0\\
\vdots & \vdots & & & & \ddots & 0\\
f(n-1) & 0 & 0 & 0 & 0 &\dots & g(n-1)\\
f(n) & 0 & 0 & 0 & 0 &\dots & 0
\end{bmatrix}$$


where $0< f(x)< 1$, $0< g(x)< 1$, and $f(x)+g(x)<1$, $f(x)$ is increasing, $g(x)$ is decreasing. Also, $g(x)$ decreases faster than $f(x)$ is increasing, so their sum $f(x)+g(x)$ is decreasing.


I want to show that by raising $P$ to any power $k$, the summation of the elements of the $i$-th row, i.e. $\sum\_{j=1}^{n+1} P^{k}\_{i,j}$ is decreasing with $i$.


I have taken the following inductive approach:


* I know that the summation of the *rows* of $P^t$ is decreasing when $t=1$, since $f(x)+g(x)$ is decreasing.
* Suppose that the summation of the rows of $P^t$ is decreasing up to $t=k-1$. I want to show that it holds for $t=k$.
* The $i$-th row of $P^k$ can be written as follows:
\begin{equation} P^k\_{i,:} = P\_{i,:} P^{k-1} \end{equation}
from which the summation is derived as
\begin{equation} \sum\_{j=1}^{n+1} P^k\_{i,j} = f(i)\sum\_{j=1}^{n+1} P^{k-1}\_{1,j} + g(i)\sum\_{j=1}^{n+1} P^{k-1}\_{i+1,j}. \end{equation}


I cannot get past this point.


Also, I haven't found any way to compute a closed-form expression of $P^k$, although it has a very specific structure: it is the concatenation of a non-zero column-row pair (the first column and the last row) and a diagonal matrix.


Edit: I do not care about the summation of the last row, only for all the $n$ first rows.



Summation of rows of a matrix P^k is decreasing with the power k
A counterexample is given by
$$P=\left(
\begin{array}{ccc}
 0 & 1 & 0 \\
 0 & 0 & \frac{1}{2} \\
 \frac{1}{2} & 0 & 0 \\
\end{array}
\right),$$
so that then
$$P^2=\left(
\begin{array}{ccc}
 0 & 0 & \frac{1}{2} \\
 \frac{1}{4} & 0 & 0 \\
 0 & \frac{1}{2} & 0 \\
\end{array}
\right).$$




---


Here the conditions on $P$ are satisfied only non-strictly. However, by approximation, this does not matter. For instance, one may replace $P$ by the close matrix  

$$P\_h=\left(
\begin{array}{ccc}
 h & 1-2h & 0 \\
 2 h & 0 & \frac{1}{2} \\
 \frac{1}{2} & 0 & 0 \\
\end{array}
\right)$$
for small enough $h>0$.
Then $P\_h$ will strictly satisfy all the conditions on $f$ and $g$, and yet $P\_h^2$ will be arbitrarily close to $P^2$, and hence your desired conclusion will fail to hold.




---


The OP has changed the question, now saying " I do not care about the summation of the last row, only for all the n first rows."


Then a counterexample is
$$P=\frac1{100}\left(
\begin{array}{ccccc}
 1 & 98 & 0 & 0 & 0 \\
 2 & 0 & 50 & 0 & 0 \\
 25 & 0 & 0 & 25 & 0 \\
 48 & 0 & 0 & 0 & 1 \\
 49 & 0 & 0 & 0 & 0 \\
\end{array}
\right),$$
so that $P^2$ is bad: the sum of the 2nd row of $P^2$ ($\approx0.27$) is less than the sum of the 3rd row of $P^2$ ($\approx0.37$).


