Centralizer of projections
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Let $H$ be a Hilbert space and $p, q$ self-adjoint projectors in $B(H)$,
i.e. $$p^2=p=p^* space text{ and } space q^2=q=q^*.$$ Suppose they have the same centralizers $C(p)=C(q)$.
Is it true that $p=pm q$?
Here $C(x)={yin B(H) : yx=xy}$.
Maybe I am asking a well known result but I couldn't find anything about this problem in the literature.
functional-analysis hilbert-spaces operator-algebras compact-operators projection
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up vote
1
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Let $H$ be a Hilbert space and $p, q$ self-adjoint projectors in $B(H)$,
i.e. $$p^2=p=p^* space text{ and } space q^2=q=q^*.$$ Suppose they have the same centralizers $C(p)=C(q)$.
Is it true that $p=pm q$?
Here $C(x)={yin B(H) : yx=xy}$.
Maybe I am asking a well known result but I couldn't find anything about this problem in the literature.
functional-analysis hilbert-spaces operator-algebras compact-operators projection
2
I think you want to say "Is it true that $p = 1 - q$ or $p = q$". Minus a projection is not a projection.
– Adrián González-Pérez
Nov 20 at 12:45
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $H$ be a Hilbert space and $p, q$ self-adjoint projectors in $B(H)$,
i.e. $$p^2=p=p^* space text{ and } space q^2=q=q^*.$$ Suppose they have the same centralizers $C(p)=C(q)$.
Is it true that $p=pm q$?
Here $C(x)={yin B(H) : yx=xy}$.
Maybe I am asking a well known result but I couldn't find anything about this problem in the literature.
functional-analysis hilbert-spaces operator-algebras compact-operators projection
Let $H$ be a Hilbert space and $p, q$ self-adjoint projectors in $B(H)$,
i.e. $$p^2=p=p^* space text{ and } space q^2=q=q^*.$$ Suppose they have the same centralizers $C(p)=C(q)$.
Is it true that $p=pm q$?
Here $C(x)={yin B(H) : yx=xy}$.
Maybe I am asking a well known result but I couldn't find anything about this problem in the literature.
functional-analysis hilbert-spaces operator-algebras compact-operators projection
functional-analysis hilbert-spaces operator-algebras compact-operators projection
edited Nov 18 at 8:05
Gaby Boy Analysis
628314
628314
asked Nov 18 at 6:00
golomorfMath
1427
1427
2
I think you want to say "Is it true that $p = 1 - q$ or $p = q$". Minus a projection is not a projection.
– Adrián González-Pérez
Nov 20 at 12:45
add a comment |
2
I think you want to say "Is it true that $p = 1 - q$ or $p = q$". Minus a projection is not a projection.
– Adrián González-Pérez
Nov 20 at 12:45
2
2
I think you want to say "Is it true that $p = 1 - q$ or $p = q$". Minus a projection is not a projection.
– Adrián González-Pérez
Nov 20 at 12:45
I think you want to say "Is it true that $p = 1 - q$ or $p = q$". Minus a projection is not a projection.
– Adrián González-Pérez
Nov 20 at 12:45
add a comment |
1 Answer
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0
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Look in Vaughan Jones notes, Exerise 2.1.13. $A$ and $A^ast$ preserve a subspace $K$, ie $A K subset K$ and $A^ast K subset K$ iff $[A,P_K] = 0$, where $P_K$ is the orthogonal projection onto $K$.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Look in Vaughan Jones notes, Exerise 2.1.13. $A$ and $A^ast$ preserve a subspace $K$, ie $A K subset K$ and $A^ast K subset K$ iff $[A,P_K] = 0$, where $P_K$ is the orthogonal projection onto $K$.
add a comment |
up vote
0
down vote
Look in Vaughan Jones notes, Exerise 2.1.13. $A$ and $A^ast$ preserve a subspace $K$, ie $A K subset K$ and $A^ast K subset K$ iff $[A,P_K] = 0$, where $P_K$ is the orthogonal projection onto $K$.
add a comment |
up vote
0
down vote
up vote
0
down vote
Look in Vaughan Jones notes, Exerise 2.1.13. $A$ and $A^ast$ preserve a subspace $K$, ie $A K subset K$ and $A^ast K subset K$ iff $[A,P_K] = 0$, where $P_K$ is the orthogonal projection onto $K$.
Look in Vaughan Jones notes, Exerise 2.1.13. $A$ and $A^ast$ preserve a subspace $K$, ie $A K subset K$ and $A^ast K subset K$ iff $[A,P_K] = 0$, where $P_K$ is the orthogonal projection onto $K$.
answered Nov 20 at 12:46
Adrián González-Pérez
856138
856138
add a comment |
add a comment |
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I think you want to say "Is it true that $p = 1 - q$ or $p = q$". Minus a projection is not a projection.
– Adrián González-Pérez
Nov 20 at 12:45