How to transform an expression into a form involving the trace of a product of two matrices












3












$begingroup$


In page 594 of Bishop's PRML, the following equation is implied:



$$
-frac{1}{2}sum(mathbf{x}_n-mathbf{bar{x}})^Tmathbf{C}^{-1}(mathbf{x}_n-mathbf{bar{x}}) = -frac{N}{2}mathrm{Tr}(mathbf{C}^{-1}mathbf{S})
$$



where



$$
mathbf{S} = frac{1}{N}sum(mathbf{x}_n-mathbf{bar{x}})(mathbf{x}_n-mathbf{bar{x}})^T
$$
,$mathbf{C}$ is a symmetric matrix and $mathbf{bar{x}} = frac{sum_{n=1}^Nmathbf{x}_n}{N}$.



I want to derive this equation myself. But I'm not sure how to do it. Could someone show why the equation holds?






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    I have taken the liberty to modify your title which was "uninformative".
    $endgroup$
    – Jean Marie
    Dec 31 '18 at 19:48
















3












$begingroup$


In page 594 of Bishop's PRML, the following equation is implied:



$$
-frac{1}{2}sum(mathbf{x}_n-mathbf{bar{x}})^Tmathbf{C}^{-1}(mathbf{x}_n-mathbf{bar{x}}) = -frac{N}{2}mathrm{Tr}(mathbf{C}^{-1}mathbf{S})
$$



where



$$
mathbf{S} = frac{1}{N}sum(mathbf{x}_n-mathbf{bar{x}})(mathbf{x}_n-mathbf{bar{x}})^T
$$
,$mathbf{C}$ is a symmetric matrix and $mathbf{bar{x}} = frac{sum_{n=1}^Nmathbf{x}_n}{N}$.



I want to derive this equation myself. But I'm not sure how to do it. Could someone show why the equation holds?






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    I have taken the liberty to modify your title which was "uninformative".
    $endgroup$
    – Jean Marie
    Dec 31 '18 at 19:48














3












3








3





$begingroup$


In page 594 of Bishop's PRML, the following equation is implied:



$$
-frac{1}{2}sum(mathbf{x}_n-mathbf{bar{x}})^Tmathbf{C}^{-1}(mathbf{x}_n-mathbf{bar{x}}) = -frac{N}{2}mathrm{Tr}(mathbf{C}^{-1}mathbf{S})
$$



where



$$
mathbf{S} = frac{1}{N}sum(mathbf{x}_n-mathbf{bar{x}})(mathbf{x}_n-mathbf{bar{x}})^T
$$
,$mathbf{C}$ is a symmetric matrix and $mathbf{bar{x}} = frac{sum_{n=1}^Nmathbf{x}_n}{N}$.



I want to derive this equation myself. But I'm not sure how to do it. Could someone show why the equation holds?






linear-algebra







share|cite|improve this question











$endgroup$




In page 594 of Bishop's PRML, the following equation is implied:



$$
-frac{1}{2}sum(mathbf{x}_n-mathbf{bar{x}})^Tmathbf{C}^{-1}(mathbf{x}_n-mathbf{bar{x}}) = -frac{N}{2}mathrm{Tr}(mathbf{C}^{-1}mathbf{S})
$$



where



$$
mathbf{S} = frac{1}{N}sum(mathbf{x}_n-mathbf{bar{x}})(mathbf{x}_n-mathbf{bar{x}})^T
$$
,$mathbf{C}$ is a symmetric matrix and $mathbf{bar{x}} = frac{sum_{n=1}^Nmathbf{x}_n}{N}$.



I want to derive this equation myself. But I'm not sure how to do it. Could someone show why the equation holds?






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 19:46









Jean Marie

31k42255




31k42255










asked Dec 31 '18 at 12:16









SandiSandi

262112




262112












  • $begingroup$
    I have taken the liberty to modify your title which was "uninformative".
    $endgroup$
    – Jean Marie
    Dec 31 '18 at 19:48


















  • $begingroup$
    I have taken the liberty to modify your title which was "uninformative".
    $endgroup$
    – Jean Marie
    Dec 31 '18 at 19:48
















$begingroup$
I have taken the liberty to modify your title which was "uninformative".
$endgroup$
– Jean Marie
Dec 31 '18 at 19:48




$begingroup$
I have taken the liberty to modify your title which was "uninformative".
$endgroup$
– Jean Marie
Dec 31 '18 at 19:48










2 Answers
2






active

oldest

votes


















4












$begingroup$

Guide:



Notice that $(x_n - bar{x})^TC^{-1}(x_n - bar{x})$ is a scalar,



hence $$(x_n - bar{x})^TC^{-1}(x_n - bar{x})= operatorname{Tr}left[(x_n - bar{x})^TC^{-1}(x_n - bar{x})right]=operatorname{Tr}left[C^{-1}(x_n - bar{x})(x_n - bar{x})^Tright]$$



since $operatorname{Tr}(AB)=operatorname{Tr}(BA)$.
Hopefully you can take it from here.






