Use the Gram-Schmidt Process to construct an orthogonal set of vectors using an inner product [closed]












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This question on my Linear Algebra test has been stumping me. Every time I finish this problem I end up with vectors which aren't orthogonal.




Use the Gram-Schmidt Process to construct an orthogonal set of vectors for the given set of vectors using the specified inner product:



Column vectors: $[1, 2]$, $[1, 1]$



Inner Product: $langle u, v rangle = u^TAv,$



where $A= begin{bmatrix}6&4\4&6end{bmatrix}$.







linear-algebra gram-schmidt







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closed as off-topic by TheGeekGreek, José Carlos Santos, Riccardo.Alestra, Namaste, The Chaz 2.0 Dec 14 '18 at 15:16


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    If you show your work instead of making everyone guess, then someone could point out where you’re going wrong.
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    – amd
    Dec 14 '18 at 8:54
















-1












$begingroup$


This question on my Linear Algebra test has been stumping me. Every time I finish this problem I end up with vectors which aren't orthogonal.




Use the Gram-Schmidt Process to construct an orthogonal set of vectors for the given set of vectors using the specified inner product:



Column vectors: $[1, 2]$, $[1, 1]$



Inner Product: $langle u, v rangle = u^TAv,$



where $A= begin{bmatrix}6&4\4&6end{bmatrix}$.







linear-algebra gram-schmidt







share|cite|improve this question











$endgroup$



closed as off-topic by TheGeekGreek, José Carlos Santos, Riccardo.Alestra, Namaste, The Chaz 2.0 Dec 14 '18 at 15:16


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." – TheGeekGreek, José Carlos Santos, Riccardo.Alestra, Namaste, The Chaz 2.0

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












  • 2




    $begingroup$
    If you show your work instead of making everyone guess, then someone could point out where you’re going wrong.
    $endgroup$
    – amd
    Dec 14 '18 at 8:54














-1












-1








-1





$begingroup$


This question on my Linear Algebra test has been stumping me. Every time I finish this problem I end up with vectors which aren't orthogonal.




Use the Gram-Schmidt Process to construct an orthogonal set of vectors for the given set of vectors using the specified inner product:



Column vectors: $[1, 2]$, $[1, 1]$



Inner Product: $langle u, v rangle = u^TAv,$



where $A= begin{bmatrix}6&4\4&6end{bmatrix}$.







linear-algebra gram-schmidt







share|cite|improve this question











$endgroup$




This question on my Linear Algebra test has been stumping me. Every time I finish this problem I end up with vectors which aren't orthogonal.




Use the Gram-Schmidt Process to construct an orthogonal set of vectors for the given set of vectors using the specified inner product:



Column vectors: $[1, 2]$, $[1, 1]$



Inner Product: $langle u, v rangle = u^TAv,$



where $A= begin{bmatrix}6&4\4&6end{bmatrix}$.







linear-algebra gram-schmidt




linear-algebra gram-schmidt






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 6:05









The Pointer

2,48731641




2,48731641










asked Dec 14 '18 at 4:34









SarahSarah

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41




closed as off-topic by TheGeekGreek, José Carlos Santos, Riccardo.Alestra, Namaste, The Chaz 2.0 Dec 14 '18 at 15:16


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." – TheGeekGreek, José Carlos Santos, Riccardo.Alestra, Namaste, The Chaz 2.0

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







closed as off-topic by TheGeekGreek, José Carlos Santos, Riccardo.Alestra, Namaste, The Chaz 2.0 Dec 14 '18 at 15:16


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." – TheGeekGreek, José Carlos Santos, Riccardo.Alestra, Namaste, The Chaz 2.0

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








  • 2




    $begingroup$
    If you show your work instead of making everyone guess, then someone could point out where you’re going wrong.
    $endgroup$
    – amd
    Dec 14 '18 at 8:54














  • 2




    $begingroup$
    If you show your work instead of making everyone guess, then someone could point out where you’re going wrong.
    $endgroup$
    – amd
    Dec 14 '18 at 8:54








2




2




$begingroup$
If you show your work instead of making everyone guess, then someone could point out where you’re going wrong.
$endgroup$
– amd
Dec 14 '18 at 8:54




$begingroup$
If you show your work instead of making everyone guess, then someone could point out where you’re going wrong.
$endgroup$
– amd
Dec 14 '18 at 8:54










