composition of finite rank projection and bounded operator












1














If $Pin B(H)$ is a finite rank projection,we assume the rank is $n$,I know the fact $PB(H)Pcong M_n(mathbb{C})$,but how to construct the isomorphism?






operator-theory operator-algebras c-star-algebras von-neumann-algebras







share|cite|improve this question






















  • I think this is pretty much the same as the result that is usually taken for granted, that there is a 1:1 corresponsance between the set of linear transformation on a finite dimensional vector space and their matrix 'representations' (set of matrices of corresponding size).
    – AnyAD
    Nov 27 '18 at 22:33
















1














If $Pin B(H)$ is a finite rank projection,we assume the rank is $n$,I know the fact $PB(H)Pcong M_n(mathbb{C})$,but how to construct the isomorphism?






operator-theory operator-algebras c-star-algebras von-neumann-algebras







share|cite|improve this question






















  • I think this is pretty much the same as the result that is usually taken for granted, that there is a 1:1 corresponsance between the set of linear transformation on a finite dimensional vector space and their matrix 'representations' (set of matrices of corresponding size).
    – AnyAD
    Nov 27 '18 at 22:33














1












1








1







If $Pin B(H)$ is a finite rank projection,we assume the rank is $n$,I know the fact $PB(H)Pcong M_n(mathbb{C})$,but how to construct the isomorphism?






operator-theory operator-algebras c-star-algebras von-neumann-algebras







share|cite|improve this question













If $Pin B(H)$ is a finite rank projection,we assume the rank is $n$,I know the fact $PB(H)Pcong M_n(mathbb{C})$,but how to construct the isomorphism?






operator-theory operator-algebras c-star-algebras von-neumann-algebras




operator-theory operator-algebras c-star-algebras von-neumann-algebras






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 27 '18 at 22:19









mathrookie

811512




811512












  • I think this is pretty much the same as the result that is usually taken for granted, that there is a 1:1 corresponsance between the set of linear transformation on a finite dimensional vector space and their matrix 'representations' (set of matrices of corresponding size).
    – AnyAD
    Nov 27 '18 at 22:33


















  • I think this is pretty much the same as the result that is usually taken for granted, that there is a 1:1 corresponsance between the set of linear transformation on a finite dimensional vector space and their matrix 'representations' (set of matrices of corresponding size).
    – AnyAD
    Nov 27 '18 at 22:33
















I think this is pretty much the same as the result that is usually taken for granted, that there is a 1:1 corresponsance between the set of linear transformation on a finite dimensional vector space and their matrix 'representations' (set of matrices of corresponding size).
– AnyAD
Nov 27 '18 at 22:33




I think this is pretty much the same as the result that is usually taken for granted, that there is a 1:1 corresponsance between the set of linear transformation on a finite dimensional vector space and their matrix 'representations' (set of matrices of corresponding size).
– AnyAD
Nov 27 '18 at 22:33










1 Answer
1






active

oldest

votes


















1














Let $x_1,ldots,x_n$ be an orthonormal basis of $PH$. Write $P=sum_{j=1}^n P_{jj}$, where $P_j$ is the orthogonal projection onto $mathbb C x_j$, i.e., $P_{jj}x=langle x,x_jrangle,x_j$. Let $P_{kj}$ be the rank-one operator that takes $x_j$ to $x_k$, i.e. $P_{kj}x=langle x,x_jrangle,x_k$.



