Derivation of Linear Regression using Normal Equations












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I was going through Andrew Ng's course on ML and had a doubt regarding one of the steps while deriving the solution for linear regression using normal equations.



Normal equation: $theta=(X^TX)^{-1}X^TY$



While deriving, there's this step:



$frac{delta}{deltatheta}theta^TX^TXtheta = X^TXfrac{delta}{deltatheta}theta^Ttheta$



But isn't matrix multiplication commutative, for us to take out $X^TX$ from inside the derivative?



Thanks






matrix-calculus linear-regression







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    0












    $begingroup$


    I was going through Andrew Ng's course on ML and had a doubt regarding one of the steps while deriving the solution for linear regression using normal equations.



    Normal equation: $theta=(X^TX)^{-1}X^TY$



    While deriving, there's this step:



    $frac{delta}{deltatheta}theta^TX^TXtheta = X^TXfrac{delta}{deltatheta}theta^Ttheta$



    But isn't matrix multiplication commutative, for us to take out $X^TX$ from inside the derivative?



    Thanks






    matrix-calculus linear-regression







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I was going through Andrew Ng's course on ML and had a doubt regarding one of the steps while deriving the solution for linear regression using normal equations.



      Normal equation: $theta=(X^TX)^{-1}X^TY$



      While deriving, there's this step:



      $frac{delta}{deltatheta}theta^TX^TXtheta = X^TXfrac{delta}{deltatheta}theta^Ttheta$



      But isn't matrix multiplication commutative, for us to take out $X^TX$ from inside the derivative?



      Thanks






      matrix-calculus linear-regression







      share|cite|improve this question









      $endgroup$




      I was going through Andrew Ng's course on ML and had a doubt regarding one of the steps while deriving the solution for linear regression using normal equations.



      Normal equation: $theta=(X^TX)^{-1}X^TY$



      While deriving, there's this step:



      $frac{delta}{deltatheta}theta^TX^TXtheta = X^TXfrac{delta}{deltatheta}theta^Ttheta$



      But isn't matrix multiplication commutative, for us to take out $X^TX$ from inside the derivative?



      Thanks






      matrix-calculus linear-regression




      matrix-calculus linear-regression






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      share|cite|improve this question











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      asked Jan 13 at 19:13









      Rish1618Rish1618

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          $begingroup$

          Given two symmetric $(A, B)$ consider these following the scalar functions and their gradients
          $$eqalign{
          alpha &= theta^TAtheta &implies frac{partialalpha}{partialtheta}=2Atheta cr
          beta &= theta^TBtheta &implies frac{partialbeta}{partialtheta}=2Btheta cr
          }$$
          It's not terribly illuminating, but you can write the second gradient in terms of the first, i.e.
          $$frac{partialbeta}{partialtheta} = BA^{-1}frac{partialalpha}{partialtheta}$$
          For the purposes of your question, $A=I$ and $B=X^TX$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Although that equality is true, it does not give insight into why it is true.



            There are many ways to compute that gradient, but here is a direct approach that simply computes all the partial derivatives individually.



            Let $A$ be a symmetric matrix. (In your context, $A= X^top X$.)
            The partial derivative of $theta^top A theta = sum_i sum_j A_{ij} theta_i theta_j$ with respect to $theta_k$ is
            $$frac{partial}{partial theta_k} theta^top A theta = sum_i sum_j A_{ij} frac{partial}{partial theta_k}(theta_i theta_j) = A_{kk} cdot 2 theta_k + sum_{i ne k} A_{ik} cdot theta_ i + sum_{j ne k} A_{kj} theta_j = 2sum_i A_{ki} theta_i = 2 (A theta)_k$$
            Stacking the partial derivatives into a vector gives you the gradient, so
            $$nabla_theta theta^top A theta = 2 A theta.$$






            share|cite|improve this answer









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              2 Answers
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              2 Answers
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              1












              $begingroup$

              Given two symmetric $(A, B)$ consider these following the scalar functions and their gradients
              $$eqalign{
              alpha &= theta^TAtheta &implies frac{partialalpha}{partialtheta}=2Atheta cr
              beta &= theta^TBtheta &implies frac{partialbeta}{partialtheta}=2Btheta cr
              }$$
              It's not terribly illuminating, but you can write the second gradient in terms of the first, i.e.
              $$frac{partialbeta}{partialtheta} = BA^{-1}frac{partialalpha}{partialtheta}$$
              For the purposes of your question, $A=I$ and $B=X^TX$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Given two symmetric $(A, B)$ consider these following the scalar functions and their gradients
                $$eqalign{
                alpha &= theta^TAtheta &implies frac{partialalpha}{partialtheta}=2Atheta cr
                beta &= theta^TBtheta &implies frac{partialbeta}{partialtheta}=2Btheta cr
                }$$
                It's not terribly illuminating, but you can write the second gradient in terms of the first, i.e.
                $$frac{partialbeta}{partialtheta} = BA^{-1}frac{partialalpha}{partialtheta}$$
                For the purposes of your question, $A=I$ and $B=X^TX$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Given two symmetric $(A, B)$ consider these following the scalar functions and their gradients
                  $$eqalign{
                  alpha &= theta^TAtheta &implies frac{partialalpha}{partialtheta}=2Atheta cr
                  beta &= theta^TBtheta &implies frac{partialbeta}{partialtheta}=2Btheta cr
                  }$$
                  It's not terribly illuminating, but you can write the second gradient in terms of the first, i.e.
                  $$frac{partialbeta}{partialtheta} = BA^{-1}frac{partialalpha}{partialtheta}$$
                  For the purposes of your question, $A=I$ and $B=X^TX$.






