Let $A$ be a symmetric (or Hermitian) matrix with size $n\times n$.

If it helps, one can consider $A$ tridiagonal (and symmetric).

By spectral theorem, there exists $\lambda_1\geq ... \geq \lambda_n$ real eigenvalues.

Are there known connection (such as inequalities etc) between its largest eigenvalue $\lambda_1$ and its largest entry $\max_{i,j}|a_{i,j}|$ ?

  • 2
    $\begingroup$ $\lambda_1 \le n \max\limits_{i,j}|a_{ij}|$. The coefficient $n$ is optimal, it can be achieved by matrix with all entries $1$. In fact, as a corollary of Gershgorin circle theorem, we have a stronger result $\lambda_1 \le \max\limits_i \left(a_{ii} +\sum\limits_{j\ne i} |a_{ij}|\right)$ $\endgroup$ – achille hui Nov 8 '16 at 16:30
  • $\begingroup$ @achillehui thanks ! and do you know any lower bound result ? $\endgroup$ – anonymus Nov 9 '16 at 13:49
  • $\begingroup$ An answer to myself, if $A$ is $n\times n$ symmetric, then, the maximum eigenvalue is bounded above by $\frac{1}{n}\sum_{i,j} a_{i,j}$. $\endgroup$ – anonymus Nov 9 '16 at 14:03

Your Answer

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

Browse other questions tagged or ask your own question.