How to show $u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$ as a matrix multiplication? [on hold]












-1














Let $u_i$'s in $mathbb{R}^m$ be a set of orthogonal vectors and $A in mathbb{R}^{n times m}$.



Consider the following matrix multiplication



$$u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$$
which is a scalar.



How can we abbreviate this matrix summation into just one matrix multiplication?






linear-algebra matrices







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put on hold as off-topic by Saad, Dando18, amWhy, Shailesh, KReiser yesterday


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    -1














    Let $u_i$'s in $mathbb{R}^m$ be a set of orthogonal vectors and $A in mathbb{R}^{n times m}$.



    Consider the following matrix multiplication



    $$u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$$
    which is a scalar.



    How can we abbreviate this matrix summation into just one matrix multiplication?






    linear-algebra matrices







    share|cite|improve this question















    put on hold as off-topic by Saad, Dando18, amWhy, Shailesh, KReiser yesterday


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, Shailesh, KReiser

    If this question can be reworded to fit the rules in the help center, please edit the question.
















      -1












      -1








      -1







      Let $u_i$'s in $mathbb{R}^m$ be a set of orthogonal vectors and $A in mathbb{R}^{n times m}$.



      Consider the following matrix multiplication



      $$u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$$
      which is a scalar.



      How can we abbreviate this matrix summation into just one matrix multiplication?






      linear-algebra matrices







      share|cite|improve this question















      Let $u_i$'s in $mathbb{R}^m$ be a set of orthogonal vectors and $A in mathbb{R}^{n times m}$.



      Consider the following matrix multiplication



      $$u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$$
      which is a scalar.



      How can we abbreviate this matrix summation into just one matrix multiplication?






      linear-algebra matrices




      linear-algebra matrices






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited yesterday









      JimmyK4542

      40.5k245105




      40.5k245105










      asked yesterday









      Saeed

      698310




      698310




      put on hold as off-topic by Saad, Dando18, amWhy, Shailesh, KReiser yesterday


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, Shailesh, KReiser

      If this question can be reworded to fit the rules in the help center, please edit the question.




      put on hold as off-topic by Saad, Dando18, amWhy, Shailesh, KReiser yesterday


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, Shailesh, KReiser

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          2 Answers
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          If you set $U = begin{bmatrix}u_1 & u_2 & cdots & u_nend{bmatrix} in mathbb{R}^{mtimes n}$, then $AU = begin{bmatrix}Au_1 & Au_2 & cdots & Au_nend{bmatrix}$.



          So $|AU|_F^2 = |Au_1|_2^2+|Au_2^2|+cdots+|Au_n|_2^2 = u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$.






          share|cite|improve this answer





























            0














            You've got
            begin{align}
            sum_{i=1}^{m}u_i^{top}A^{top}Au_i &= sum_{i=1}^{m}operatorname{Tr}(u_iu_i^{T}A^{T}A)=operatorname{Tr}left(sum_{i=1}^m u_i u_i^{top} cdot A^{top}Aright)
            end{align}






            share|cite|improve this answer




























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2














              If you set $U = begin{bmatrix}u_1 & u_2 & cdots & u_nend{bmatrix} in mathbb{R}^{mtimes n}$, then $AU = begin{bmatrix}Au_1 & Au_2 & cdots & Au_nend{bmatrix}$.



              So $|AU|_F^2 = |Au_1|_2^2+|Au_2^2|+cdots+|Au_n|_2^2 = u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$.






              share|cite|improve this answer


























                2














                If you set $U = begin{bmatrix}u_1 & u_2 & cdots & u_nend{bmatrix} in mathbb{R}^{mtimes n}$, then $AU = begin{bmatrix}Au_1 & Au_2 & cdots & Au_nend{bmatrix}$.



                So $|AU|_F^2 = |Au_1|_2^2+|Au_2^2|+cdots+|Au_n|_2^2 = u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$.






                share|cite|improve this answer
























                  2












                  2








                  2






                  If you set $U = begin{bmatrix}u_1 & u_2 & cdots & u_nend{bmatrix} in mathbb{R}^{mtimes n}$, then $AU = begin{bmatrix}Au_1 & Au_2 & cdots & Au_nend{bmatrix}$.



                  So $|AU|_F^2 = |Au_1|_2^2+|Au_2^2|+cdots+|Au_n|_2^2 = u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$.






                  share|cite|improve this answer












                  If you set $U = begin{bmatrix}u_1 & u_2 & cdots & u_nend{bmatrix} in mathbb{R}^{mtimes n}$, then $AU = begin{bmatrix}Au_1 & Au_2 & cdots & Au_nend{bmatrix}$.



                  So $|AU|_F^2 = |Au_1|_2^2+|Au_2^2|+cdots+|Au_n|_2^2 = u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  JimmyK4542

                  40.5k245105




                  40.5k245105























                      0














                      You've got
                      begin{align}
                      sum_{i=1}^{m}u_i^{top}A^{top}Au_i &= sum_{i=1}^{m}operatorname{Tr}(u_iu_i^{T}A^{T}A)=operatorname{Tr}left(sum_{i=1}^m u_i u_i^{top} cdot A^{top}Aright)
                      end{align}






                      share|cite|improve this answer


























                        0














                        You've got
                        begin{align}
                        sum_{i=1}^{m}u_i^{top}A^{top}Au_i &= sum_{i=1}^{m}operatorname{Tr}(u_iu_i^{T}A^{T}A)=operatorname{Tr}left(sum_{i=1}^m u_i u_i^{top} cdot A^{top}Aright)
                        end{align}






                        share|cite|improve this answer
























                          0












                          0








                          0






                          You've got
                          begin{align}
                          sum_{i=1}^{m}u_i^{top}A^{top}Au_i &= sum_{i=1}^{m}operatorname{Tr}(u_iu_i^{T}A^{T}A)=operatorname{Tr}left(sum_{i=1}^m u_i u_i^{top} cdot A^{top}Aright)
                          end{align}






                          share|cite|improve this answer












                          You've got
                          begin{align}
                          sum_{i=1}^{m}u_i^{top}A^{top}Au_i &= sum_{i=1}^{m}operatorname{Tr}(u_iu_i^{T}A^{T}A)=operatorname{Tr}left(sum_{i=1}^m u_i u_i^{top} cdot A^{top}Aright)
                          end{align}







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered yesterday









                          Nadiels

                          2,385413




                          2,385413















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