                  share|cite|improve this answer









                  $endgroup$



                  Given two symmetric $(A, B)$ consider these following the scalar functions and their gradients
                  $$eqalign{
                  alpha &= theta^TAtheta &implies frac{partialalpha}{partialtheta}=2Atheta cr
                  beta &= theta^TBtheta &implies frac{partialbeta}{partialtheta}=2Btheta cr
                  }$$
                  It's not terribly illuminating, but you can write the second gradient in terms of the first, i.e.
                  $$frac{partialbeta}{partialtheta} = BA^{-1}frac{partialalpha}{partialtheta}$$
                  For the purposes of your question, $A=I$ and $B=X^TX$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 13 at 21:30









                  greggreg

                  8,2751823




                  8,2751823























                      0












                      $begingroup$

                      Although that equality is true, it does not give insight into why it is true.



                      There are many ways to compute that gradient, but here is a direct approach that simply computes all the partial derivatives individually.



                      Let $A$ be a symmetric matrix. (In your context, $A= X^top X$.)
                      The partial derivative of $theta^top A theta = sum_i sum_j A_{ij} theta_i theta_j$ with respect to $theta_k$ is
                      $$frac{partial}{partial theta_k} theta^top A theta = sum_i sum_j A_{ij} frac{partial}{partial theta_k}(theta_i theta_j) = A_{kk} cdot 2 theta_k + sum_{i ne k} A_{ik} cdot theta_ i + sum_{j ne k} A_{kj} theta_j = 2sum_i A_{ki} theta_i = 2 (A theta)_k$$
                      Stacking the partial derivatives into a vector gives you the gradient, so
                      $$nabla_theta theta^top A theta = 2 A theta.$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Although that equality is true, it does not give insight into why it is true.



                        There are many ways to compute that gradient, but here is a direct approach that simply computes all the partial derivatives individually.



                        Let $A$ be a symmetric matrix. (In your context, $A= X^top X$.)
                        The partial derivative of $theta^top A theta = sum_i sum_j A_{ij} theta_i theta_j$ with respect to $theta_k$ is
                        $$frac{partial}{partial theta_k} theta^top A theta = sum_i sum_j A_{ij} frac{partial}{partial theta_k}(theta_i theta_j) = A_{kk} cdot 2 theta_k + sum_{i ne k} A_{ik} cdot theta_ i + sum_{j ne k} A_{kj} theta_j = 2sum_i A_{ki} theta_i = 2 (A theta)_k$$
                        Stacking the partial derivatives into a vector gives you the gradient, so
                        $$nabla_theta theta^top A theta = 2 A theta.$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Although that equality is true, it does not give insight into why it is true.



                          There are many ways to compute that gradient, but here is a direct approach that simply computes all the partial derivatives individually.



                          Let $A$ be a symmetric matrix. (In your context, $A= X^top X$.)
                          The partial derivative of $theta^top A theta = sum_i sum_j A_{ij} theta_i theta_j$ with respect to $theta_k$ is
                          $$frac{partial}{partial theta_k} theta^top A theta = sum_i sum_j A_{ij} frac{partial}{partial theta_k}(theta_i theta_j) = A_{kk} cdot 2 theta_k + sum_{i ne k} A_{ik} cdot theta_ i + sum_{j ne k} A_{kj} theta_j = 2sum_i A_{ki} theta_i = 2 (A theta)_k$$
                          Stacking the partial derivatives into a vector gives you the gradient, so
                          $$nabla_theta theta^top A theta = 2 A theta.$$






                          share|cite|improve this answer









                          $endgroup$



                          Although that equality is true, it does not give insight into why it is true.



                          There are many ways to compute that gradient, but here is a direct approach that simply computes all the partial derivatives individually.



                          Let $A$ be a symmetric matrix. (In your context, $A= X^top X$.)
                          The partial derivative of $theta^top A theta = sum_i sum_j A_{ij} theta_i theta_j$ with respect to $theta_k$ is
                          $$frac{partial}{partial theta_k} theta^top A theta = sum_i sum_j A_{ij} frac{partial}{partial theta_k}(theta_i theta_j) = A_{kk} cdot 2 theta_k + sum_{i ne k} A_{ik} cdot theta_ i + sum_{j ne k} A_{kj} theta_j = 2sum_i A_{ki} theta_i = 2 (A theta)_k$$
                          Stacking the partial derivatives into a vector gives you the gradient, so
                          $$nabla_theta theta^top A theta = 2 A theta.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 13 at 19:29









                          angryavianangryavian

                          41k23380




                          41k23380






























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