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    With the help of Siong Thye Goh, I did the following:



    begin{align}
    -frac{1}{2}sum_{n=1}^N(mathbf{x}_n - mathbf{bar{x}})^Tmathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}}) &= -frac{1}{2}sum_{n=1}^Nmathrm{Tr}[(mathbf{x}_n - mathbf{bar{x}})^Tmathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})]\
    &= -frac{1}{2}sum_{n=1}^Nmathrm{Tr}[mathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
    &= -frac{1}{2}mathrm{Tr}[sum^N_{n=1}mathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
    &= -frac{1}{2}mathrm{Tr}[mathbf{C}^{-1}sum_{n=1}^N(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
    &= -frac{1}{2}mathrm{Tr}[mathbf{C}^{-1}Nmathbf{S}] = -frac{N}{2}mathrm{Tr}[mathbf{C}^{-1}mathbf{S}]
    end{align}






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      +1 great job! just a minor careless step at the very beginning, there's a negative there. ;)
      $endgroup$
      – Siong Thye Goh
      Dec 31 '18 at 13:00











    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%2f3057651%2fhow-to-transform-an-expression-into-a-form-involving-the-trace-of-a-product-of-t%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Guide:



    Notice that $(x_n - bar{x})^TC^{-1}(x_n - bar{x})$ is a scalar,



    hence $$(x_n - bar{x})^TC^{-1}(x_n - bar{x})= operatorname{Tr}left[(x_n - bar{x})^TC^{-1}(x_n - bar{x})right]=operatorname{Tr}left[C^{-1}(x_n - bar{x})(x_n - bar{x})^Tright]$$



    since $operatorname{Tr}(AB)=operatorname{Tr}(BA)$.
    Hopefully you can take it from here.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      Guide:



      Notice that $(x_n - bar{x})^TC^{-1}(x_n - bar{x})$ is a scalar,



      hence $$(x_n - bar{x})^TC^{-1}(x_n - bar{x})= operatorname{Tr}left[(x_n - bar{x})^TC^{-1}(x_n - bar{x})right]=operatorname{Tr}left[C^{-1}(x_n - bar{x})(x_n - bar{x})^Tright]$$



      since $operatorname{Tr}(AB)=operatorname{Tr}(BA)$.
      Hopefully you can take it from here.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        Guide:



        Notice that $(x_n - bar{x})^TC^{-1}(x_n - bar{x})$ is a scalar,



        hence $$(x_n - bar{x})^TC^{-1}(x_n - bar{x})= operatorname{Tr}left[(x_n - bar{x})^TC^{-1}(x_n - bar{x})right]=operatorname{Tr}left[C^{-1}(x_n - bar{x})(x_n - bar{x})^Tright]$$



        since $operatorname{Tr}(AB)=operatorname{Tr}(BA)$.
        Hopefully you can take it from here.






        share|cite|improve this answer











        $endgroup$



        Guide:



        Notice that $(x_n - bar{x})^TC^{-1}(x_n - bar{x})$ is a scalar,



        hence $$(x_n - bar{x})^TC^{-1}(x_n - bar{x})= operatorname{Tr}left[(x_n - bar{x})^TC^{-1}(x_n - bar{x})right]=operatorname{Tr}left[C^{-1}(x_n - bar{x})(x_n - bar{x})^Tright]$$



        since $operatorname{Tr}(AB)=operatorname{Tr}(BA)$.
        Hopefully you can take it from here.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 31 '18 at 12:28









        Bernard

        123k741117




        123k741117










        answered Dec 31 '18 at 12:19









        Siong Thye GohSiong Thye Goh

        103k1468119




        103k1468119























            4












            $begingroup$

            With the help of Siong Thye Goh, I did the following:



            begin{align}
            -frac{1}{2}sum_{n=1}^N(mathbf{x}_n - mathbf{bar{x}})^Tmathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}}) &= -frac{1}{2}sum_{n=1}^Nmathrm{Tr}[(mathbf{x}_n - mathbf{bar{x}})^Tmathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})]\
            &= -frac{1}{2}sum_{n=1}^Nmathrm{Tr}[mathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
            &= -frac{1}{2}mathrm{Tr}[sum^N_{n=1}mathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
            &= -frac{1}{2}mathrm{Tr}[mathbf{C}^{-1}sum_{n=1}^N(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
            &= -frac{1}{2}mathrm{Tr}[mathbf{C}^{-1}Nmathbf{S}] = -frac{N}{2}mathrm{Tr}[mathbf{C}^{-1}mathbf{S}]
            end{align}






