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‎1:let ‎‎$ S‎ ,‎ T‎ ‎\in B‎ (‎ H‎ ) $ ‎‎be positive operators‎‎. ( ‎‎‎‎$ B‎ (‎ H‎ ) ‎‎‎‎$ ‎is ‎bounded ‎operator ‎on ‎‎$‎H‎$‎).‎‎

please ‎help ‎me ‎to ‎prove:

‎$ r ( S + T ) ‎\geq ‎max ‎‎\{‎r( S‎ )‎ ,‎ r‎ (‎ ‎T) ‎\}‎ $‎?‎ ‎

(‎$ r ( S ) ‎‎$ ‎is ‎‎‎Spectral ‎radius ‎‎$ ‎S‎$‎.)‎

2: let $ T \in K (H ) $ be normal. ( $ K( H ) $is compact operator on Hilbert space)

is ‎it ‎right ‎to ‎say:‎

$‎T ‎\geq 0‎$ ‎if ‎only ‎if All its eigenvalues are ‎nonzero‎؟

thanks.

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1) Since $S,T\geq 0$ we have in particular $S,T$ normal and hence $r(S)=\|S\|$ as well as $r(T)=\|T\|$. Further it holds in general in a c*-algebra that for $0\leq x \leq y$ holds $\|x\| \leq \|y\|$, hence $r(S)=\|S\| \leq \|S+T\| = r(S+T)$ and similar for $r(T)$.

2) Do you mean self-adjoint instead of normal. For normal operators this is clearly wrong, of course there do exist operators, which are compact, normal but not self-adjoint and in particular not positive.

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  • $\begingroup$ Both sides are wrong? can you give me an example for "there do exist operators, which are compact, normal but not self-ad joint and in particular not positive." ? $\endgroup$ – M.O Jan 24 '17 at 13:58
  • $\begingroup$ E.g. $i$ is normal, compact but not self adjoint. $0$ is a compact operator with zero eigenvalue and the given example $i$ is compact, has non-zero eigenvalues but is not positive, so both directions are wrong $\endgroup$ – Sebastian Bechtel Jan 24 '17 at 14:48

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