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Let $H$ be a $n \times n$ hermitian matrix which satisfies

$$ \sum_i^n |H_{ij}|^2 \leq 1,~~ \sum_j^n |H_{ij}|^2 \leq 1 . $$ This means that the norm of each column or row vector of $H$ is less than 1.

Then my question is, "Does this condition imply the maximum modulus of its eigenvalues $|\lambda_i|$ to be less than some constant?"

For similar case, the modulus of all the eigenvalues of Stochastic matrix $P$ which satisfies $\sum_i P_{ij}=1$ is less than 1.

2 by 2 case

I first tried $2 \times 2$ matrix. The general $2 \times 2$ hermitian matrix $M$ can be written as

$$ M= \begin{bmatrix} a+d & b+ic\\ b-ic & a-d \end{bmatrix} $$ where $a,b,c,d$ are real numbers. Then above condition implies

$$ (a+d)^2+b^2+c^2 \leq 1 \\ (a-d)^2+b^2+c^2 \leq 1 \\ $$

Also, the two eigenvalues for $M $ is

$$ \lambda_{\pm} = a \pm \sqrt{b^2+c^2+d^2} $$ Since $\pm a$ gives same maximum modulus (if $a>0$, $\lambda_{+}$ will give the maximum modulus, and vise versa), I only consider the maximum value of $\lambda_{+}$ with $a>0$. Squaring both side, we get

$$ (\lambda_+ -a)^2 = b^2+c^2+d^2 $$ Using the two inequality, I get

$$ (\lambda_+ -a)^2 = b^2+c^2+d^2 \leq 1-a^2-2ad\\ (\lambda_+ -a)^2 = b^2+c^2+d^2 \leq 1-a^2+2ad\\ $$ Both condition gives the same lower bound for $\lambda_{+}$ (if the sign of $a,d$ are same the first inequality gives lower bound and for opposite sign, the second inequality gives lower bound, and they gives the same lower bound)

So without loss of generality, I choose $a,d \geq 0$. Then tabulating the first inequality, I get

$$ \lambda_+^2 - 2a\lambda_+ - 1+2a^2+2ad \leq 0 $$

This means that the maximum value of $\lambda_+^{max}$ occurs at the solution of

$$ \lambda_+^2 - 2a\lambda_+ - 1+2a^2+2ad = 0 \\ \Rightarrow \lambda_+ = a \pm \sqrt{1-a^2-2ad} $$

So the maximum value of $\lambda_+^{max}$ would occur at $d=0$ and $\frac{d \lambda}{da}=0 \Rightarrow$ $\lambda_+^{max}=\sqrt{2}$ with $a=\frac{1}{\sqrt{2}}$

So there is a modulus maximum for all the eigenvalues. Does this generally hold for larger Hermitian matrix? If there's any flaw in my derivation, please let me know.

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For a square matrix $A$ over $\Bbb C$ let $\|A\|=\left(\sum_{|a_{ij}^2|} \right)^{1/2} $ be its Euclidean norm. If $A=\|a_{ij}\|$ put $A^*=\|\bar a_{ji}\|$. Matrix $A$ is Hermitian iff $A^*=A$.

Theorem. Let $A$ be an $n\times n$ square matrix over $\Bbb C$ with eigenvalues $\mu_1,\dots,\mu_n$. Then

$\sum_{r=1}^n |\mu_r|^2\le \|A\|$, $\sum_{r=1}^n |\operatorname{Re}(\mu_r)|^2\le \|B\|$, $\sum_{r=1}^n |\operatorname{Im}(\mu_r)|^2\le \|B\|$,

where $B=\frac 12 \left(A+A^*\right)$ and $B=\frac 12 \left(A-A^*\right)$. The equality in any one of these inequalities implies the equality in all three and is realized iff $A$ is normal.

See P. Lankaster, “Theory of Matrices”, Academic Press, 1969 (Theorem 7.3.1 in Russian translation, Moskow, Nauka, 1978). Attributed, I expect, to I. Schur, Math. Ann 66 (1909), 488-510.

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