Is the Maximal eigenvalue of $A+B $ greater than that of $A$? $A$ and $B$ are nonegative n×n matrices, $A$ is irreducible and $B$ is nonzero matrix. 
I want to prove that the maximal eigenvalue of $A+B$ is greater than that of $A$.
We know that the maximal eigenvalue of a reducible matrix is nonegative and that of a irreducible matrix is positive,allowing us to compare them.
 A: Since $A$ is irreducible, let $y$ be the positive eigenvector associated to the eigenvalue $\rho(A)$.
$\rho(A+B)$ is an eigenvalue of $A+B$ and also an eigenvalue of the non-negative matrix $(A+B)^T$. Therefore there is a non-zero vector $z\geq 0$ s.t. $z^T(A+B)=\rho(A+B)z^T$.
Thus $0< \rho(A)y=Ay$ and $z^T\rho(A)y=z^TAy\leq z^T(A+B)y=\rho(A+B)z^Ty$.
Since $z^Ty>0$, we deduce that $\rho(A)\leq\rho(A+B)$.
EDIT. I forget... The last inequality is strict when $A+B$ is irreducible and $B>0$; otherwise it is not (in general). 
A: Wielandt's theorem says that if $|B_{i,j}|\leq A_{i,j}$ for all $i$ and $j,$ and $A$ is irreducible, then $\rho(B)\leq\rho(A).$
Applying this to the present case, since $A$ is irreducible, $A+B$ is irreducible, and $0\leq A_{i,j}\leq (A+B)_{i,j}$, so $\rho(A)\leq\rho(A+B)$, and since $A$ and $A+B$ are nonnegative matrices, we know that each has its spectral radius as an eigenvalue, which is necessarily the largest in absolute value and simple, because they are irreducible.
