Find two vectors X, Y [closed] The Next CEO of Stack OverflowTrouble applying gram-schmidt to two vectorsFinding vector orthogonal to two vectorsHow to get the two eigen vectors for eigen =1Find unit singular vectors for two known singular values.Finding a vector that is orthogonal to two other vectors3 dimensional matrices and vectorsHow to complete a set of independent vectors to a basis of a subspace?Spans of linearly independent vectorsfind a basis for $R^4$ given the vectors.Find orthonormal vectors for given vectors using Gram-Schmidt

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Find two vectors X, Y [closed]



The Next CEO of Stack OverflowTrouble applying gram-schmidt to two vectorsFinding vector orthogonal to two vectorsHow to get the two eigen vectors for eigen =1Find unit singular vectors for two known singular values.Finding a vector that is orthogonal to two other vectors3 dimensional matrices and vectorsHow to complete a set of independent vectors to a basis of a subspace?Spans of linearly independent vectorsfind a basis for $R^4$ given the vectors.Find orthonormal vectors for given vectors using Gram-Schmidt










1












$begingroup$



Let $$C=beginpmatrix
1 & 2 & 3 \
-1 & 1 & 1 \
1 & 0 & 1 \
endpmatrix.$$



Find two vectors $X, Y in R^3$ such that $X^TCY neq Y^TCX$.







linear-algebra matrices







share|cite|improve this question











$endgroup$



closed as off-topic by dantopa, Alex Provost, Leucippus, YiFan, Cesareo Mar 20 at 6:59


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." – dantopa, Alex Provost, Leucippus, YiFan, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 2




    $begingroup$
    $e_i^TCe_j$ is the $i,j$ entry of $C$. Choose $i$ and $j$ conveniently and taking advantage that $C$ is not symmetric.
    $endgroup$
    – user647486
    Mar 19 at 20:02











  • $begingroup$
    What result? 2 vectors?
    $endgroup$
    – Sarah
    Mar 19 at 20:16










  • $begingroup$
    Welcome to MSE. In the future please include your own thoughts, the effort made so far, and the specific difficulties that got you stuck.
    $endgroup$
    – Lee David Chung Lin
    Mar 20 at 0:24















1












$begingroup$



Let $$C=beginpmatrix
1 & 2 & 3 \
-1 & 1 & 1 \
1 & 0 & 1 \
endpmatrix.$$



Find two vectors $X, Y in R^3$ such that $X^TCY neq Y^TCX$.







linear-algebra matrices







share|cite|improve this question











$endgroup$



closed as off-topic by dantopa, Alex Provost, Leucippus, YiFan, Cesareo Mar 20 at 6:59


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." – dantopa, Alex Provost, Leucippus, YiFan, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 2




    $begingroup$
    $e_i^TCe_j$ is the $i,j$ entry of $C$. Choose $i$ and $j$ conveniently and taking advantage that $C$ is not symmetric.
    $endgroup$
    – user647486
    Mar 19 at 20:02











  • $begingroup$
    What result? 2 vectors?
    $endgroup$
    – Sarah
    Mar 19 at 20:16










  • $begingroup$
    Welcome to MSE. In the future please include your own thoughts, the effort made so far, and the specific difficulties that got you stuck.
    $endgroup$
    – Lee David Chung Lin
    Mar 20 at 0:24













1












1








1


0



$begingroup$



Let $$C=beginpmatrix
1 & 2 & 3 \
-1 & 1 & 1 \
1 & 0 & 1 \
endpmatrix.$$



Find two vectors $X, Y in R^3$ such that $X^TCY neq Y^TCX$.







linear-algebra matrices







share|cite|improve this question











$endgroup$





Let $$C=beginpmatrix
1 & 2 & 3 \
-1 & 1 & 1 \
1 & 0 & 1 \
endpmatrix.$$



Find two vectors $X, Y in R^3$ such that $X^TCY neq Y^TCX$.







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 Mar 19 at 20:05









Maria Mazur

49.3k1360123




49.3k1360123










asked Mar 19 at 20:01









SarahSarah

175




175




closed as off-topic by dantopa, Alex Provost, Leucippus, YiFan, Cesareo Mar 20 at 6:59


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." – dantopa, Alex Provost, Leucippus, YiFan, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by dantopa, Alex Provost, Leucippus, YiFan, Cesareo Mar 20 at 6:59


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." – dantopa, Alex Provost, Leucippus, YiFan, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 2




    $begingroup$
    $e_i^TCe_j$ is the $i,j$ entry of $C$. Choose $i$ and $j$ conveniently and taking advantage that $C$ is not symmetric.
    $endgroup$
    – user647486
    Mar 19 at 20:02











  • $begingroup$
    What result? 2 vectors?
    $endgroup$
    – Sarah
    Mar 19 at 20:16










