TITLE: Theorem about positive matrices
QUESTION [15 upvotes]: We will call a matrix positive matrix if all elements in the matrix are positive, and we will denote the largest eigenvalue with $\lambda_{\max}$, what is exist because of the Perron–Frobenius theorem.
Theorem. Let $A$ be a positive square matrix. If any element increases in the matrix then $\lambda_{\max}$ increases.
My questions. 

Is there a name for this theorem and can anybody say books or papers what refer to it?
How to prove it?

REPLY [2 votes]: I recently came across the following (beautiful) result which is slightly improving the answer of @user1551. The following results come from the book Nonnegative Matrices in the Mathematical Sciences of Abraham Berman and Robert J. Plemmons:

Corollary 1.5 (p.27):
  Let $A,B$ be nonnegative square matrices.
   - If $0\leq A \leq B$, then $\rho(A)\leq \rho(B)$.
   - If $0\leq A \leq B$, $A\neq B$ and $A+B$ is irreducible, then $\rho(A)<\rho(B)$.

Note 1: Here $A\leq B$ means $A_{i,j}\leq B_{i,j}$ for every $i,j$.
Note 2: If $0\leq A \leq B$, $A\neq B$ and $A+B$ is reducible, then both can occur, i.e. $\rho(A)=\rho(B)$ or $\rho(A)<\rho(B)$. Indeed, let $U,V,W\in\Bbb R^{2\times 2}$ with
$$ U =\begin{pmatrix} 1& 0 \\ 0 & 1 \end{pmatrix}, \quad  V =\begin{pmatrix} 1& 1 \\ 0 & 1 \end{pmatrix}, 
\quad  W =\begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}.$$
Then $U\leq V \leq W$, $U\neq V$, $V\neq W$, $U+V$ is reducible, $V+W$ is reducible and $\rho(U)=\rho(V)<\rho(W)$.
Note 3: The following result shows that the answer of @user1551 is implied by Corollary 1.5.

Corollary 1.10 (p.28):
  Let $A,B$ be nonnegative square matrices. If $A$ is irreducible, then $A+B$ is irreducible.

Finally, to see how this also improves the answer to your question, note that the matrix $V$ above is reducible as well as the matrix $$V'=\begin{pmatrix}0 & 0 \\ \epsilon & 0 \end{pmatrix}.$$
However, Corollary 1.5 implies that $\rho(V)<\rho(V+V')$ be cause $V+V'$ is strictly positive (and thus irreducible).