Given any $Tin B(H)$, the operator $P_kTP_j$ is rank-one, and maps $x_j$ to $x_k$. We have $$
P_kTP_jx=langle TP_jx,x_krangle,x_k=langle Tx_j,x_krangle ,langle x,x_jrangle,x_k=langle Tx_j,x_krangle,P_{kj}x.
$$
So
$$tag1
PTP=sum_{k,j}P_kTP_j=sum_{k,j}langle Tx_j,x_krangle,P_{kj}.
$$
We also have
$$tag2
P_{kj}P_{ab}x=langle x,x_brangle P_{kj}x_a=langle x,x_brangle,langle x_a,x_jrangle,x_k=delta_{a,j},langle x,x_brangle, x_k=delta_{a,j},P_{kb}x.
$$
Given $Ain M_n(mathbb C)$, we write $A=sum_{k,j} A_{kj},E_{kj}$, where ${E_{kj}}$ are the canonical matrix units. Define $Gamma:M_n(mathbb C)to PB(H)P$ by
$$
Gamma(A)=sum_{k,j} A_{kj},P_{kj}.
$$
It is easy to see that this is one-to-one, $*$-preserving, and linear. It is onto by $(1)$. And, using $(2)$,
begin{align}
Gamma(AB)&=sum_{k,j} (AB)_{kj},P_{kj}=sum_{k,j}sum_{ell} A_{kell}B_{ell j},P_{kj}=sum_{k,j}sum_{ell,m} A_{kell}B_{m j},P_{kell}P_{m j}\ \
&=left(sum_{k,ell} A_{kell} ,P_{kell}right)
left(sum_{j,m} B_{mj} ,P_{mj}right)
=Gamma(A)Gamma(B).
end{align}






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3016383%2fcomposition-of-finite-rank-projection-and-bounded-operator%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Let $x_1,ldots,x_n$ be an orthonormal basis of $PH$. Write $P=sum_{j=1}^n P_{jj}$, where $P_j$ is the orthogonal projection onto $mathbb C x_j$, i.e., $P_{jj}x=langle x,x_jrangle,x_j$. Let $P_{kj}$ be the rank-one operator that takes $x_j$ to $x_k$, i.e. $P_{kj}x=langle x,x_jrangle,x_k$.



    Given any $Tin B(H)$, the operator $P_kTP_j$ is rank-one, and maps $x_j$ to $x_k$. We have $$
    P_kTP_jx=langle TP_jx,x_krangle,x_k=langle Tx_j,x_krangle ,langle x,x_jrangle,x_k=langle Tx_j,x_krangle,P_{kj}x.
    $$
    So
    $$tag1
    PTP=sum_{k,j}P_kTP_j=sum_{k,j}langle Tx_j,x_krangle,P_{kj}.
    $$
    We also have
    $$tag2
    P_{kj}P_{ab}x=langle x,x_brangle P_{kj}x_a=langle x,x_brangle,langle x_a,x_jrangle,x_k=delta_{a,j},langle x,x_brangle, x_k=delta_{a,j},P_{kb}x.
    $$
    Given $Ain M_n(mathbb C)$, we write $A=sum_{k,j} A_{kj},E_{kj}$, where ${E_{kj}}$ are the canonical matrix units. Define $Gamma:M_n(mathbb C)to PB(H)P$ by
    $$
    Gamma(A)=sum_{k,j} A_{kj},P_{kj}.
    $$
    It is easy to see that this is one-to-one, $*$-preserving, and linear. It is onto by $(1)$. And, using $(2)$,
    begin{align}
    Gamma(AB)&=sum_{k,j} (AB)_{kj},P_{kj}=sum_{k,j}sum_{ell} A_{kell}B_{ell j},P_{kj}=sum_{k,j}sum_{ell,m} A_{kell}B_{m j},P_{kell}P_{m j}\ \
    &=left(sum_{k,ell} A_{kell} ,P_{kell}right)
    left(sum_{j,m} B_{mj} ,P_{mj}right)
    =Gamma(A)Gamma(B).
    end{align}






    share|cite|improve this answer


























      1














      Let $x_1,ldots,x_n$ be an orthonormal basis of $PH$. Write $P=sum_{j=1}^n P_{jj}$, where $P_j$ is the orthogonal projection onto $mathbb C x_j$, i.e., $P_{jj}x=langle x,x_jrangle,x_j$. Let $P_{kj}$ be the rank-one operator that takes $x_j$ to $x_k$, i.e. $P_{kj}x=langle x,x_jrangle,x_k$.