  • $begingroup$
    Welcome to MSE. In the future please include your own thoughts, the effort made so far, and the specific difficulties that got you stuck.
    $endgroup$
    – Lee David Chung Lin
    Mar 20 at 0:24












  • 2




    $begingroup$
    $e_i^TCe_j$ is the $i,j$ entry of $C$. Choose $i$ and $j$ conveniently and taking advantage that $C$ is not symmetric.
    $endgroup$
    – user647486
    Mar 19 at 20:02











  • $begingroup$
    What result? 2 vectors?
    $endgroup$
    – Sarah
    Mar 19 at 20:16










  • $begingroup$
    Welcome to MSE. In the future please include your own thoughts, the effort made so far, and the specific difficulties that got you stuck.
    $endgroup$
    – Lee David Chung Lin
    Mar 20 at 0:24







2




2




$begingroup$
$e_i^TCe_j$ is the $i,j$ entry of $C$. Choose $i$ and $j$ conveniently and taking advantage that $C$ is not symmetric.
$endgroup$
– user647486
Mar 19 at 20:02





$begingroup$
$e_i^TCe_j$ is the $i,j$ entry of $C$. Choose $i$ and $j$ conveniently and taking advantage that $C$ is not symmetric.
$endgroup$
– user647486
Mar 19 at 20:02













$begingroup$
What result? 2 vectors?
$endgroup$
– Sarah
Mar 19 at 20:16




$begingroup$
What result? 2 vectors?
$endgroup$
– Sarah
Mar 19 at 20:16












$begingroup$
Welcome to MSE. In the future please include your own thoughts, the effort made so far, and the specific difficulties that got you stuck.
$endgroup$
– Lee David Chung Lin
Mar 20 at 0:24




$begingroup$
Welcome to MSE. In the future please include your own thoughts, the effort made so far, and the specific difficulties that got you stuck.
$endgroup$
– Lee David Chung Lin
Mar 20 at 0:24










2 Answers
2






active

oldest

votes


















0












$begingroup$

Let $X=(1,0,0)$ and $Y=(0,0,1)$



Then $$X^TCY = (1,0,0)^T(3,1,1) = beginpmatrix
3 & 1 & 1 \
0 & 0 & 0 \
0 & 0 & 0 \
endpmatrix$$



and
$$Y^TCX = (0,0,1)^T(1,-1,1) = beginpmatrix
0 & 0 & 0 \
0 & 0 & 0 \
1 & -1 & 1 \
endpmatrix$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Those are not vectors in $R^3$ though, right?
    $endgroup$
    – Sarah
    Mar 19 at 20:06










  • $begingroup$
    Ahh...................
    $endgroup$
    – Maria Mazur
    Mar 19 at 20:07










  • $begingroup$
    Did you just guess to get these? Or was there a way you found these matrices?
    $endgroup$
    – Sarah
    Mar 19 at 20:23










  • $begingroup$
    Almost anything should work. You must be very unluckly to find something not working.
    $endgroup$
    – Maria Mazur
    Mar 19 at 20:24






  • 1




    $begingroup$
    Shouldn't the product be a scalar?
    $endgroup$
    – EuklidAlexandria
    Mar 19 at 22:54


















1












$begingroup$

I hope this can help:



Let X and Y be vectors in $R^3$



begineqnarray
X&=&beginpmatrix x_1 \ x_2 \ x_3 endpmatrix\
Y&=&beginpmatrix y_1 \ y_2 \ y_3 endpmatrix
endeqnarray



beginequation
X^TCY = beginpmatrix x_1 & x_2 & x_3 endpmatrixbeginpmatrix 1 & 2 & 3\ -1 & 1 & 1\ 1 & 0 & 0endpmatrixbeginpmatrix y_1 \ y_2 \ y_3 endpmatrix
endequation



beginequation
X^TCY = (x_1-x_2+x_3)y_1+(2x_1+x_2)y_2+(3x_1+x_2)y_3
endequation



NOTE THAT $X^TCY$ IS AN SCALAR. Similarly:
beginequation
Y^TCX = beginpmatrix y_1 & y_2 & y_3 endpmatrixbeginpmatrix 1 & 2 & 3\ -1 & 1 & 1\ 1 & 0 & 0endpmatrixbeginpmatrix x_1 \ x_2 \ x_3 endpmatrix
endequation



beginequation
Y^TCX = (x_1+2x_2+3x_3)y_1+(-x_1+x_2+x_3)y_2+x_1y_3
endequation



Now suppose that we want to know when $X^TCY = Y^TCX$, then:



begineqnarray
X^TCY &=& Y^TCX\
(x_1-x_2+x_3)y_1+(2x_1&+&x_2)y_2+(3x_1+x_2)y_3=\(x_1+2x_2+3x_3)y_1+(-x_1+x_2&+&x_3)y_2+x_1y_3
endeqnarray



begincases x_1-x_2+x_3= x_1+2x_2+3x_3 \ 2x_1+x_2=-x_1+x_2+x_3\ 3x_1+x_2=x_1 endcases



begincases 3x_2+2x_3= 0\ 3x_1-x_3=0\ 2x_1+x_2=0 endcases



This is an homogeneous system of equations and it has either 1 solution (the trivial solution $X=0$) or infinite solutions. It's easy to prove that the only solution is the trivial solution. Then $X^TCY neq Y^TCX$ if $Yneq0$ and $Xneq0$






share|cite|improve this answer











$endgroup$



















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Let $X=(1,0,0)$ and $Y=(0,0,1)$



    Then $$X^TCY = (1,0,0)^T(3,1,1) = beginpmatrix
    3 & 1 & 1 \
    0 & 0 & 0 \
    0 & 0 & 0 \
    endpmatrix$$



    and
    $$Y^TCX = (0,0,1)^T(1,-1,1) = beginpmatrix
    0 & 0 & 0 \
    0 & 0 & 0 \
    1 & -1 & 1 \
    endpmatrix$$






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Those are not vectors in $R^3$ though, right?
      $endgroup$
      – Sarah
      Mar 19 at 20:06










    • $begingroup$
      Ahh...................
      $endgroup$
      – Maria Mazur
      Mar 19 at 20:07










    • $begingroup$
      Did you just guess to get these? Or was there a way you found these matrices?
      $endgroup$
      – Sarah
      Mar 19 at 20:23










    • $begingroup$
      Almost anything should work. You must be very unluckly to find something not working.
      $endgroup$
      – Maria Mazur
      Mar 19 at 20:24






    • 1




      $begingroup$
      Shouldn't the product be a scalar?
      $endgroup$
      – EuklidAlexandria
      Mar 19 at 22:54















    0












    $begingroup$

    Let $X=(1,0,0)$ and $Y=(0,0,1)$



    Then $$X^TCY = (1,0,0)^T(3,1,1) = beginpmatrix
    3 & 1 & 1 \
    0 & 0 & 0 \
    0 & 0 & 0 \
    endpmatrix$$



    and
    $$Y^TCX = (0,0,1)^T(1,-1,1) = beginpmatrix
    0 & 0 & 0 \
    0 & 0 & 0 \
    1 & -1 & 1 \
    endpmatrix$$






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Those are not vectors in $R^3$ though, right?
      $endgroup$
      – Sarah
      Mar 19 at 20:06










    • $begingroup$
      Ahh...................
      $endgroup$
      – Maria Mazur
      Mar 19 at 20:07










    • $begingroup$
      Did you just guess to get these? Or was there a way you found these matrices?
      $endgroup$
      – Sarah
      Mar 19 at 20:23










    • $begingroup$
      Almost anything should work. You must be very unluckly to find something not working.
      $endgroup$
      – Maria Mazur
      Mar 19 at 20:24






    • 1




      $begingroup$
      Shouldn't the product be a scalar?
      $endgroup$
      – EuklidAlexandria
      Mar 19 at 22:54













    0












    0








    0





    $begingroup$

    Let $X=(1,0,0)$ and $Y=(0,0,1)$



    Then $$X^TCY = (1,0,0)^T(3,1,1) = beginpmatrix
    3 & 1 & 1 \
    0 & 0 & 0 \
    0 & 0 & 0 \
    endpmatrix$$



    and
    $$Y^TCX = (0,0,1)^T(1,-1,1) = beginpmatrix
    0 & 0 & 0 \
    0 & 0 & 0 \
    1 & -1 & 1 \
    endpmatrix$$






    share|cite|improve this answer











    $endgroup$



    Let $X=(1,0,0)$ and $Y=(0,0,1)$



    Then $$X^TCY = (1,0,0)^T(3,1,1) = beginpmatrix
    3 & 1 & 1 \
    0 & 0 & 0 \
    0 & 0 & 0 \
    endpmatrix$$



    and
    $$Y^TCX = (0,0,1)^T(1,-1,1) = beginpmatrix
    0 & 0 & 0 \
    0 & 0 & 0 \
    1 & -1 & 1 \
    endpmatrix$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 19 at 20:17

























    answered Mar 19 at 20:03









    Maria MazurMaria Mazur

    49.3k1360123




    49.3k1360123











    • $begingroup$
      Those are not vectors in $R^3$ though, right?
      $endgroup$
      – Sarah
      Mar 19 at 20:06










    • $begingroup$
      Ahh...................
      $endgroup$
      – Maria Mazur
      Mar 19 at 20:07










