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I want to show that the numerical radius of the complex valued matrix $$ A := \begin{pmatrix} x & 0 \\ y & 0 \end{pmatrix} $$ is strictly larger than $|x|$, where $y \ne 0$ and $|x| = 1$. Brute force substitution $w= (w_1,w_2)$ with $|w|=1$ into $wAw^*$ seemed to yield nothing.

The numerical radius of $A \in \mathbb{C}^{n \times n}$ is defined as $$ W(A) :=\sup_{ww^* =1 } | wAw^* |. $$

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  • $\begingroup$ Why didn't you accept the answer? $\endgroup$ – Viktor Glombik Oct 8 '19 at 16:45
  • $\begingroup$ This answer shows how this problem can be solved without the theorem mentioned in the existing answer. $\endgroup$ – Viktor Glombik Oct 8 '19 at 19:36
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The numerical radius of a $2\times 2$ matrix $A$ is the closed elliptical disc with foci in the eigenvalues $\lambda_1, \lambda_2$ of $A$ and semi-major axis $$ a =\frac{1}{2}\sqrt{\text{tr}\big(A^* A\big)-2\Re\left(\lambda_1 \overline{\lambda}_2\right)}. $$ See, for instance, the book Numerical Range by K. E. Gustafson and D. K. M. Rao, Springer-Verlag, New York, 1997 or here.

For the given matrix $A$, we have $\lambda_1=0$, $\lambda_2=x$ and $a=\frac{1}{2}\sqrt{|x|^2+|y|^2}$. Since $y\ne 0$ the numerical range is a non-degenerated elliptical disc (if it were degenerated, then it would be the line segment $[0,x]$ with $a=\frac{1}{2}|x|$ and semi-minor axis $b=0$). It follows that $x$ is an interior point of the numerical range. Hence, there are numbers in the numerical range of $A$ whose absolute value is strictly larger than $|x|$.

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  • $\begingroup$ Do the eigenvalues have to be different for this theorem to apply? I tried to use it to calculate the numerical range of $A := \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and ended up with foci (= eigenvalues ) $\lambda_{1,2} = 0$ and minor axis $= 1$. But an ellipse with both foci equal to zero seems to be a point, contradiction the minor axis of length 1. $\endgroup$ – Viktor Glombik Oct 13 '19 at 21:48
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    $\begingroup$ @Viktor Glombik An ellipse with both foci at $0$ and the semi minor axis $\frac{1}{2}$ is the circle with center at $0$ and radius $\frac{1}{2}$. In the above result elliptical disc can be degenerated to a circular disc if matrix has only one eigenvalue, or to a line segment if the matrix is normal. In the case of scalar multiple of the identity matrix the numerical range is only one point. $\endgroup$ – Janko Bracic Oct 21 '19 at 4:05

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