Modulus of eigenvalues of Hermitian matrix under some conditions

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.

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.