2 Answers
2






active

oldest

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1












$begingroup$

Let $v_1=(1,2)$, $v_2=(1,1)$. We can let the first vector in our orthogonal set be $u_1=v_1=(1,2)$. The second vector $u_2$ is found by subtracting the projection of $v_2$ onto $u_1$ from $v_2$:
$$
u_2 = v_2 - mathsf{proj}_{u_1}(v_2) = v_2 - frac{langle v_2,u_1rangle}{langle u_1,u_1rangle}u_1.
$$
We compute
$$
langle v_2,u_1rangle = pmatrix{1&1}pmatrix{6&4\4&6}pmatrix{1\2}= 30
$$
and
$$
langle u_1,u_1rangle = pmatrix{1&2}pmatrix{6&4\4&6}pmatrix{1\2}= 46,
$$
so
$$
u_2 = (1,1) - frac{15}{23}cdot(1,2) = left(frac8{23}, -frac7{23}right).
$$






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    0












    $begingroup$

    Let $V$ be an inner product space, with the inner product defined as $langle u, v rangle = u^T A v$, where $A= begin{bmatrix}6&4\4&6end{bmatrix}$. The set ${ (1, 2), (1, 1) }$ is a basis for $V$, since the vectors are linearly independent and span $V$.



    Let $v_1 = (1, 2)$.



    $v_2 = (1, 1) - dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} v_1 = dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} (1, 2) = (1,1) - dfrac{30}{46}(1, 2) = left( dfrac{8}{23} , -dfrac{7}{23} right)$



    Therefore, $left { (1, 2), left( dfrac{8}{23} , -dfrac{7}{23} right) right }$ is an orthogonal basis for $V$.



    Check:



    $left langle (1, 2), left( dfrac{8}{23}, dfrac{-7}{23} right) right rangle = dfrac{112}{23} - dfrac{112}{23} = 0$, as required.



    Reference: http://www.math.tamu.edu/~yvorobet/MATH304-503/Lect3-07web.pdf






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Let $v_1=(1,2)$, $v_2=(1,1)$. We can let the first vector in our orthogonal set be $u_1=v_1=(1,2)$. The second vector $u_2$ is found by subtracting the projection of $v_2$ onto $u_1$ from $v_2$:
      $$
      u_2 = v_2 - mathsf{proj}_{u_1}(v_2) = v_2 - frac{langle v_2,u_1rangle}{langle u_1,u_1rangle}u_1.
      $$
      We compute
      $$
      langle v_2,u_1rangle = pmatrix{1&1}pmatrix{6&4\4&6}pmatrix{1\2}= 30
      $$
      and
      $$
      langle u_1,u_1rangle = pmatrix{1&2}pmatrix{6&4\4&6}pmatrix{1\2}= 46,
      $$
      so
      $$
      u_2 = (1,1) - frac{15}{23}cdot(1,2) = left(frac8{23}, -frac7{23}right).
      $$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Let $v_1=(1,2)$, $v_2=(1,1)$. We can let the first vector in our orthogonal set be $u_1=v_1=(1,2)$. The second vector $u_2$ is found by subtracting the projection of $v_2$ onto $u_1$ from $v_2$:
        $$
        u_2 = v_2 - mathsf{proj}_{u_1}(v_2) = v_2 - frac{langle v_2,u_1rangle}{langle u_1,u_1rangle}u_1.
        $$
        We compute
        $$
        langle v_2,u_1rangle = pmatrix{1&1}pmatrix{6&4\4&6}pmatrix{1\2}= 30
        $$
        and
        $$
        langle u_1,u_1rangle = pmatrix{1&2}pmatrix{6&4\4&6}pmatrix{1\2}= 46,
        $$
        so
        $$
        u_2 = (1,1) - frac{15}{23}cdot(1,2) = left(frac8{23}, -frac7{23}right).
        $$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Let $v_1=(1,2)$, $v_2=(1,1)$. We can let the first vector in our orthogonal set be $u_1=v_1=(1,2)$. The second vector $u_2$ is found by subtracting the projection of $v_2$ onto $u_1$ from $v_2$:
          $$
          u_2 = v_2 - mathsf{proj}_{u_1}(v_2) = v_2 - frac{langle v_2,u_1rangle}{langle u_1,u_1rangle}u_1.
          $$
          We compute
          $$
          langle v_2,u_1rangle = pmatrix{1&1}pmatrix{6&4\4&6}pmatrix{1\2}= 30
          $$
          and
          $$
          langle u_1,u_1rangle = pmatrix{1&2}pmatrix{6&4\4&6}pmatrix{1\2}= 46,
          $$
          so
          $$
          u_2 = (1,1) - frac{15}{23}cdot(1,2) = left(frac8{23}, -frac7{23}right).
          $$






          share|cite|improve this answer









          $endgroup$



          Let $v_1=(1,2)$, $v_2=(1,1)$. We can let the first vector in our orthogonal set be $u_1=v_1=(1,2)$. The second vector $u_2$ is found by subtracting the projection of $v_2$ onto $u_1$ from $v_2$:
          $$
          u_2 = v_2 - mathsf{proj}_{u_1}(v_2) = v_2 - frac{langle v_2,u_1rangle}{langle u_1,u_1rangle}u_1.
          $$
          We compute
          $$
          langle v_2,u_1rangle = pmatrix{1&1}pmatrix{6&4\4&6}pmatrix{1\2}= 30
          $$
          and
          $$
          langle u_1,u_1rangle = pmatrix{1&2}pmatrix{6&4\4&6}pmatrix{1\2}= 46,
          $$
          so
          $$
          u_2 = (1,1) - frac{15}{23}cdot(1,2) = left(frac8{23}, -frac7{23}right).
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 7:33









