While trying to find the numerical range ($W(T) := \{\langle Tx, x \rangle: \| x \| = 1\}$) of the matrix $T := \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, I encountered this problem:

Find all $x = (x_1, x_2) \in \mathbb{C}^2$ that lie in $\{ \overline{x_1} x_2: \| x \| = 1\}$, where $\overline{\ \cdot \ }$ denotes complex conjugation.

I found here that $\overline{W(T)} = \frac{1}{2} \mathbb{D}$, where $\mathbb{D}$ probably denotes the unit disk.

Any hints are welcome.

My ideas

  • T is nilpontent with impotence index 2, but I haven't found a connection between nilpontence and numerical range. Denoting by $r(T) := \sup_{\lambda \in W(T)} | \lambda |$ the numerical radius of $T$, I know that $r(T^n) \le r(T)^n$, maybe something similar holds for the numerical range.
  • I tried writing $x_1 := e^{i \alpha} \sin\left(\frac{\phi}{2}\right)$ and $x_2 := x_1 := e^{i \beta} \cos\left(\frac{\phi}{2}\right)$ with $\phi \in [0, \phi]$. Then $\overline{x_1} x_2 = \frac{\sin(\phi)}{2} e^{i(\beta - \alpha)}$ but I don't know how to continue from there.
  • We also have $W(A + B) \subset W(A) + W(B)$, so a nice decomposition of $T$ might be helpful.
  • By the Toeplitz-Hausdorff theorem, $W(T)$ is convex and compact but I have't been able to use this, too.
  • The first theorem in this paper shows that $W(T)$ is an elliptical disk where the foci are the eigenvalues and the minor axis is $z :=\sqrt{\text{tr}(A^* A) - | \lambda_1 |^2 + | \lambda_2 |^2}$. In this case $\lambda_1 = \lambda_2 = 0$ and therefore $z = 1$. But how would this ellipse look like?
  • Lastly, multiplying out (with $a := \Re(x_1)$, $b := \Im(x_1)$, $c := \Re(x_2)$, $d := \Im(x_2)$ we have $$ W(T) = \{ ac + bd + i (ad - bc): a^2 + b^2 + c^2 + d^2 = 1\} , $$ but I can't deduce anything from that.

Any hints are appreciated.

Update 1 The first remark from here claims that the numerical range of any nilpotent $2 \times 2$ matrix is circular, centred at 0 with radius $\frac{\sqrt{\text{tr}(A^*A)}}{2}$, which shows in our case that $W(T) = \frac{\mathbb{D}}{2}$. Can anybody point me to a proof of this please?

  • 1
    $\begingroup$ If $|x_1|^2+|x_2|^2=1$ then write $x_1= e^{i\varphi_1}\cos(\theta) + e^{i\varphi_2}\sin(\theta)$ to get that $\overline{x_1}x_2 = e^{-i\varphi_1+i\varphi_2}\cos(\theta)\sin(\theta)$, ie the set is the the image of $\cos\cdot\sin$ times the complex unit circle. Note that this image is $[-\frac12,\frac12]$. $\endgroup$ – s.harp Oct 13 at 23:04
  • $\begingroup$ @s.harp Never mind, I get it now. Feel free to post this as an answer. But please include where $\varphi_{1,2}$ and $\theta$ come from and why we can restrict ourselves to those values. $\endgroup$ – Viktor Glombik Oct 14 at 0:00

Well, the fact that $W(T)$ is circular and centered around $0$ is easy enough for your particular $T$. If $|x_1|^2+|x_2|^2=1$, then the same holds for $|x_1|^2+|e^{i\theta}x_2|^2$ for any $\theta\in \mathbb{R}$. However, $\overline{x}_1 e^{i\theta}x_2=e^{i\theta} \overline{x}_1 x_2$, implying that $e^{i\theta}W(T)=W(T)$. Thus, $W(T)$ is circular around $0$.

Now, $|\overline{x}_1x_2|=|x_1|\sqrt{1-|x_1|^2},$ so we can just analyse this to get the radius of your set. So let $f(r)=r\sqrt{1-r^2}$ and note that $f$ is differentiable with $$ f'(r)=\sqrt{1-r^2}-\frac{2r^2}{2\sqrt{1-r^2}}=\frac{1-2r^2}{\sqrt{1-r^2}}, $$ which is $0$ if and only if $r=\frac{1}{\sqrt{2}}$. Since $f$ is positive and maps both $0$ and $1$ to $0$, this must be a maximum, and $f(\frac{1}{\sqrt{2}})=\frac{1}{2}$.

Adding this together, we get that $W(T)=\frac{1}{2} \bar{\mathbb{D}}$.

  • $\begingroup$ But the complexity of the optimisation task should explode as the number variables goes to $\infty$, right? $\endgroup$ – WoolierThanThou Oct 26 at 13:12
  • $\begingroup$ Yes, I'd reckon we would have $f: \ell_p \to \ell_p$ for some $p$, and $x_k \in [0,1]$ for all $k$. I thought I could "find a pattern" and prove it by induction. $\endgroup$ – Viktor Glombik Oct 26 at 13:13
  • 1
    $\begingroup$ A good guess is probably, that the optimal solution on $\mathbb{C}^n$ is $x_j=x_k=1/\sqrt{n}$, yielding a numerical range of $\frac{n-1}{n}$. In this case, it's clear that the numerical range in the infinite dimensional case contains all of the unit disk, but then you'd have to ask if your disk is closed or not. $\endgroup$ – WoolierThanThou Oct 26 at 13:15
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Viktor Glombik Oct 26 at 13:17

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