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
    1 hour ago
















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
    1 hour ago














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 1 hour ago









Jean Marie

28.8k41949




28.8k41949










asked 8 hours ago









Sandi

243112




243112












  • I have taken the liberty to modify your title which was "uninformative".
    – Jean Marie
    1 hour ago


















  • I have taken the liberty to modify your title which was "uninformative".
    – Jean Marie
    1 hour ago
















I have taken the liberty to modify your title which was "uninformative".
– Jean Marie
1 hour ago




I have taken the liberty to modify your title which was "uninformative".
– Jean Marie
1 hour ago










2 Answers
2






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oldest

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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
      8 hours ago











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    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 8 hours ago









        Bernard

        118k639112




        118k639112










        answered 8 hours ago









        Siong Thye Goh

        99.3k1464117




        99.3k1464117























            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
              8 hours ago
















            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
              8 hours ago














            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










            answered 8 hours ago









            Sandi

            243112




            243112












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


















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
















            +1 great job! just a minor careless step at the very beginning, there's a negative there. ;)
            – Siong Thye Goh
            8 hours ago




            +1 great job! just a minor careless step at the very beginning, there's a negative there. ;)
            – Siong Thye Goh
            8 hours ago


















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