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.










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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

















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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.










share|cite|improve this question




















  • 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















up vote
1
down vote

favorite
2









up vote
1
down vote

favorite
2






2





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.










share|cite|improve this question















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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edited Nov 18 at 8:05









Gaby Boy Analysis

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asked Nov 18 at 6:00









golomorfMath

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  • 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




    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












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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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    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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      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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        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 $K$.






        share|cite|improve this answer












        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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        share|cite|improve this answer










        answered Nov 20 at 12:46









        Adrián González-Pérez

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