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$T$ is hermitian operator from space $V$ to $V$, show that $S$ operator that uphold $S^3=T$ and $S=S^*$ exist and unique.

From what I'm understanding, $S$ operator must be unitary and hermitian operator, but I don't know how to prove that his existence and unique.

Thanks for help!

1 Answers 1

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No, $S$ is not unitary. For existence, use the continuous functional calculus or the Spectral Theorem.

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    I still don't understand how to do it. Cab you show me the solution with the spectral theorem? – 2017-02-10
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    It depends on which version of the spectral theorem you use. If you use the multiplication operator version, where $T$ is unitarily equivalent to multiplication by $f$, then $S$ corresponds (with the same unitary operator) to multiplication by $f^{1/3}$. If you use the projection-valued measure version where $T = \int \lambda \; dE_\lambda$, then $S = \int \lambda^{1/3}\; dE_\lambda$. – 2017-02-10
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    I appreciate your help, but I think this solution is not for my level. This is my first course in linear algebra and we never used Integrals or things that close to that. – 2017-02-10
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    OK, I thought this was on a Hilbert space. Just do a diagonalization by unitary operator: $T = U D U^*$, and $S = U D^{1/3} U^*$ where $D^{1/3}$ has the cube roots of the diagonal elements of the diagonal matrix $D$. – 2017-02-12