Consider the matrix $$D=\begin{pmatrix}1&0\\0&e^{i\theta}\end{pmatrix}.$$ For the commonly used norms $\|\cdot\|$ on $\mathbb{C}^2$ or for $\theta=0$ the associated subordinate norm is $1$. Is it always true ? can a subordinate norm be strictly bigger than one ?
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2$\begingroup$ If $\Vert\cdot\Vert_2$ denotes the usual norm on ${\bf C}^2$ and we pick an invertible $A\in {\rm GL}_2({\mathbb C})$ then we can define a new norm by $\Vert x\Vert_A = \Vert Ax\Vert_2$. I suspect, but have not checked, that "most" choices of $A$ will lead to examples where there exist $y\in {\bf C}^2$ such that $\Vert Dy\Vert_A > \Vert y\Vert$. $\endgroup$– Yemon ChoiCommented Feb 13, 2023 at 14:50
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3$\begingroup$ I just learned (from Wikipedia), that the subordinate norm is another word for operator norm. Which also explains the answer below. $\endgroup$– Willie WongCommented Feb 13, 2023 at 15:46
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1 Answer
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Consider the norm $\Vert\cdot\Vert$ on $\mathbb{C}^2$ defined by $\Vert z \Vert^2 := |z_1|^2+|z_1+z_2|^2$. Then $$\left\Vert \left(\begin{array}{c}1 \\ -1 \end{array}\right) \right\Vert^2 = 1,$$ $$\left\Vert D\left(\begin{array}{c}1 \\ -1 \end{array}\right) \right\Vert^2 = \left\Vert \left(\begin{array}{c}1 \\ -e^{i\theta} \end{array}\right) \right\Vert^2 = 1 + |1-e^{i\theta}|^2.$$ If $e^{i\theta} \ne 1$, we derive that the subordinate norm of $D$ is $>1$.
answered Feb 13, 2023 at 14:59