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












3














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
























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
















3














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
























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














3












3








3







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















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

28.9k41949




28.9k41949










asked Dec 31 '18 at 12:16









SandiSandi

255112




255112












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


















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
















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




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










2 Answers
2






active

oldest

votes


















4














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































    4














    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























    • +1 great job! just a minor careless step at the very beginning, there's a negative there. ;)
      – 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














    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




























      4














      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


























        4












        4








        4






        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














        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

        118k639112




        118k639112










        answered Dec 31 '18 at 12:19









        Siong Thye GohSiong Thye Goh

        99.8k1464117




        99.8k1464117























            4














            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























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
















            4














            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























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














            4












            4








            4






            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














            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

            255112




            255112












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


















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
















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




            +1 great job! just a minor careless step at the very beginning, there's a negative there. ;)
            – 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.





            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%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

            mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

            Image
            0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
            Read more

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more