# Max eigenvalue of symmetric matrix and its relation to diagonal values

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?

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

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

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 $$\lambda_n\ge a_{ii} \qquad \forall i$$
• 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? Sep 25, 2019 at 12:29