matrix algebra and idempotent matrix












0












$begingroup$


I'm having a little trouble understanding a few derivations in my book for least squares regression.



$textbf{Question 1}$:



If $textbf{M}^0 textbf{i} = [textbf{I} - frac{1}{n}textbf{ii'}]textbf{i} = textbf{i}-frac{1}{n}textbf{i(i'i)}=textbf{0}$



Then how does this imply that $textbf{i'M}^0 = textbf{0'}$? Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?



and moreover how do we get that



$$sum_{i=1}^n(x_i -bar x)=textbf{i}'[textbf{M}^0textbf{x}]$$



It's not explicitly stated, but i'm pretty sure $textbf{i}$ is a row vector of 1's.



And also, evidently $textbf{M}^0$ is being defined as $[textbf{I}-frac{1}{n}textbf{i}textbf{i}']$



$textbf{Question 2}$:



Why is it the case that for the expansion of $textbf{e}'_0textbf{e}_0 = (textbf{y-Xb})'(textbf{y - xb})= $
$$textbf{y'y - b'X'y - y'Xb + b'X'Xb} = textbf{y'y - 2y'Xb + b'X'Xb}$$



In particular, how can we group the $textbf{b'X'y}$ and $textbf{y'Xb}$ as the same?



Q1 is from the appendix of William H. Greene - Econometric analysis 7th ed.






linear-algebra matrix-equations least-squares idempotents







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I'm having a little trouble understanding a few derivations in my book for least squares regression.



    $textbf{Question 1}$:



    If $textbf{M}^0 textbf{i} = [textbf{I} - frac{1}{n}textbf{ii'}]textbf{i} = textbf{i}-frac{1}{n}textbf{i(i'i)}=textbf{0}$



    Then how does this imply that $textbf{i'M}^0 = textbf{0'}$? Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?



    and moreover how do we get that



    $$sum_{i=1}^n(x_i -bar x)=textbf{i}'[textbf{M}^0textbf{x}]$$



    It's not explicitly stated, but i'm pretty sure $textbf{i}$ is a row vector of 1's.



    And also, evidently $textbf{M}^0$ is being defined as $[textbf{I}-frac{1}{n}textbf{i}textbf{i}']$



    $textbf{Question 2}$:



    Why is it the case that for the expansion of $textbf{e}'_0textbf{e}_0 = (textbf{y-Xb})'(textbf{y - xb})= $
    $$textbf{y'y - b'X'y - y'Xb + b'X'Xb} = textbf{y'y - 2y'Xb + b'X'Xb}$$



    In particular, how can we group the $textbf{b'X'y}$ and $textbf{y'Xb}$ as the same?



    Q1 is from the appendix of William H. Greene - Econometric analysis 7th ed.






    linear-algebra matrix-equations least-squares idempotents







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I'm having a little trouble understanding a few derivations in my book for least squares regression.



      $textbf{Question 1}$:



      If $textbf{M}^0 textbf{i} = [textbf{I} - frac{1}{n}textbf{ii'}]textbf{i} = textbf{i}-frac{1}{n}textbf{i(i'i)}=textbf{0}$



      Then how does this imply that $textbf{i'M}^0 = textbf{0'}$? Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?



      and moreover how do we get that



      $$sum_{i=1}^n(x_i -bar x)=textbf{i}'[textbf{M}^0textbf{x}]$$



      It's not explicitly stated, but i'm pretty sure $textbf{i}$ is a row vector of 1's.



      And also, evidently $textbf{M}^0$ is being defined as $[textbf{I}-frac{1}{n}textbf{i}textbf{i}']$



      $textbf{Question 2}$:



      Why is it the case that for the expansion of $textbf{e}'_0textbf{e}_0 = (textbf{y-Xb})'(textbf{y - xb})= $
      $$textbf{y'y - b'X'y - y'Xb + b'X'Xb} = textbf{y'y - 2y'Xb + b'X'Xb}$$



      In particular, how can we group the $textbf{b'X'y}$ and $textbf{y'Xb}$ as the same?



      Q1 is from the appendix of William H. Greene - Econometric analysis 7th ed.






      linear-algebra matrix-equations least-squares idempotents







      share|cite|improve this question











      $endgroup$




      I'm having a little trouble understanding a few derivations in my book for least squares regression.



