How to prove the largest eigenvalue of a matrix is smaller or equal to the max of absolute sum of rows?

Question

Namely, I need to prove ${\max\limits_i} |\lambda_i| \leq {\max\limits_i}{\sum\limits_j} |M_{ij}|\mid$, where $M$ is the matrix and $\lambda_i$ are its eigenvalues.

I'm not sure if there is any helpful theorem or lemma. Is there any hint on how to solve this?

Answer 1

In fact a stronger statement is true. Let $\rho(M) = \max_i|\lambda_i|$ be the spectral radius of $M$. Then for each positive integer $k$, we have $\rho(M)\leqslant \|M^k\|^{\frac1k}$. For if $v$ is an eigenvector of $M$ with associated eigenvalue $\lambda$, then $$ |\lambda|^k\|v\| = \|\lambda^k v\| = \|A^k v\| \leqslant \|A^k\|\|v\|, $$ and dividing by $\|v\|$ (which is positive since $v\ne0$) yields $$ |\lambda|^k\leqslant\|A^k\|. $$ Raising both sides to $\frac1k$ and taking the maximum over $|\lambda_i|$ yields the statement.

In fact it is true that $\rho(M) = \lim_{k\to\infty}\|M^k\|^{\frac1k}$, a result known as Gelfand's formula, but that is beyond the scope of this question.

Answer 2

Hint: Take an eigenvector $v$ corresponding to the maximum eigenvalue, and scale it so that the largest entry (in absolute value) is $1$, at the $j$th row. Look at the $j$th row of $Mv$.