How to prove this matrix equation












1














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



























    1














    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

























      1












      1








      1







      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






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      asked 1 hour ago









      Sandi

      213111




      213111






















          2 Answers
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          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































            2














            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
              57 mins ago











            Your Answer





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            2 Answers
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            active

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            2 Answers
            2






            active

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            active

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            active

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            2














            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




























              2














              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


























                2












                2








                2






                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 1 hour ago









                Bernard

                118k639112




                118k639112










                answered 1 hour ago









                Siong Thye Goh

                99.1k1464117




                99.1k1464117























                    2














                    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
                      57 mins ago
















                    2














                    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
                      57 mins ago














                    2












                    2








                    2






                    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 1 hour ago









                    Sandi

                    213111




                    213111












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


















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
















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




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


















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