Prove that the interval contains at least one eigenvalue

Question

Let $A = (a_{ij})_{1 \leq i,j \leq n}$ be a symmetric matrix whose entries are real numbers.For each $i \in \{1,2,\cdots,n\}$, show that the set $$E_i := \left\{z : |z - a_{ii}| \leq \left( \sum_{j \neq i} a_{ij}^2 \right)^\frac{1}{2} \right\}$$ contains at least one eigenvalue of A.

Answer 1

Let $A=X\operatorname{diag}(\lambda_1,\ldots,\lambda_n)X^T$ be an orthogonal diagonalisation, $\mathbf x_1,\ldots,\mathbf x_n$ be the columns of $X$, so that the $j$-th coordinate of $\mathbf x_k$ is $x_{jk}$, the $(j,k)$-th coordinate of $X$. Let $D$ and $F$ be respectively the diagonal and off-diagonal parts of $A$, so that $A=D+F$. Then $(\lambda_kI-D)\mathbf x_k=F\mathbf x_k$ for each $k$. Taking the squared moduli of the $i$-th coordinates on both sides, we get $$ (\lambda_k-a_{ii})^2x_{ik}^2=\sum_{j\ne i}a_{ij}^2x_{jk}^2. $$ Sum over $k$, we obtain $$ \sum_k(\lambda_k-a_{ii})^2x_{ik}^2=\sum_{j\ne i}\left(a_{ij}^2\sum_kx_{jk}^2\right)=\sum_{j\ne i}a_{ij}^2.\tag{1} $$ But we also have $$ \min_k(\lambda_k-a_{ii})^2=\left(\min_k(\lambda_k-a_{ii})^2\right)\left(\sum_kx_{ik}^2\right)\le\sum_k(\lambda_k-a_{ii})^2x_{ik}^2.\tag{2} $$ Hence the result follows from $(1)$ and $(2)$.