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Robert Israel
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$\|C(k)\|$ can be arbitrarily large, since e.g. you can add some large integer multiple of $2\pi i I$ without changing $e^{C(k)}$. If you want to try to avoid this, you might specify that $C(k)$ is the principal branch of the logarithm of $e^{kA+B} e^{-kA}$ (note that although $e^{kA+B} e^{-kA}$ is not hermitian, it has the same eigenvalues as $e^{-kA/2} e^{kA+B} e^{-kA/2}$ which is positive definite). Then the eigenvalues of $C(k)$ will be $O(k)$; unfortunately, since $C(k)$ is not hermitian or normal this does not imply $\|C(k)\| = O(k)$.

$\|C(k)\|$ can be arbitrarily large, since e.g. you can add some large integer multiple of $2\pi i I$ without changing $e^{C(k)}$. If you want to avoid this, you might specify that $C(k)$ is the principal branch of the logarithm of $e^{kA+B} e^{-kA}$ (note that although $e^{kA+B} e^{-kA}$ is not hermitian, it has the same eigenvalues as $e^{-kA/2} e^{kA+B} e^{-kA/2}$ which is positive definite)

$\|C(k)\|$ can be arbitrarily large, since e.g. you can add some large integer multiple of $2\pi i I$ without changing $e^{C(k)}$. If you want to try to avoid this, you might specify that $C(k)$ is the principal branch of the logarithm of $e^{kA+B} e^{-kA}$ (note that although $e^{kA+B} e^{-kA}$ is not hermitian, it has the same eigenvalues as $e^{-kA/2} e^{kA+B} e^{-kA/2}$ which is positive definite). Then the eigenvalues of $C(k)$ will be $O(k)$; unfortunately, since $C(k)$ is not hermitian or normal this does not imply $\|C(k)\| = O(k)$.

Source Link
Robert Israel
  • 54.2k
  • 1
  • 76
  • 152

$\|C(k)\|$ can be arbitrarily large, since e.g. you can add some large integer multiple of $2\pi i I$ without changing $e^{C(k)}$. If you want to avoid this, you might specify that $C(k)$ is the principal branch of the logarithm of $e^{kA+B} e^{-kA}$ (note that although $e^{kA+B} e^{-kA}$ is not hermitian, it has the same eigenvalues as $e^{-kA/2} e^{kA+B} e^{-kA/2}$ which is positive definite)