      Given any $Tin B(H)$, the operator $P_kTP_j$ is rank-one, and maps $x_j$ to $x_k$. We have $$
      P_kTP_jx=langle TP_jx,x_krangle,x_k=langle Tx_j,x_krangle ,langle x,x_jrangle,x_k=langle Tx_j,x_krangle,P_{kj}x.
      $$
      So
      $$tag1
      PTP=sum_{k,j}P_kTP_j=sum_{k,j}langle Tx_j,x_krangle,P_{kj}.
      $$
      We also have
      $$tag2
      P_{kj}P_{ab}x=langle x,x_brangle P_{kj}x_a=langle x,x_brangle,langle x_a,x_jrangle,x_k=delta_{a,j},langle x,x_brangle, x_k=delta_{a,j},P_{kb}x.
      $$
      Given $Ain M_n(mathbb C)$, we write $A=sum_{k,j} A_{kj},E_{kj}$, where ${E_{kj}}$ are the canonical matrix units. Define $Gamma:M_n(mathbb C)to PB(H)P$ by
      $$
      Gamma(A)=sum_{k,j} A_{kj},P_{kj}.
      $$
      It is easy to see that this is one-to-one, $*$-preserving, and linear. It is onto by $(1)$. And, using $(2)$,
      begin{align}
      Gamma(AB)&=sum_{k,j} (AB)_{kj},P_{kj}=sum_{k,j}sum_{ell} A_{kell}B_{ell j},P_{kj}=sum_{k,j}sum_{ell,m} A_{kell}B_{m j},P_{kell}P_{m j}\ \
      &=left(sum_{k,ell} A_{kell} ,P_{kell}right)
      left(sum_{j,m} B_{mj} ,P_{mj}right)
      =Gamma(A)Gamma(B).
      end{align}






      share|cite|improve this answer
























        1












        1








        1






        Let $x_1,ldots,x_n$ be an orthonormal basis of $PH$. Write $P=sum_{j=1}^n P_{jj}$, where $P_j$ is the orthogonal projection onto $mathbb C x_j$, i.e., $P_{jj}x=langle x,x_jrangle,x_j$. Let $P_{kj}$ be the rank-one operator that takes $x_j$ to $x_k$, i.e. $P_{kj}x=langle x,x_jrangle,x_k$.



        Given any $Tin B(H)$, the operator $P_kTP_j$ is rank-one, and maps $x_j$ to $x_k$. We have $$
        P_kTP_jx=langle TP_jx,x_krangle,x_k=langle Tx_j,x_krangle ,langle x,x_jrangle,x_k=langle Tx_j,x_krangle,P_{kj}x.
        $$
        So
        $$tag1
        PTP=sum_{k,j}P_kTP_j=sum_{k,j}langle Tx_j,x_krangle,P_{kj}.
        $$
        We also have
        $$tag2
        P_{kj}P_{ab}x=langle x,x_brangle P_{kj}x_a=langle x,x_brangle,langle x_a,x_jrangle,x_k=delta_{a,j},langle x,x_brangle, x_k=delta_{a,j},P_{kb}x.
        $$
        Given $Ain M_n(mathbb C)$, we write $A=sum_{k,j} A_{kj},E_{kj}$, where ${E_{kj}}$ are the canonical matrix units. Define $Gamma:M_n(mathbb C)to PB(H)P$ by
        $$
        Gamma(A)=sum_{k,j} A_{kj},P_{kj}.
        $$
        It is easy to see that this is one-to-one, $*$-preserving, and linear. It is onto by $(1)$. And, using $(2)$,
        begin{align}
        Gamma(AB)&=sum_{k,j} (AB)_{kj},P_{kj}=sum_{k,j}sum_{ell} A_{kell}B_{ell j},P_{kj}=sum_{k,j}sum_{ell,m} A_{kell}B_{m j},P_{kell}P_{m j}\ \
        &=left(sum_{k,ell} A_{kell} ,P_{kell}right)
        left(sum_{j,m} B_{mj} ,P_{mj}right)
        =Gamma(A)Gamma(B).
        end{align}






        share|cite|improve this answer












        Let $x_1,ldots,x_n$ be an orthonormal basis of $PH$. Write $P=sum_{j=1}^n P_{jj}$, where $P_j$ is the orthogonal projection onto $mathbb C x_j$, i.e., $P_{jj}x=langle x,x_jrangle,x_j$. Let $P_{kj}$ be the rank-one operator that takes $x_j$ to $x_k$, i.e. $P_{kj}x=langle x,x_jrangle,x_k$.



