Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a Hermitian matrix, its eigenvalues can be determined from its Rayleigh quotient via the min-max theorem.

Are there generalizations of such relation to a non-Hermitian matrix? Note that the Rayleigh quotient itself does not require the matrix to be Hermitian, but the eigenvalues of a square matrix and of its Hermitian part may not be the same.

Thanks and regards!

share|cite|improve this question
up vote 2 down vote accepted

You are looking at the numerical range of a matrix. For any (complex) linear transformation $T$, one defines the numerical range of a matrix as \begin{align} W(T)=\{x^HTx\mid x^Hx=1\} \end{align} This is a mapping from the unit sphere to the complex plane. In general, $W(T)$ is a subset of the complex plane. If it is a subset of the real line, then $T$ should be a hermitian operator. Some well known results on $W(T)$ are

  • $W(T)$ lies in a disc with radius $||T||$.
  • $W(T)$ contains all eigenvalues of $T$.
  • $W(T)$ is the convex hull of eigenvalues of $T$.
  • $W(T)$ is a closed compact convex set for finite-dimensional $T$.

The last result is the well-known Toeplitz-hausdorff theorem. You can learn more on this beautiful theory by searching for this theorem.

share|cite|improve this answer

Unfortunately, it is not true that $W(T)$ is the convex hull of the eigenvalues of $T$. It is however true that it contains such a convex hull, and in fact these two sets agree if $T$ is normal.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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