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

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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• What book is this from? Commented Dec 1, 2013 at 6:01
• @Potato,I have edit.Thank you
– user94270
Commented Dec 1, 2013 at 6:10

## 1 Answer

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 $i$.

• Why is $x^*Ax \ge \lambda_{n-1} \|x \|^2$? $\lambda_{n-1}$ is not the smallest eigenvalue.
– user27126
Commented Dec 1, 2013 at 7:18
• oh,That's my ask it,Thank you
– user94270
Commented Dec 1, 2013 at 7:19
• @Sanchez This is because $x\perp e$. That is, $x$ lies inside the subspace spanned by the eigenvectors corresponding to $\lambda_1,\ldots,\lambda_{n-1}$. (Every Hermitian matrix is a normal matrix; it possesses a complete set of orthogonal eigenvectors.) Commented Dec 1, 2013 at 7:21
• Oh,I kown,Thank you @user1551
– user94270
Commented Dec 1, 2013 at 7:26
• @user1551, thanks!
– user27126
Commented Dec 1, 2013 at 7:27