          Math1000Math1000

          19.4k31746




          19.4k31746























              0












              $begingroup$

              Let $V$ be an inner product space, with the inner product defined as $langle u, v rangle = u^T A v$, where $A= begin{bmatrix}6&4\4&6end{bmatrix}$. The set ${ (1, 2), (1, 1) }$ is a basis for $V$, since the vectors are linearly independent and span $V$.



              Let $v_1 = (1, 2)$.



              $v_2 = (1, 1) - dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} v_1 = dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} (1, 2) = (1,1) - dfrac{30}{46}(1, 2) = left( dfrac{8}{23} , -dfrac{7}{23} right)$



              Therefore, $left { (1, 2), left( dfrac{8}{23} , -dfrac{7}{23} right) right }$ is an orthogonal basis for $V$.



              Check:



              $left langle (1, 2), left( dfrac{8}{23}, dfrac{-7}{23} right) right rangle = dfrac{112}{23} - dfrac{112}{23} = 0$, as required.



              Reference: http://www.math.tamu.edu/~yvorobet/MATH304-503/Lect3-07web.pdf






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Let $V$ be an inner product space, with the inner product defined as $langle u, v rangle = u^T A v$, where $A= begin{bmatrix}6&4\4&6end{bmatrix}$. The set ${ (1, 2), (1, 1) }$ is a basis for $V$, since the vectors are linearly independent and span $V$.



                Let $v_1 = (1, 2)$.



                $v_2 = (1, 1) - dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} v_1 = dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} (1, 2) = (1,1) - dfrac{30}{46}(1, 2) = left( dfrac{8}{23} , -dfrac{7}{23} right)$



                Therefore, $left { (1, 2), left( dfrac{8}{23} , -dfrac{7}{23} right) right }$ is an orthogonal basis for $V$.



                Check:



                $left langle (1, 2), left( dfrac{8}{23}, dfrac{-7}{23} right) right rangle = dfrac{112}{23} - dfrac{112}{23} = 0$, as required.



                Reference: http://www.math.tamu.edu/~yvorobet/MATH304-503/Lect3-07web.pdf






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let $V$ be an inner product space, with the inner product defined as $langle u, v rangle = u^T A v$, where $A= begin{bmatrix}6&4\4&6end{bmatrix}$. The set ${ (1, 2), (1, 1) }$ is a basis for $V$, since the vectors are linearly independent and span $V$.



                  Let $v_1 = (1, 2)$.



                  $v_2 = (1, 1) - dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} v_1 = dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} (1, 2) = (1,1) - dfrac{30}{46}(1, 2) = left( dfrac{8}{23} , -dfrac{7}{23} right)$



                  Therefore, $left { (1, 2), left( dfrac{8}{23} , -dfrac{7}{23} right) right }$ is an orthogonal basis for $V$.



                  Check:



                  $left langle (1, 2), left( dfrac{8}{23}, dfrac{-7}{23} right) right rangle = dfrac{112}{23} - dfrac{112}{23} = 0$, as required.



                  Reference: http://www.math.tamu.edu/~yvorobet/MATH304-503/Lect3-07web.pdf






                  share|cite|improve this answer











                  $endgroup$



                  Let $V$ be an inner product space, with the inner product defined as $langle u, v rangle = u^T A v$, where $A= begin{bmatrix}6&4\4&6end{bmatrix}$. The set ${ (1, 2), (1, 1) }$ is a basis for $V$, since the vectors are linearly independent and span $V$.



                  Let $v_1 = (1, 2)$.



                  $v_2 = (1, 1) - dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} v_1 = dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} (1, 2) = (1,1) - dfrac{30}{46}(1, 2) = left( dfrac{8}{23} , -dfrac{7}{23} right)$



                  Therefore, $left { (1, 2), left( dfrac{8}{23} , -dfrac{7}{23} right) right }$ is an orthogonal basis for $V$.



                  Check:



                  $left langle (1, 2), left( dfrac{8}{23}, dfrac{-7}{23} right) right rangle = dfrac{112}{23} - dfrac{112}{23} = 0$, as required.



                  Reference: http://www.math.tamu.edu/~yvorobet/MATH304-503/Lect3-07web.pdf







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 14 '18 at 23:27

























                  answered Dec 14 '18 at 7:46









                  The PointerThe Pointer

                  2,48731641




                  2,48731641















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