If number of zeros is increased in below diagonal entries of $A$, the largest eigenvalue of $B$ decreases. Is it right? Let $A_i$ be a $n\times n$ lower triangular matrix with diagonal entries $1$ and below diagonal entries are $0$ or $1$. Let O be the zero matrix of order $n$. Consider
$$B_i=\left[ {\begin{array}{cc}
\text{O} & A_i \\
A_i^T & \text{O}
\end{array} } \right] \text{    };i=1,2$$
 Suppose $A_1$ has $a_1$ number of zero entries below diagonal and $A_2$ has $a_2$ number of zero entries below diagonal.
I observed that when $a_2>a_1$, the largest eigenvalue of $B_2$ is less than that of $B_1$. Is it right? if so, any hint to prove it?
 A: Of course, it is false; yet, it is true in a certain sense!
$\det(B-\lambda I_{2n})=\det(\lambda^2I_n-AA^T)$; then $\rho(B)^2=\rho(AA^T)$ and it suffices to consider $\rho(AA^T)$. 
i) Consider $E=I_{10}+{J_{10}}^T$ where $J_k$ is the nilpotent Jordan block of dimension $k$. Then $E$ contains $9$ non-zero entries under the diagonal and $\rho(EE^T)\approx 3.91$. Consider $C=I_{10}+R$ where $r_{i,j}=0$ except $r_{8,1}=r_{9,1}=r_{9,2}=r_{10,1}=r_{10,2}=r_{10,3}=1$. Then $C$ contains $6$ non-zero entries under the diagonal and $\rho(CC^T)\approx 6.90$. That implies that your conjecture is false. 
ii) Note that $A$ and $AA^T$ are non-negative matrices (in the sense $A\geq 0$ iff for every $i,j$, $a_{i,j}\geq 0$). Now, we transform one $0$ under the diagonal of $A$ in $1$; if $A'$ is the obtained matrix (it has a zero less than $A$), then $A'A'^T-AA^T\geq 0$ and, consequently (it is a theorem about the non-negative matrices!), $\rho(A'A'^T)\geq \rho(AA^T)$.
EDIT. What is important is the location of the $1$'s. If we replace some $0$'s by $1$'s far from (resp. near) the diagonal of $A$, then $\rho(AA^T)$ increases quickly (resp. slowly).
