1
$\begingroup$

Let $A \in \mathbb{R}^{n \times n}$ be a symmetric positive definite matrix, and let $B \in \mathbb{R}^{n \times n}$ be an arbitrary matrix.

Define the numerical range or field of values of $B$ as \begin{align} W(B) = \left\{\frac{(Bv,v)}{(v,v)}, 0 \ne v \in \mathbb{C}^n \right\} \end{align} where $(\cdot,\cdot)$ is the euclidean inner product.

Would the numerical range \begin{align} W_A(B) = \left\{\frac{(ABv,v)}{(Av,v)}, 0 \ne v \in \mathbb{C}^n \right\} \end{align} have all the same properties than $W(B)$ since the norms are equivalent?

$\endgroup$
  • 2
    $\begingroup$ "all the same properties" what are you referring to ? $\endgroup$ – Surb Sep 1 '17 at 13:52
  • 1
    $\begingroup$ @Surb First, thanks for your answer! I am thinking of properties like convexity, closedness, infinite curvature in the boundary if eigenvalues are in it, numerical range of a sum included in the sum of numerical ranges. Let $0 \notin W_A(B)$, the spectrum of $B^{-1}C$ or of $CB^{-1}$ is in $\frac{W_A(B)}{W_A(C)}$. Let $B$ be hermitian positive definite, then the spectrum of $BC$ is in $W_A(C) W_A(B)$... $\endgroup$ – Astor Sep 1 '17 at 13:57
3
$\begingroup$

If $A=R^*R$ is the Cholesky factorization of $A$, then $$\frac{(ABv,v)}{(Av,v)} = \frac{(R^*RBv,v)}{(R^*Rv,v)} = \frac{(RBv,Rv)}{(Rv,Rv)} = \frac{(RBR^{-1}w,w)}{(w,w)}$$ for $w=Rv$, hence $W_A(B) = W(RBR^{-1})$ and has "all the properties" of a numerical range.

$\endgroup$
  • $\begingroup$ Simple and frankly beautiful answer. Sorry if the question was a bit stupid. $\endgroup$ – Astor Sep 1 '17 at 14:06

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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