Calculate an orthonormal basis












0












$begingroup$


We have the following Gram-Schmidt algorithm:



enter image description here



I want to calculate for the following vectors an orthonormal basis, with an accuracy of $epsilon=5cdot 10^{-3}$.
begin{equation*}a_1=begin{pmatrix}1 \ 10^{-3} \ 10^{-3}end{pmatrix}, a_2=begin{pmatrix}1 \ 10^{-3} \ 0end{pmatrix}, a_3=begin{pmatrix}1 \ 0 \ 10^{-3}end{pmatrix}end{equation*}



I have done the following:





  • $i=1$ :


begin{equation*}q_1=a_1=begin{pmatrix}1 \ 10^{-3} \ 10^{-3}end{pmatrix}end{equation*}



We don't enter the inner for loop.



Then $r_{11}=|q_1|_2=sqrt{1+2cdot 10^{-6}}$ and so $q_1=frac{q_1}{r_1}=frac{1}{sqrt{1+2cdot 10^{-6}}}begin{pmatrix}1 \ 10^{-3} \ 10^{-3}end{pmatrix}$.





  • $i=2$ :


begin{equation*}q_2=a_2=begin{pmatrix}1 \ 10^{-3} \ 0end{pmatrix}end{equation*}



At the inner loop we have for $j=1$ : begin{equation*}r_{12}=q_1^Tcdot a_2=frac{sqrt{1+ 10^{-6}}}{sqrt{1+2cdot 10^{-6}}} text{ and } q_2=q_2-r_{12}cdot q_1=frac{1}{1+2cdot 10^{-6}}begin{pmatrix}10^{-6} \ 10^{-9} \ -10^{-3}-10^{-9}end{pmatrix}end{equation*}
After the inner for loop we have begin{equation*}r_{22}=|q_2|_2=frac{sqrt{3cdot 10^{-2}+2cdot 10^{-18}+10^{-6}}}{1+2cdot 10^{-6}}end{equation*} and begin{equation*}q_2=frac{q_2}{r_{22}}=frac{1}{sqrt{3cdot 10^{-2}+2cdot 10^{-18}+10^{-6}}}begin{pmatrix}10^{-6} \ 10^{-9} \ -10^{-3}-10^{-9}end{pmatrix}end{equation*}





  • $i=3$ :


begin{equation*}q_3=a_3=begin{pmatrix}1 \ 0 \ 10^{-3}end{pmatrix}end{equation*}



At the inner for loop we have $j=1$ and $j=2$.



For $j=1$ we get:



begin{equation*}r_{13}=q_1^Ta_3=frac{1+10^{-6}}{sqrt{1+2cdot 10^{-6}}} text{ and } q_3=q_3-r_{13}q_1=frac{1}{1+2cdot 10^{-6}}begin{pmatrix}10^{-6} \ -10^{-3}-10^{-9} \ 10^{-9}end{pmatrix}end{equation*}



For $j=2$ we get:



begin{equation*}r_{23}=q_2^Ta_3=frac{-10^{-12}}{sqrt{3cdot 10^{-12}+2cdot 10^{-18}+10^{-6}}}end{equation*} and begin{equation*}q_3=q_3-r_{23}q_2=frac{1}{(1+2cdot 10^{-6})(3+2cdot 10^{-6}+10^6)}begin{pmatrix}4cdot 10^{-6}+4cdot 10^{-12}+1\ -4cdot 10^{-3}-4cdot 10^{-9}-10^{-3} \ 0end{pmatrix}end{equation*}



After the inner for loop we have begin{equation*}r_{33}=|q_3|_2=frac{[4cdot 10^{-6}+4cdot 10^{-12}+1]^2+[ -4cdot 10^{-3}-4cdot 10^{-9}-10^{-3}]^2}{(1+2cdot 10^{-6})(3+2cdot 10^{-6}+10^6)}end{equation*} and begin{equation*}q_3=frac{q_3}{r_{33}}end{equation*}



$$$$



Is everything correct?