      $textbf{Question 1}$:



      If $textbf{M}^0 textbf{i} = [textbf{I} - frac{1}{n}textbf{ii'}]textbf{i} = textbf{i}-frac{1}{n}textbf{i(i'i)}=textbf{0}$



      Then how does this imply that $textbf{i'M}^0 = textbf{0'}$? Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?



      and moreover how do we get that



      $$sum_{i=1}^n(x_i -bar x)=textbf{i}'[textbf{M}^0textbf{x}]$$



      It's not explicitly stated, but i'm pretty sure $textbf{i}$ is a row vector of 1's.



      And also, evidently $textbf{M}^0$ is being defined as $[textbf{I}-frac{1}{n}textbf{i}textbf{i}']$



      $textbf{Question 2}$:



      Why is it the case that for the expansion of $textbf{e}'_0textbf{e}_0 = (textbf{y-Xb})'(textbf{y - xb})= $
      $$textbf{y'y - b'X'y - y'Xb + b'X'Xb} = textbf{y'y - 2y'Xb + b'X'Xb}$$



      In particular, how can we group the $textbf{b'X'y}$ and $textbf{y'Xb}$ as the same?



      Q1 is from the appendix of William H. Greene - Econometric analysis 7th ed.






      linear-algebra matrix-equations least-squares idempotents




      linear-algebra matrix-equations least-squares idempotents






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 21 at 7:27







      elbarto

















      asked Jan 21 at 6:52









      elbartoelbarto

      1,555827




      1,555827






















          1 Answer
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          $begingroup$

          For Question 1,




          Then how does this imply that $textbf{i'M}^0 = textbf{0'}$?
          Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?




          $i'M^0 = (M^0 i)^{'}$, since $M^0$ is symmetric ($(M^0)^T=M^0$).



          For Question 2, both $b'X'y$ and $y'Xb$ are scalars and taking the transpose of a scalar doesn't change its value, hence $b'X'y = y'Xb$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for that :)
            $endgroup$
            – elbarto
            Jan 21 at 21:42











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          $begingroup$

          For Question 1,




          Then how does this imply that $textbf{i'M}^0 = textbf{0'}$?
          Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?




          $i'M^0 = (M^0 i)^{'}$, since $M^0$ is symmetric ($(M^0)^T=M^0$).



          For Question 2, both $b'X'y$ and $y'Xb$ are scalars and taking the transpose of a scalar doesn't change its value, hence $b'X'y = y'Xb$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for that :)
            $endgroup$
            – elbarto
            Jan 21 at 21:42
















          1












          $begingroup$

          For Question 1,




          Then how does this imply that $textbf{i'M}^0 = textbf{0'}$?
          Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?




          $i'M^0 = (M^0 i)^{'}$, since $M^0$ is symmetric ($(M^0)^T=M^0$).



          For Question 2, both $b'X'y$ and $y'Xb$ are scalars and taking the transpose of a scalar doesn't change its value, hence $b'X'y = y'Xb$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for that :)
            $endgroup$
            – elbarto
            Jan 21 at 21:42














          1












          1








          1





          $begingroup$

          For Question 1,




          Then how does this imply that $textbf{i'M}^0 = textbf{0'}$?
          Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?




          $i'M^0 = (M^0 i)^{'}$, since $M^0$ is symmetric ($(M^0)^T=M^0$).



          For Question 2, both $b'X'y$ and $y'Xb$ are scalars and taking the transpose of a scalar doesn't change its value, hence $b'X'y = y'Xb$.






          share|cite|improve this answer









          $endgroup$



          For Question 1,




          Then how does this imply that $textbf{i'M}^0 = textbf{0'}$?
          Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?




          $i'M^0 = (M^0 i)^{'}$, since $M^0$ is symmetric ($(M^0)^T=M^0$).



          For Question 2, both $b'X'y$ and $y'Xb$ are scalars and taking the transpose of a scalar doesn't change its value, hence $b'X'y = y'Xb$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 21 at 7:40









          pointguard0pointguard0

          1,50211021




          1,50211021












          • $begingroup$
            Thank you for that :)
            $endgroup$
            – elbarto
            Jan 21 at 21:42


















          • $begingroup$
            Thank you for that :)
            $endgroup$
            – elbarto
            Jan 21 at 21:42
















          $begingroup$
          Thank you for that :)
          $endgroup$
          – elbarto
          Jan 21 at 21:42




          $begingroup$
          Thank you for that :)
          $endgroup$
          – elbarto
          Jan 21 at 21:42


















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