Is $A+D$ an irreducible matrix? 
Suppose that $A$ is an irreducible matrix and all its entries are non-negative.
  Let $D$ be a diagonal matrix whose all entries are positive.
Is $A+D$ an irreducible matrix?

Since $A$ is an irreducible matrix  so for each $i,j$ there exists $m\in \Bbb N$ such that $A_{ij}^m>0$.
But how can I show that the same is true if we consider $A+D$?
Also I did not find any examples to show that the result is false.
I guess the result is true.
How to prove it?
Any help.
 A: Given a pair $i, j$ choose $m$ such that
$(A^m)_{ij} > 0; \tag 1$
now consider $(A + D)^m$; though somewhat difficult to express in $LaTeX$, it may be written as a sum of the form
$(A + D)^m = A^m + B_m, \tag 2$
where $B_m$ is the sum of every degree $m$ product of $A$s and $D$s which contain at least one factor of $D$; since every entry of every factor in these products is non-negative, so is every entry of each product, hence every entry of their sum; in particular, 
$(B_m)_{ij} \ge 0; \tag 3$
it follows that
$((A + D)^m)_{ij} = (A^m)_{ij} + (B_m)_{ij} > 0 \tag 4$
by virtue of (1).  Thus $A + D$ is irreducible.
We can see this with greater rigor by looking at it inductively; note that
$(A + D)^2 = A^2 + AD + DA + D^2 = A^2 + B_2, \tag 5$
$(A + D)^3 = (A + D)(A^2 + AD + DA + D^2)$$
= A^3 + A^2D + ADA + AD^2 + DA^2 + DAD + D^2 A + D^3 = A^3 + B_3; \tag 6$
$B_2$ and $B_3$ are defined by these equations and it is clear by inspection that each has only nonnegative entries.  Now indeed if
$(A + D)^k = A^k + B_k, \tag 7$
then
$A^{k + 1} + B_{k + 1} = (A + D)^{k + 1} = (A + D)(A^k + B_k) = A^{k + 1} + AB_k + DA^k + DB_k \tag 8$
also has only non-negative entries since every summand on the right-hand side does; by induction, it follows that $(A + D)^n$ has only non-negative entries for every $n$, and the same holds for every $B_n$; it is now easy to see that if (1) binds for some $m$, we have
$((A + D)^m)_{ij} = (A^m)_{ij} + (B_m)_{ij} > 0; \tag{9}$
since this holds for every pair $ij$ for some $m$, we see that $A + D$ is irreducible.
