Eigenvalue bounded by sum of row and column Let $A =(a_{ij})$ be a matrix and $R_i=\sum_j |a_{ij}|$ and $C_j=\sum_i |a_{ij}|$
Let $\lambda$ be an eigen value of $A$.
Prove that there exists a $k$ such that $$|\lambda| \leq \sqrt{R_k C_k}$$
Any hints?
Edit: What I tried so far
we have $AX=\lambda X$ where $||X||=  X^tX=1$
$$|\lambda|^2 = X^tA^tAX = \sum_i\sum_j x_i(A^tA)_{i,j}x_j = \sum_p\sum_ix_ia_{ip}\sum_jx_ja_{pj} \leq \sum_p\sqrt{\sum_i|x_i|^2|a_{ip}|}\sqrt{R_p}\sqrt{\sum_j|x_j|^2|a_{pj}|}\sqrt{C_p} \text{     (Cauchy schwartz)}$$
Continuing: Let $R_kC_k=\max_p{R_pL_p}$
$$\begin{align} |\lambda|^2 &\leq \sqrt{R_kC_k} \sum_p\sqrt{\sum_i|x_i|^2|a_{ip}|}\sqrt{\sum_j|x_j|^2|a_{pj}|} \\
 &\leq \sqrt{R_kC_k}\sqrt{\sum_p\sum_i|x_i|^2|a_{ip}|\sum_p\sum_j|x_j|^2|a_{pj}|}\\ &=\sqrt{R_kC_k}\sqrt{\sum_i|x_i|^2\sum_p|a_{ip}|\sum_j|x_j|^2\sum_p|a_{pj}|} \\
&\leq...
\end{align}$$
I'm still stuck here...
 A: From the Ostrowski theorem all eigenvalues are contained in the union of discs
$$
O_i = \left\{z \mid |z - a_{ii}| \leq (R_i - |a_{ii}|)^\alpha (C_i - |a_{ii}|)^{1-\alpha}\right\}, \quad \alpha \in [0, 1],\\
\lambda(A) \in \cup_{i=1}^n O_i.
$$
Here I've used $R_i$ and $C_i$ from question definition, not the ones from referred page.
Consider $\alpha = 1/2$, and let $k$ be the disc where the eigenvalue belongs:
$$
|\lambda - a_{kk}| \leq \sqrt{(R_k - |a_{kk}|)(C_k - |a_{kk}|)}
$$
and
$$
|\lambda| \leq |a_{kk}| + |\lambda - a_{kk}| \leq 
|a_{kk}| + \sqrt{(R_k - |a_{kk}|)(C_k - |a_{kk}|)}.
$$
Let's prove
$$
|a_{kk}| + \sqrt{(R_k - |a_{kk}|)(C_k - |a_{kk}|)} \leq \sqrt{R_kC_k}\\
\sqrt{(R_k - |a_{kk}|)(C_k - |a_{kk}|)} \leq \sqrt{R_kC_k} - |a_{kk}|\\
(R_k - |a_{kk}|)(C_k - |a_{kk}|) \leq R_kC_k - 2 |a_{kk}|\sqrt{R_kC_k}+ |a_{kk}|^2\\
-|a_{kk}| (R_k + C_k) \leq -2|a_{kk}| \sqrt{R_k C_k}\\
\frac{R_k + C_k}{2} \geq \sqrt{R_k C_k}.
$$
The last line is obvious and all operations were equivalent. Hence,
$$
|\lambda| \leq \sqrt{R_k C_k}.
$$