orthonormal gram-schmidt







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    We have the following Gram-Schmidt algorithm:



    enter image description here



    I want to calculate for the following vectors an orthonormal basis, with an accuracy of $epsilon=5cdot 10^{-3}$.
    begin{equation*}a_1=begin{pmatrix}1 \ 10^{-3} \ 10^{-3}end{pmatrix}, a_2=begin{pmatrix}1 \ 10^{-3} \ 0end{pmatrix}, a_3=begin{pmatrix}1 \ 0 \ 10^{-3}end{pmatrix}end{equation*}



    I have done the following:





    • $i=1$ :


    begin{equation*}q_1=a_1=begin{pmatrix}1 \ 10^{-3} \ 10^{-3}end{pmatrix}end{equation*}



    We don't enter the inner for loop.



    Then $r_{11}=|q_1|_2=sqrt{1+2cdot 10^{-6}}$ and so $q_1=frac{q_1}{r_1}=frac{1}{sqrt{1+2cdot 10^{-6}}}begin{pmatrix}1 \ 10^{-3} \ 10^{-3}end{pmatrix}$.





    • $i=2$ :


    begin{equation*}q_2=a_2=begin{pmatrix}1 \ 10^{-3} \ 0end{pmatrix}end{equation*}



    At the inner loop we have for $j=1$ : begin{equation*}r_{12}=q_1^Tcdot a_2=frac{sqrt{1+ 10^{-6}}}{sqrt{1+2cdot 10^{-6}}} text{ and } q_2=q_2-r_{12}cdot q_1=frac{1}{1+2cdot 10^{-6}}begin{pmatrix}10^{-6} \ 10^{-9} \ -10^{-3}-10^{-9}end{pmatrix}end{equation*}
    After the inner for loop we have begin{equation*}r_{22}=|q_2|_2=frac{sqrt{3cdot 10^{-2}+2cdot 10^{-18}+10^{-6}}}{1+2cdot 10^{-6}}end{equation*} and begin{equation*}q_2=frac{q_2}{r_{22}}=frac{1}{sqrt{3cdot 10^{-2}+2cdot 10^{-18}+10^{-6}}}begin{pmatrix}10^{-6} \ 10^{-9} \ -10^{-3}-10^{-9}end{pmatrix}end{equation*}





    • $i=3$ :


    begin{equation*}q_3=a_3=begin{pmatrix}1 \ 0 \ 10^{-3}end{pmatrix}end{equation*}



    At the inner for loop we have $j=1$ and $j=2$.



    For $j=1$ we get:



    begin{equation*}r_{13}=q_1^Ta_3=frac{1+10^{-6}}{sqrt{1+2cdot 10^{-6}}} text{ and } q_3=q_3-r_{13}q_1=frac{1}{1+2cdot 10^{-6}}begin{pmatrix}10^{-6} \ -10^{-3}-10^{-9} \ 10^{-9}end{pmatrix}end{equation*}



    For $j=2$ we get:



    begin{equation*}r_{23}=q_2^Ta_3=frac{-10^{-12}}{sqrt{3cdot 10^{-12}+2cdot 10^{-18}+10^{-6}}}end{equation*} and begin{equation*}q_3=q_3-r_{23}q_2=frac{1}{(1+2cdot 10^{-6})(3+2cdot 10^{-6}+10^6)}begin{pmatrix}4cdot 10^{-6}+4cdot 10^{-12}+1\ -4cdot 10^{-3}-4cdot 10^{-9}-10^{-3} \ 0end{pmatrix}end{equation*}



    After the inner for loop we have begin{equation*}r_{33}=|q_3|_2=frac{[4cdot 10^{-6}+4cdot 10^{-12}+1]^2+[ -4cdot 10^{-3}-4cdot 10^{-9}-10^{-3}]^2}{(1+2cdot 10^{-6})(3+2cdot 10^{-6}+10^6)}end{equation*} and begin{equation*}q_3=frac{q_3}{r_{33}}end{equation*}



    $$$$



    Is everything correct?






