On normal operators and its adjoint and the Putnam-Fuglede theorem. (Chinese. English summary) Zbl 0614.47020
For linear bounded operators on a Hilbert space we prove the following theorems.
Theorem 1. Let M and N be normal operators, and \(\sigma\) (M)\(\cap \sigma (N)\subset {\mathbb{R}}\), then for any operator A, \(AM=NA\) implies \(AM=N^*A\) and \(AM^*=NA.\)
Theorem 2. Let M and N be normal operators and \(\sigma\) (M)\(\cap (1/(\sigma (N)\setminus \{0\})\subset {\mathbb{R}}\), if there is an operator A such that \(MAN=A\), then \(MAN^*=A\) and \(M^*AN=A\), where \(1/(\sigma (N)\setminus \{0\})=\{1/\lambda:\lambda \in \sigma (N)\setminus \{0\}\}.\)
Theorem 3. Let \(\{M_ 1,M_ 2\}\) and \(\{N_ 1,N_ 2\}\) be commuative pairs of normal operators, and \((\sigma (M_ 2)/(\sigma (M_ 1)\setminus \{0\}))\cup (\sigma (N_ 1)/(\sigma (N_ 2)\setminus \{0\})\subset {\mathbb{R}}\), if there is an operator A such that \(M_ 1AN_ 1=M_ 2AN_ 2\), then \(M_ 1AN^*_ 1=M_ 2AN^*_ 2\) and \(M^*_ 1AN_ 1=M^*_ 2AN_ 2\).
Theorem 1. Let M and N be normal operators, and \(\sigma\) (M)\(\cap \sigma (N)\subset {\mathbb{R}}\), then for any operator A, \(AM=NA\) implies \(AM=N^*A\) and \(AM^*=NA.\)
Theorem 2. Let M and N be normal operators and \(\sigma\) (M)\(\cap (1/(\sigma (N)\setminus \{0\})\subset {\mathbb{R}}\), if there is an operator A such that \(MAN=A\), then \(MAN^*=A\) and \(M^*AN=A\), where \(1/(\sigma (N)\setminus \{0\})=\{1/\lambda:\lambda \in \sigma (N)\setminus \{0\}\}.\)
Theorem 3. Let \(\{M_ 1,M_ 2\}\) and \(\{N_ 1,N_ 2\}\) be commuative pairs of normal operators, and \((\sigma (M_ 2)/(\sigma (M_ 1)\setminus \{0\}))\cup (\sigma (N_ 1)/(\sigma (N_ 2)\setminus \{0\})\subset {\mathbb{R}}\), if there is an operator A such that \(M_ 1AN_ 1=M_ 2AN_ 2\), then \(M_ 1AN^*_ 1=M_ 2AN^*_ 2\) and \(M^*_ 1AN_ 1=M^*_ 2AN_ 2\).
MSC:
| 47B15 | Hermitian and normal operators (spectral measures, functional calculus, etc.) |
| 47B20 | Subnormal operators, hyponormal operators, etc. |
| 47A10 | Spectrum, resolvent |