        Given any $Tin B(H)$, the operator $P_kTP_j$ is rank-one, and maps $x_j$ to $x_k$. We have $$
        P_kTP_jx=langle TP_jx,x_krangle,x_k=langle Tx_j,x_krangle ,langle x,x_jrangle,x_k=langle Tx_j,x_krangle,P_{kj}x.
        $$
        So
        $$tag1
        PTP=sum_{k,j}P_kTP_j=sum_{k,j}langle Tx_j,x_krangle,P_{kj}.
        $$
        We also have
        $$tag2
        P_{kj}P_{ab}x=langle x,x_brangle P_{kj}x_a=langle x,x_brangle,langle x_a,x_jrangle,x_k=delta_{a,j},langle x,x_brangle, x_k=delta_{a,j},P_{kb}x.
        $$
        Given $Ain M_n(mathbb C)$, we write $A=sum_{k,j} A_{kj},E_{kj}$, where ${E_{kj}}$ are the canonical matrix units. Define $Gamma:M_n(mathbb C)to PB(H)P$ by
        $$
        Gamma(A)=sum_{k,j} A_{kj},P_{kj}.
        $$
        It is easy to see that this is one-to-one, $*$-preserving, and linear. It is onto by $(1)$. And, using $(2)$,
        begin{align}
        Gamma(AB)&=sum_{k,j} (AB)_{kj},P_{kj}=sum_{k,j}sum_{ell} A_{kell}B_{ell j},P_{kj}=sum_{k,j}sum_{ell,m} A_{kell}B_{m j},P_{kell}P_{m j}\ \
        &=left(sum_{k,ell} A_{kell} ,P_{kell}right)
        left(sum_{j,m} B_{mj} ,P_{mj}right)
        =Gamma(A)Gamma(B).
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 1:08









        Martin Argerami

        124k1176174




        124k1176174






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3016383%2fcomposition-of-finite-rank-projection-and-bounded-operator%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Index of /

            This page is only for reference, If you need detailed information, please check here
            Read more

            Tribalistas

            Image
              Nota: Para o álbum homônimo, veja Tribalistas (álbum de 2002). Para o segundo álbum homônimo, veja Tribalistas (álbum de 2017). Para o tipo de organização tribal, veja Tribalismo. Tribalistas Informação geral Origem Salvador e Rio de Janeiro País   Brasil Gênero(s) MPB Pop Samba Período em atividade 2002–2004 2017–atualmente Gravadora(s) EMI Phonomotor Records Universal Music Afiliação(ões) Dadi Pretinho da Serrinha Margareth Menezes Integrantes Arnaldo Antunes Carlinhos Brown Marisa Monte Página oficial www2.uol.com.br/tribalistas/ ...
            Read more

            Listed building

            Image
            Listed building (literalmente: prédio listado ) é toda estrutura protegida por lei pelo complexo sistema de tombamento do Reino Unido. Um listed building é qualquer estrutura oficialmente designada como tendo especial interesse arquitetônico, histórico ou cultural. É um estatuto amplamente usado, aplicado a centenas de milhares de estruturas. Estruturas tombadas não podem ser alteradas sem permissão especial, e desfrutam de isenções fiscais e outros benefícios. Cada um dos países do Reino Unido trata tombamento de forma diferente: Índice 1 Inglaterra 2 País de Gales 3 Irlanda do Norte 4 Escócia 5 Referências Inglaterra | O Palácio de Buckingham, Londres. As estruturas tombadas são listadas no Statutory List of Buildings of Special Architectural or Historic Interest , e incluem campos, sinais de trânsito, piers e áreas geográficas. A principal legislação pertinente é o Ancient Monuments and Archaeological Areas Act 1979 , [ 1 ] embora legisla...
            Read more