    orthonormal gram-schmidt







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      We have the following Gram-Schmidt algorithm:



      enter image description here



      I want to calculate for the following vectors an orthonormal basis, with an accuracy of $epsilon=5cdot 10^{-3}$.
      begin{equation*}a_1=begin{pmatrix}1 \ 10^{-3} \ 10^{-3}end{pmatrix}, a_2=begin{pmatrix}1 \ 10^{-3} \ 0end{pmatrix}, a_3=begin{pmatrix}1 \ 0 \ 10^{-3}end{pmatrix}end{equation*}



      I have done the following:





      • $i=1$ :


      begin{equation*}q_1=a_1=begin{pmatrix}1 \ 10^{-3} \ 10^{-3}end{pmatrix}end{equation*}



      We don't enter the inner for loop.



      Then $r_{11}=|q_1|_2=sqrt{1+2cdot 10^{-6}}$ and so $q_1=frac{q_1}{r_1}=frac{1}{sqrt{1+2cdot 10^{-6}}}begin{pmatrix}1 \ 10^{-3} \ 10^{-3}end{pmatrix}$.





      • $i=2$ :


      begin{equation*}q_2=a_2=begin{pmatrix}1 \ 10^{-3} \ 0end{pmatrix}end{equation*}



      At the inner loop we have for $j=1$ : begin{equation*}r_{12}=q_1^Tcdot a_2=frac{sqrt{1+ 10^{-6}}}{sqrt{1+2cdot 10^{-6}}} text{ and } q_2=q_2-r_{12}cdot q_1=frac{1}{1+2cdot 10^{-6}}begin{pmatrix}10^{-6} \ 10^{-9} \ -10^{-3}-10^{-9}end{pmatrix}end{equation*}
      After the inner for loop we have begin{equation*}r_{22}=|q_2|_2=frac{sqrt{3cdot 10^{-2}+2cdot 10^{-18}+10^{-6}}}{1+2cdot 10^{-6}}end{equation*} and begin{equation*}q_2=frac{q_2}{r_{22}}=frac{1}{sqrt{3cdot 10^{-2}+2cdot 10^{-18}+10^{-6}}}begin{pmatrix}10^{-6} \ 10^{-9} \ -10^{-3}-10^{-9}end{pmatrix}end{equation*}





      • $i=3$ :


      begin{equation*}q_3=a_3=begin{pmatrix}1 \ 0 \ 10^{-3}end{pmatrix}end{equation*}



      At the inner for loop we have $j=1$ and $j=2$.



      For $j=1$ we get:



      begin{equation*}r_{13}=q_1^Ta_3=frac{1+10^{-6}}{sqrt{1+2cdot 10^{-6}}} text{ and } q_3=q_3-r_{13}q_1=frac{1}{1+2cdot 10^{-6}}begin{pmatrix}10^{-6} \ -10^{-3}-10^{-9} \ 10^{-9}end{pmatrix}end{equation*}



      For $j=2$ we get:



      begin{equation*}r_{23}=q_2^Ta_3=frac{-10^{-12}}{sqrt{3cdot 10^{-12}+2cdot 10^{-18}+10^{-6}}}end{equation*} and begin{equation*}q_3=q_3-r_{23}q_2=frac{1}{(1+2cdot 10^{-6})(3+2cdot 10^{-6}+10^6)}begin{pmatrix}4cdot 10^{-6}+4cdot 10^{-12}+1\ -4cdot 10^{-3}-4cdot 10^{-9}-10^{-3} \ 0end{pmatrix}end{equation*}



      After the inner for loop we have begin{equation*}r_{33}=|q_3|_2=frac{[4cdot 10^{-6}+4cdot 10^{-12}+1]^2+[ -4cdot 10^{-3}-4cdot 10^{-9}-10^{-3}]^2}{(1+2cdot 10^{-6})(3+2cdot 10^{-6}+10^6)}end{equation*} and begin{equation*}q_3=frac{q_3}{r_{33}}end{equation*}



      $$$$



      Is everything correct?






