There exists a linear operator with no proper invariant subspaces
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Let $A$ be a bounded operator on a Hilbert space $H$ with two invariant subspaces $M$ and $N$ s.t. $N subset M$, dim$(M cap N^{perp})> 1$, and have no invariant subspaces between $N$ and $M$. Then, show that, there exists an operator $B$ on $H$ which has no proper invariant subspace.
All I want a hint for constructing $B$ with the help of $A$ and given conditions, even a little hint will be appreciated. Thanks in advance.
linear-algebra functional-analysis operator-algebras
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$begingroup$
Let $A$ be a bounded operator on a Hilbert space $H$ with two invariant subspaces $M$ and $N$ s.t. $N subset M$, dim$(M cap N^{perp})> 1$, and have no invariant subspaces between $N$ and $M$. Then, show that, there exists an operator $B$ on $H$ which has no proper invariant subspace.
All I want a hint for constructing $B$ with the help of $A$ and given conditions, even a little hint will be appreciated. Thanks in advance.
linear-algebra functional-analysis operator-algebras
$endgroup$
add a comment |
$begingroup$
Let $A$ be a bounded operator on a Hilbert space $H$ with two invariant subspaces $M$ and $N$ s.t. $N subset M$, dim$(M cap N^{perp})> 1$, and have no invariant subspaces between $N$ and $M$. Then, show that, there exists an operator $B$ on $H$ which has no proper invariant subspace.
All I want a hint for constructing $B$ with the help of $A$ and given conditions, even a little hint will be appreciated. Thanks in advance.
linear-algebra functional-analysis operator-algebras
$endgroup$
Let $A$ be a bounded operator on a Hilbert space $H$ with two invariant subspaces $M$ and $N$ s.t. $N subset M$, dim$(M cap N^{perp})> 1$, and have no invariant subspaces between $N$ and $M$. Then, show that, there exists an operator $B$ on $H$ which has no proper invariant subspace.
All I want a hint for constructing $B$ with the help of $A$ and given conditions, even a little hint will be appreciated. Thanks in advance.
linear-algebra functional-analysis operator-algebras
linear-algebra functional-analysis operator-algebras
asked Jan 8 at 11:38
splifesplife
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The hint is to think about the invariant subspaces of $A$ the lie in $Mcap N^perp$.
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1 Answer
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$begingroup$
The hint is to think about the invariant subspaces of $A$ the lie in $Mcap N^perp$.
$endgroup$
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$begingroup$
The hint is to think about the invariant subspaces of $A$ the lie in $Mcap N^perp$.
$endgroup$
add a comment |
$begingroup$
The hint is to think about the invariant subspaces of $A$ the lie in $Mcap N^perp$.
$endgroup$
The hint is to think about the invariant subspaces of $A$ the lie in $Mcap N^perp$.
answered Jan 8 at 13:53
Martin ArgeramiMartin Argerami
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