Suppose I have an arbitrary real matrix $\mathbf{A}\in\mathbb{R}^{n\times n}$ such that the sum of absolute values in each row is greater than $1$:

$$\sum_{j=1}^n |A_{ij}|>1,\quad \forall i=1,\ldots,n.$$

(I guess this means that the infinity norm $||\mathbf{A}||_{\infty}$ of $\mathbf{A}$ is greater than $1$.)

Intuitively, this should imply that the spectral radius (largest eigenvalue in absolute terms) of $\mathbf{A}$ is greater than $1$. Is this true? And if so, how to prove it?


Try $$ A=\pmatrix{1&-1\\1&-1}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.