About the $1$-norm and an M-matrix I'm studying a proof about PageRank and I have some doubts. The proof states the following:

If $\beta < 1$, and $P$ is a stochastic matrix for column (i.e. $ \lVert P \rVert=1$) then $I-\beta P$ is a $M$-matrix, diagonally-dominant and so its inverse is not negative. Moreover $\lVert (I-\beta P)^{-1} \rVert= \frac1 {1-\beta}$.

I proved that $I−\beta P$ is an $M$−matrix and diagonally-dominant. Does there exist a theorem that states that a matrix of this kind has an inverse that is not negative?. What about the last request? Could someone please help?
 A: For the first part note that diagonally dominant matrices are invertible and see the previous Question How to prove that an M-matrix is inverse-non-negative?
Throughout we use the $1$-norm for vectors without explicit annotation:
$$ ||(v_1,v_2,\ldots,v_n)|| = \sum_{i=1}^n |v_i| $$
The final claim you need help with is the $1$-norm of the matrix $||(I-\beta P)^{-1}|| = \frac{1}{1-\beta}$.
Since $||\beta P|| = \beta < 1$, we have a convergent "geometric series" for the inverse of $I-\beta P$:
$$ (I-\beta P)^{-1} = I + \beta P + (\beta P)^2 + (\beta P)^3 + \ldots $$
Now we get by the triangle inequality (taking $1$-norms) that an upper bound on $||(I-\beta P)^{-1}||$ is the fraction $\frac{1}{1-\beta}$:
$$ ||(I-\beta P)^{-1}|| \le \frac{1}{1-\beta} $$
All that remains is to show this upper bound is exact, and for that it suffices to given an example.
In the Question you have written that "$P$ is a stochastic matrix for column".  Possibly by this you mean that the matrix $P$ is left stochastic as described in the Wikipedia article Stochastic matrix, i.e. that each column sums to $1$.  What this affects is the direction of the multiplication involved in defining the $1$-norm of the matrix $P$.  Since you skipped over that part quickly, you are apparently satisfied with the assignment $||P|| = 1$ based on the left multiplication, e.g.
$$ (1,1,1,\ldots,1) P = (1,1,1,\ldots,1) $$
If so, we only need to note that for $(I-\beta P)^{-1}$ we can say the following:
$$ (1,1,1,\ldots,1) (I-\beta P) = (1-\beta) (1,1,1,\ldots,1) $$
It follows that $1$-norm of $(I-\beta P)^{-1}$ being $\frac{1}{1-\beta}$ is attained by the vector $(1,1,1,\ldots,1)$ when the above equation is solved:
$$ (1,1,1,\ldots,1) (I-\beta P)^{-1} = \frac{1}{1-\beta} (1,1,1,\ldots,1) $$
That is, since $(1,1,1,\ldots,1) $ is a left eigenvector of $(I-\beta P)^{-1}$ corresponding to eigenvalue $\frac{1}{1-\beta}$, the $1$-norm is attained here:
$$ || (1,1,1,\ldots,1) (I-\beta P)^{-1} || = \frac{1}{1-\beta} ||(1,1,1,\ldots,1)|| $$
