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.
This isn't true. E.g. equalities hold when $A$ is a diagonal matrix.
However, the statements are true in almost all generic cases. By relabelling the rows and columns of $A$ if necessary, we may assume without loss of generality that $d_{11}\le d_{22}\le\cdots\le d_{nn}$.
Suppose that the first column of $A$ contains a nonzero off-diagonal entry. Let $e_1=(1,0,\ldots,0)^T$. Then $$ \lambda_\min(A)=\min_{\|u\|_2=1}u^TAu\le e_1^TAe_1=a_{11}=d_{11}=\lambda_\min(D). $$ Since the first column of $A$ contains a nonzero off-diagonal entry, $e_1$ is not an eigenvector of $A$. The inequality above must therefore be strict. It follows that \begin{aligned} \lambda_\min(A) &=t\lambda_\min(A)+(1-t)\lambda_\min(A)\\ &<t\lambda_\min(A)+(1-t)\lambda_\min(D)\\ &=\lambda_\min(tA)+\lambda_\min((1-t)D)\\ &\le\lambda_\min(tA+(1-t)D)\\ &=\lambda_\min(D+tB). \end{aligned} Similarly, if there is a nonzero off-diagonal entry on the last column of $A$, then $\lambda_\max(A)>\lambda_\max(D+tB)$.