I'm reading a document where it is said that if $$A=\begin{pmatrix}0 & 1\\0 & 0 \end{pmatrix}$$ then the norm of the resolvent for $z \neq 0$ is given by $$\|R(z,A)\|= \frac{\sqrt{2}}{\sqrt{1+2|z|^2-\sqrt{1+4|z|^2}}}.$$ I think that if $z\neq 0$ then $$ R(z,A)=(A-zI)^{-1}=\begin{pmatrix}-1/z & -1/z^2\\0 & -1/z \end{pmatrix}$$ and because of that $$\|R(z,A)\|=\frac{\sqrt{2|z|^2+1}}{|z|^2}.$$

Am I wrong?.


You have computed the Frobenius norm of the resolvent, whereas the first formula uses the spectral norm of the resolvent.

  • 1
    $\begingroup$ Thank you. But, how can I find the spectral norm of the resolvent?. $\endgroup$ – Kanmat Apr 8 at 11:35
  • $\begingroup$ I don't know how can I calculate the spectral norm. Can you tell me what I have to do, please? $\endgroup$ – Kanmat Apr 8 at 14:52
  • $\begingroup$ The spectral norm of $A$ coincides with the largest spectral value of $A$, i.e., the largest eigenvalue of $A^\top A$. $\endgroup$ – gerw Apr 8 at 19:50

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