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I am stuck with the following question.

Question: Let $A \in \mathbb{C}^{m \times m}$ be arbitrary. Let $W(A)$ be the numerical range i.e. the set of all Rayleigh quotients of $A$ corresponding to a all nonzero vectors $x \in \mathbb{C}^m$. Show that $W(A)$ contains the convex hull of the eigenvalues of $A$.

Progress: We can write $W(A) = \{ x^TAx : \| x \| = 1 \}$. If $(\lambda_i, q_i)$ is an eigenpair of $A$, we know that the Rayleigh quotient is $r(q_i) = \frac{q_i^TAq_i}{q_i^Tq_i} = \lambda_i$. It is enough to show that every convex combination $\alpha_1\lambda_1 + \alpha_2\lambda_2 \in W(A)$ (i.e. $\alpha_1 + \alpha_2 = 1$).

Because $A \in \mathbb{C}^{m \times m}$ we don't have an orthogonal basis of eigenvectors. Since we can, I thought it would be helpfull to choose $\| q_i \| = 1$. I tried taking $x = \sqrt{\alpha_1}q_1 + \sqrt{\alpha_2}q_2$, then $$ r(x) = \frac{\alpha_1\lambda_1 + \alpha_2\lambda_2 + \sqrt{\alpha_1\alpha_2}(\lambda_1 + \lambda_2)q_1^Tq_2}{\| \text{norm} \|^2}, $$ but then the problem with the $q$'s not being orthogonal arises.

Can somebody help me solve this problem?

Math Student 020
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  • Why is no answer accepted? – ViktorStein Oct 07 '19 at 11:49
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    Your result is known as Toeplitz-Hausdorff Theorem. Two alternative proofs can be done : https://www.ams.org/journals/proc/1970-025-01/S0002-9939-1970-0262849-9/S0002-9939-1970-0262849-9.pdf and https://math.stackexchange.com/questions/2974633/proof-explanation-convexity-of-the-numerical-range-of-an-operator-toeplitz-hau – R. W. Prado Aug 16 '21 at 21:49

1 Answers1

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The numerical range $W(A)$ is convex. This is a non-trivial fact, as far as I know, so maybe you are supposed to use this fact and not prove it.

Martin Argerami
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