Numerical radius of $2 \times 2$ matrix

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 $$W(A) :=\sup_{ww^* =1 } | wAw^* |.$$

• Why didn't you accept the answer? – Viktor Glombik Oct 8 '19 at 16:45
• This answer shows how this problem can be solved without the theorem mentioned in the existing answer. – Viktor Glombik Oct 8 '19 at 19:36

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 $$|x|$$.
• 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. – Viktor Glombik Oct 13 '19 at 21:48
• @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. – Janko Bracic Oct 21 '19 at 4:05