      orthonormal gram-schmidt







      share|cite|improve this question









      $endgroup$




      We have the following Gram-Schmidt algorithm:



      enter image description here



      I want to calculate for the following vectors an orthonormal basis, with an accuracy of $epsilon=5cdot 10^{-3}$.
      begin{equation*}a_1=begin{pmatrix}1 \ 10^{-3} \ 10^{-3}end{pmatrix}, a_2=begin{pmatrix}1 \ 10^{-3} \ 0end{pmatrix}, a_3=begin{pmatrix}1 \ 0 \ 10^{-3}end{pmatrix}end{equation*}



      I have done the following:





      • $i=1$ :


      begin{equation*}q_1=a_1=begin{pmatrix}1 \ 10^{-3} \ 10^{-3}end{pmatrix}end{equation*}



      We don't enter the inner for loop.



      Then $r_{11}=|q_1|_2=sqrt{1+2cdot 10^{-6}}$ and so $q_1=frac{q_1}{r_1}=frac{1}{sqrt{1+2cdot 10^{-6}}}begin{pmatrix}1 \ 10^{-3} \ 10^{-3}end{pmatrix}$.





      • $i=2$ :


      begin{equation*}q_2=a_2=begin{pmatrix}1 \ 10^{-3} \ 0end{pmatrix}end{equation*}



      At the inner loop we have for $j=1$ : begin{equation*}r_{12}=q_1^Tcdot a_2=frac{sqrt{1+ 10^{-6}}}{sqrt{1+2cdot 10^{-6}}} text{ and } q_2=q_2-r_{12}cdot q_1=frac{1}{1+2cdot 10^{-6}}begin{pmatrix}10^{-6} \ 10^{-9} \ -10^{-3}-10^{-9}end{pmatrix}end{equation*}
      After the inner for loop we have begin{equation*}r_{22}=|q_2|_2=frac{sqrt{3cdot 10^{-2}+2cdot 10^{-18}+10^{-6}}}{1+2cdot 10^{-6}}end{equation*} and begin{equation*}q_2=frac{q_2}{r_{22}}=frac{1}{sqrt{3cdot 10^{-2}+2cdot 10^{-18}+10^{-6}}}begin{pmatrix}10^{-6} \ 10^{-9} \ -10^{-3}-10^{-9}end{pmatrix}end{equation*}





      • $i=3$ :


      begin{equation*}q_3=a_3=begin{pmatrix}1 \ 0 \ 10^{-3}end{pmatrix}end{equation*}



      At the inner for loop we have $j=1$ and $j=2$.



      For $j=1$ we get:



      begin{equation*}r_{13}=q_1^Ta_3=frac{1+10^{-6}}{sqrt{1+2cdot 10^{-6}}} text{ and } q_3=q_3-r_{13}q_1=frac{1}{1+2cdot 10^{-6}}begin{pmatrix}10^{-6} \ -10^{-3}-10^{-9} \ 10^{-9}end{pmatrix}end{equation*}



      For $j=2$ we get:



      begin{equation*}r_{23}=q_2^Ta_3=frac{-10^{-12}}{sqrt{3cdot 10^{-12}+2cdot 10^{-18}+10^{-6}}}end{equation*} and begin{equation*}q_3=q_3-r_{23}q_2=frac{1}{(1+2cdot 10^{-6})(3+2cdot 10^{-6}+10^6)}begin{pmatrix}4cdot 10^{-6}+4cdot 10^{-12}+1\ -4cdot 10^{-3}-4cdot 10^{-9}-10^{-3} \ 0end{pmatrix}end{equation*}



      After the inner for loop we have begin{equation*}r_{33}=|q_3|_2=frac{[4cdot 10^{-6}+4cdot 10^{-12}+1]^2+[ -4cdot 10^{-3}-4cdot 10^{-9}-10^{-3}]^2}{(1+2cdot 10^{-6})(3+2cdot 10^{-6}+10^6)}end{equation*} and begin{equation*}q_3=frac{q_3}{r_{33}}end{equation*}



      $$$$



      Is everything correct?






      orthonormal gram-schmidt




      orthonormal gram-schmidt






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 27 at 13:41









      Mary StarMary Star

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