Leave $ H $ be a $ n times n $ Hermitian matrix that 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

So my question is: "Does this condition imply the maximum module of its own values? $ | lambda_i | $ be less than some constant? "

For the similar case, the module of all the eigenvalues of the stochastic matrix $ P $ what satisfies $ sum_i P_ {ij} = 1 $ is less than 1

## 2 for 2 case

First I tried $ 2 times 2 $ matrix.

The general $ 2 times 2 $ Hermitian matrix $ M $ It can be written as

$$

M =

begin {bmatrix}

a + d & b + ic \

b-id and a-d

end {bmatrix}

$$

where $ a, b, c, d $ they are real numbers

So the previous condition implies

$$

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

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

$$

In addition, the two proper values for $ M $ is

$$

lambda { pm = a pm sqrt {b ^ 2 + c ^ 2 + d ^ 2}

$$

As $ ± a $ gives the same maximum module (if $ a> 0 $, $ lambda _ {+} $ will give the maximum module, and vice versa), I only consider the maximum value of $ lambda _ {+} $ with $ a> 0 $.

Squared both sides, we get

$$

( lambda_ + -a) ^ 2 = b ^ 2 + c ^ 2 + d ^ 2

$$

Using the two inequalities, 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 conditions give the same lower limit for $ lambda _ {+} $ (if the sign of $ a, d $ the first inequality is equal to a lower limit and for the opposite sign, the second inequality gives a lower limit, and gives the same lower limit

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 in 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 happen in $ d = 0 $ Y $ frac {d lambda} {da} = 0 Rightarrow $ $ lambda _ + ^ {max} = sqrt {2} $ with $ a = frac {1} { sqrt {2}} $

Then there is a maximum of module for all eigenvalues. Is this generally valid for a larger Hermitian matrix? If there is a defect in my referral, please let me know.