Suppose we have a $n\times n$ symmetric positive semi-definite matrix $\mathbf{A}$.
Based on Gershgorin circles theorem all the eigenvalues of the, $\mathbf{A}=[a\_{ij}]$, are located in the union of $n$ circles: 
\begin{equation\*}
 \bigcup\_{i=1}^{p}\bigg\{r\in \mathbb{R}:|r-a\_{ii}|\leq R\_{i}(\mathbf{A})\bigg\}
\end{equation\*}
where $R\_{i}(\mathbf{A})=\sum\_{j,j\neq i}^{n}|a\_{ij}|$. Therefore, bounds of Gershgorin:
\begin{equation\*}
[\lambda^{\geq}(\mathbf{A}) = min\_{i} (a\_{ii}-R\_{i}(\mathbf{A})), \lambda^{\leq}(\mathbf{A}) = max\_{i} (a\_{ii}+R\_{i}(\mathbf{A}))]
\end{equation\*}
Now, we create a family of matrices:
\begin{equation}
 \mathbf{A}(t) = t\mathbf{B}+\mathbf{D},
\end{equation}
where $\mathbf{D}$ is the same as $\mathbf{A}$ with all the off-diagonal entries reduced to zero and $\mathbf{B}$ is the same as $\mathbf{A}$ with all the diagonal entries reduced to zero all
along the interval $0<t\leq1$.


Are the following statements correct?
\begin{equation}
λ\_1(A(t))>λ\_1(A) 
\\
λ\_n(A(t))<λ\_n(A)
\end{equation}
where $\lambda\_{1}(\mathbf{A}(t))$ and $\lambda\_{n}(\mathbf{A}(t))$ are the smallest and largest eigenvalue repectively. 


Thanks.



Shifted eigenvalues and Gershgorin theorem
We cannot have strict inequalities in all cases since you could have $B=0$. After this adjustment, we can obtain the claim as follows.


Let me slightly change notations and consider $A(s)=A-sB$ (so $s=1-t$). We can assume that $\lambda\_1(A)=0$. Let $v$ be a normalized eigenvector, so $Av=(D+B)v=0$ and hence $\langle v, Bv\rangle \le 0$. In fact, the case $\langle v, Bv \rangle = 0$ is trivial because then $Dv=0$, so $D\_{jj}=0$ for all $j$ with $v\_j\not= 0$. It then follows that $B\_{jk}=0$ as well for these components, so $\lambda\_1(A(s))=0$ for all $s$.


If $\langle v, Bv \rangle < 0$, then first order perturbation theory gives
$$
\lambda\_1(A(s))=-s\langle v, Bv \rangle +o(s) > \lambda\_1(A)
$$
for all small $s>0$ (strictly speaking, the argument in this form requires the lowest eigenvalue to be non-degenerate, but we can adapt the argument to the general situation by working in the corresponding eigenspace). So we have the inequality locally, and of course this is good enough since any $A(s\_0)$ can take over the role of $A$.