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              +1 great job! just a minor careless step at the very beginning, there's a negative there. ;)
              $endgroup$
              – Siong Thye Goh
              Dec 31 '18 at 13:00
















            4












            $begingroup$

            With the help of Siong Thye Goh, I did the following:



            begin{align}
            -frac{1}{2}sum_{n=1}^N(mathbf{x}_n - mathbf{bar{x}})^Tmathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}}) &= -frac{1}{2}sum_{n=1}^Nmathrm{Tr}[(mathbf{x}_n - mathbf{bar{x}})^Tmathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})]\
            &= -frac{1}{2}sum_{n=1}^Nmathrm{Tr}[mathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
            &= -frac{1}{2}mathrm{Tr}[sum^N_{n=1}mathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
            &= -frac{1}{2}mathrm{Tr}[mathbf{C}^{-1}sum_{n=1}^N(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
            &= -frac{1}{2}mathrm{Tr}[mathbf{C}^{-1}Nmathbf{S}] = -frac{N}{2}mathrm{Tr}[mathbf{C}^{-1}mathbf{S}]
            end{align}






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              +1 great job! just a minor careless step at the very beginning, there's a negative there. ;)
              $endgroup$
              – Siong Thye Goh
              Dec 31 '18 at 13:00














            4












            4








            4





            $begingroup$

            With the help of Siong Thye Goh, I did the following:



            begin{align}
            -frac{1}{2}sum_{n=1}^N(mathbf{x}_n - mathbf{bar{x}})^Tmathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}}) &= -frac{1}{2}sum_{n=1}^Nmathrm{Tr}[(mathbf{x}_n - mathbf{bar{x}})^Tmathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})]\
            &= -frac{1}{2}sum_{n=1}^Nmathrm{Tr}[mathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
            &= -frac{1}{2}mathrm{Tr}[sum^N_{n=1}mathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
            &= -frac{1}{2}mathrm{Tr}[mathbf{C}^{-1}sum_{n=1}^N(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
            &= -frac{1}{2}mathrm{Tr}[mathbf{C}^{-1}Nmathbf{S}] = -frac{N}{2}mathrm{Tr}[mathbf{C}^{-1}mathbf{S}]
            end{align}






            share|cite|improve this answer











            $endgroup$



            With the help of Siong Thye Goh, I did the following:



            begin{align}
            -frac{1}{2}sum_{n=1}^N(mathbf{x}_n - mathbf{bar{x}})^Tmathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}}) &= -frac{1}{2}sum_{n=1}^Nmathrm{Tr}[(mathbf{x}_n - mathbf{bar{x}})^Tmathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})]\
            &= -frac{1}{2}sum_{n=1}^Nmathrm{Tr}[mathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
            &= -frac{1}{2}mathrm{Tr}[sum^N_{n=1}mathbf{C}^{-1}(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
            &= -frac{1}{2}mathrm{Tr}[mathbf{C}^{-1}sum_{n=1}^N(mathbf{x}_n - mathbf{bar{x}})(mathbf{x}_n - mathbf{bar{x}})^T]\
            &= -frac{1}{2}mathrm{Tr}[mathbf{C}^{-1}Nmathbf{S}] = -frac{N}{2}mathrm{Tr}[mathbf{C}^{-1}mathbf{S}]
            end{align}







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 1 at 14:34

























            answered Dec 31 '18 at 12:43









            SandiSandi

            262112




            262112












            • $begingroup$
              +1 great job! just a minor careless step at the very beginning, there's a negative there. ;)
              $endgroup$
              – Siong Thye Goh
              Dec 31 '18 at 13:00


















            • $begingroup$
              +1 great job! just a minor careless step at the very beginning, there's a negative there. ;)
              $endgroup$
              – Siong Thye Goh
              Dec 31 '18 at 13:00
















            $begingroup$
            +1 great job! just a minor careless step at the very beginning, there's a negative there. ;)
            $endgroup$
            – Siong Thye Goh
            Dec 31 '18 at 13:00




            $begingroup$
            +1 great job! just a minor careless step at the very beginning, there's a negative there. ;)
            $endgroup$
            – Siong Thye Goh
            Dec 31 '18 at 13:00


















            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057651%2fhow-to-transform-an-expression-into-a-form-involving-the-trace-of-a-product-of-t%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

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more