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I know result is true for complex inner product space because we can diagonalize Tand S. but in real inner product space T and S cannot be diagonalized.then is this result true ?how I proceed I don not know. Give met any hint

Riya
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This is true for finite dimensional inner product spaces. On such spaces, we note that for normal operators, $ST = TS \implies ST^* = T^*S$, as proven here.

I'm not sure whether this is the case for arbitrary Hilbert spaces.

Community
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Ben Grossmann
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  • To prove this I need TS^=ST^ – Riya Nov 05 '14 at 15:19
  • If that is the case, you could use this result. Note that while the statement uses diagonalizations over the complex numbers, it holds for any normal operators on a real (finite dimensional) inner product space. – Ben Grossmann Nov 05 '14 at 15:21
  • this case is not given can you suggest me that if ST=TS then IS ST^=TS^ – Riya Nov 05 '14 at 15:23
  • "This case is not given": do you mean that the space might be infinite-dimensional? – Ben Grossmann Nov 05 '14 at 15:26
  • But normal operator on real inner product space may not have any non zero characterstic vectors – Riya Nov 05 '14 at 15:28
  • space is finite dimensional – Riya Nov 05 '14 at 15:28
  • If $ST$ is normal as an operator over a complex inner-product space, then it is certainly normal as an operator over the restricted real inner-product space. If you want an approach that doesn't use the complex extension to the real space, then that is another matter. – Ben Grossmann Nov 05 '14 at 15:31