How prove this matrix inequality $\lambda_{n-1}\le\frac{n}{n-1}\min{\{a_{jj}:1\le j\le n\}}$

Question

let this Positive semi-definite matrix $A=(a_{ij})_{n\times n}$,and the Characteristic values is $\lambda_{1},\lambda_{2},\cdots,\lambda_{n}$,such $\lambda_{1}\ge \lambda_{2}\ge\cdots\ge \lambda_{n-1}\ge\lambda_{n}\ge 0$, and the matrix $A=(a_{ij})_{n\times n}$ such $$a_{11}+a_{21}+a_{31}+\cdots+a_{n1}=0$$ $$a_{12}+a_{22}+a_{32}+\cdots+a_{n2}=0$$ $$a_{13}+a_{23}+a_{33}+\cdots+a_{n3}=0$$ $$\cdots\cdots\cdots\cdots$$ $$a_{1n}+a_{2n}+a_{3n}+\cdots+a_{nn}=0$$

show that $$\lambda_{n-1}\le\dfrac{n}{n-1}\min{\{a_{jj}:1\le j\le n\}}$$

My try: this book Hint:

note this symmetry matrix $$A-\lambda_{n-1}\left(I-\dfrac{1}{n}J\right)$$

where $J=(a_{ij}),a_{ij}=1$. and I can't,Thank you very much

can you help me? Thank you

This book is from china famous author.

it is well know this book problem is very very hard!and it is said this book is hardest in china linear algebra problem.

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Answer 1

Since $A$ is positive semidefinite and its column sums are zero, $e=(1,\ldots,1)^\top$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda_n=0$. Therefore $e$ is also an eigenvector of $B=A - \lambda_{n-1} (I-\frac1nJ)$ corresponding to the eigenvalue $0$. Now, for all $x\perp e$, we have $Jx=0$ and $x^\ast Ax\ge \lambda_{n-1}\|x\|^2$. Hence $x^\ast Bx\ge0$ and $B$ is positive semidefinite.

Therefore $e_i^\ast Be_i\ge0$ for each $i$, where $\{e_1,\ldots,e_n\}$ is the canonical basis of $\mathbb{C}^n$. That is, $a_{ii} - \lambda_{n-1}(1-\frac1n)\ge0$. Hence $\lambda_{n-1}\le\frac{n}{n-1}a_{ii}$ for every $i$.