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












-1












$begingroup$


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











$endgroup$



closed as off-topic by Saad, Dando18, Namaste, Shailesh, KReiser Jan 4 at 0:09


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, Namaste, Shailesh, KReiser

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





















    -1












    $begingroup$


    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











    $endgroup$



    closed as off-topic by Saad, Dando18, Namaste, Shailesh, KReiser Jan 4 at 0:09


    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, Namaste, Shailesh, KReiser

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



















      -1












      -1








      -1





      $begingroup$


      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











      $endgroup$




      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 Jan 3 at 15:17









      JimmyK4542

      41.3k245107




      41.3k245107










      asked Jan 3 at 15:01









      SaeedSaeed

      1,124310




      1,124310




      closed as off-topic by Saad, Dando18, Namaste, Shailesh, KReiser Jan 4 at 0:09


      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, Namaste, Shailesh, KReiser

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







      closed as off-topic by Saad, Dando18, Namaste, Shailesh, KReiser Jan 4 at 0:09


      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, Namaste, Shailesh, KReiser

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






















          2 Answers
          2






          active

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          2












          $begingroup$

          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









          $endgroup$





















            0












            $begingroup$

            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









            $endgroup$




















              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              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









              $endgroup$


















                2












                $begingroup$

                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









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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









                  $endgroup$



                  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 Jan 3 at 15:14









                  JimmyK4542JimmyK4542

                  41.3k245107




                  41.3k245107























                      0












                      $begingroup$

                      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









                      $endgroup$


















                        0












                        $begingroup$

                        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









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          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









                          $endgroup$



                          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 Jan 3 at 15:14









                          NadielsNadiels

                          2,385413




                          2,385413















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