I saw few questions about it, but still can't understand.

Let $A$ be a symmetric matrix and $\lambda_{\max}$ its largest eigenvalue. Is the following true for all $A$?

$$ \lambda_{\max} \ge a_{ii} \forall i $$

That is, is the largest eigenvalue of a symmetric matrix always greater than any of its diagonal entries?

Is it somehow related to spectral radius and the following equation?

$$ \rho(A)=\max|\lambda_i|. $$

  • $\begingroup$ Did you mean to write $a_{ii} > \lambda$? $\endgroup$
    – Leo
    Sep 25, 2019 at 11:45
  • $\begingroup$ @Leo No, I mean max eigenvalue bigger than any diagonal entity $\endgroup$ Sep 25, 2019 at 12:11

2 Answers 2


Let $\lambda_1,\cdots, \lambda_n$ be eigenvalues of $A $ in increasing order,i.e. $\lambda_n$ is the maximum eigenvalue.

By Min-max theorem $$\lambda_n=\max_{x\in \mathbb{R^n},||{x}||=1 }x^TAx=\sum x_ix_j a_{ij} $$ while on the other hand$$a_{ii}=e_i^TAe_i $$ It follows that $$\lambda_n\ge a_{ii} \qquad \forall i$$


"No, I mean max eigenvalue bigger than any diagonal entity"

Yes. Take the adjacency matrix of any undirected simple graph. It's diagonal is all zeros, and each off diagonal entry is either zero or one. It's largest eigenvalue is always positive due to the Perron-Frobenius theorem and therefore larger than any diagonal entry. For slot of graphs (and random graphs in particular), the largest eigenvalue is about the largest degree in the graph, so you can find matrices where the difference between the largest eigenvalue and the largest diagonal entry (in this case, zero) is arbitrarily large.

  • $\begingroup$ Thank you, Sorry maybe I didn't explain myself the right way. I know that for some matrix $\lambda_{max} \ge a_{ii} \forall i $. My question is it true for all symmetric matrix? $\endgroup$ Sep 25, 2019 at 12:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .