Can anyone help me with the following problem? Suppose $u=(u_1,u_2,...u_n)^T$, $e=(1,1,...1)^T$, and we have $u\geq e$. Now for any symmetric matrix $A\in S^n$ with $diag(A)=0$, can we claim the following $$ \lambda_{max}(uu^T+A)\geq\lambda_{max}(ee^T+A)? $$ where $\lambda_{max}$ represent the largest eigenvalue of a symmetric matrix.
Thanks!
I think we cannot. Consider $$ A=\left( \begin{array}{ccc} 0 & -2 & 2 \\ -2 & 0 & -2 \\ 2 & -2 & 0 \\ \end{array} \right)$$ and $u=(1,2,1)^T$. Then $\lambda_{\rm max}(uu^T+A)=4$, but $\lambda_{\rm max}(1+A)=(5+\sqrt{17})/2\approx4.6$.
OLD ANSWER
I am assuming that the inequality you want to prove is term-by-term also. I do not understand why you cannot simplify the $A$ term on both sides of the inequality.
If ${\rm diag(A)=0}$, ${\rm tr}(A)=0$. Therefore $\sum \lambda=0$, which implies that $\lambda_{\rm max}\ge0$. If $\lambda_{\rm max}=0$, the inequality is trivial. If $\lambda_{\rm max}>0$, we can simplify it from the inequality, leaving to prove that $$uu^T\ge ee^T,$$ which is clearly true since for any $u_ku_{k'}\ge 1$.