    • $begingroup$
      Did you just guess to get these? Or was there a way you found these matrices?
      $endgroup$
      – Sarah
      Mar 19 at 20:23










    • $begingroup$
      Almost anything should work. You must be very unluckly to find something not working.
      $endgroup$
      – Maria Mazur
      Mar 19 at 20:24






    • 1




      $begingroup$
      Shouldn't the product be a scalar?
      $endgroup$
      – EuklidAlexandria
      Mar 19 at 22:54
















    • $begingroup$
      Those are not vectors in $R^3$ though, right?
      $endgroup$
      – Sarah
      Mar 19 at 20:06










    • $begingroup$
      Ahh...................
      $endgroup$
      – Maria Mazur
      Mar 19 at 20:07










    • $begingroup$
      Did you just guess to get these? Or was there a way you found these matrices?
      $endgroup$
      – Sarah
      Mar 19 at 20:23










    • $begingroup$
      Almost anything should work. You must be very unluckly to find something not working.
      $endgroup$
      – Maria Mazur
      Mar 19 at 20:24






    • 1




      $begingroup$
      Shouldn't the product be a scalar?
      $endgroup$
      – EuklidAlexandria
      Mar 19 at 22:54















    $begingroup$
    Those are not vectors in $R^3$ though, right?
    $endgroup$
    – Sarah
    Mar 19 at 20:06




    $begingroup$
    Those are not vectors in $R^3$ though, right?
    $endgroup$
    – Sarah
    Mar 19 at 20:06












    $begingroup$
    Ahh...................
    $endgroup$
    – Maria Mazur
    Mar 19 at 20:07




    $begingroup$
    Ahh...................
    $endgroup$
    – Maria Mazur
    Mar 19 at 20:07












    $begingroup$
    Did you just guess to get these? Or was there a way you found these matrices?
    $endgroup$
    – Sarah
    Mar 19 at 20:23




    $begingroup$
    Did you just guess to get these? Or was there a way you found these matrices?
    $endgroup$
    – Sarah
    Mar 19 at 20:23












    $begingroup$
    Almost anything should work. You must be very unluckly to find something not working.
    $endgroup$
    – Maria Mazur
    Mar 19 at 20:24




    $begingroup$
    Almost anything should work. You must be very unluckly to find something not working.
    $endgroup$
    – Maria Mazur
    Mar 19 at 20:24




    1




    1




    $begingroup$
    Shouldn't the product be a scalar?
    $endgroup$
    – EuklidAlexandria
    Mar 19 at 22:54




    $begingroup$
    Shouldn't the product be a scalar?
    $endgroup$
    – EuklidAlexandria
    Mar 19 at 22:54











    1












    $begingroup$

    I hope this can help:



    Let X and Y be vectors in $R^3$



    begineqnarray
    X&=&beginpmatrix x_1 \ x_2 \ x_3 endpmatrix\
    Y&=&beginpmatrix y_1 \ y_2 \ y_3 endpmatrix
    endeqnarray



    beginequation
    X^TCY = beginpmatrix x_1 & x_2 & x_3 endpmatrixbeginpmatrix 1 & 2 & 3\ -1 & 1 & 1\ 1 & 0 & 0endpmatrixbeginpmatrix y_1 \ y_2 \ y_3 endpmatrix
    endequation



    beginequation
    X^TCY = (x_1-x_2+x_3)y_1+(2x_1+x_2)y_2+(3x_1+x_2)y_3
    endequation



    NOTE THAT $X^TCY$ IS AN SCALAR. Similarly:
    beginequation
    Y^TCX = beginpmatrix y_1 & y_2 & y_3 endpmatrixbeginpmatrix 1 & 2 & 3\ -1 & 1 & 1\ 1 & 0 & 0endpmatrixbeginpmatrix x_1 \ x_2 \ x_3 endpmatrix
    endequation



    beginequation
    Y^TCX = (x_1+2x_2+3x_3)y_1+(-x_1+x_2+x_3)y_2+x_1y_3
    endequation



    Now suppose that we want to know when $X^TCY = Y^TCX$, then:



    begineqnarray
    X^TCY &=& Y^TCX\
    (x_1-x_2+x_3)y_1+(2x_1&+&x_2)y_2+(3x_1+x_2)y_3=\(x_1+2x_2+3x_3)y_1+(-x_1+x_2&+&x_3)y_2+x_1y_3
    endeqnarray



    begincases x_1-x_2+x_3= x_1+2x_2+3x_3 \ 2x_1+x_2=-x_1+x_2+x_3\ 3x_1+x_2=x_1 endcases



    begincases 3x_2+2x_3= 0\ 3x_1-x_3=0\ 2x_1+x_2=0 endcases



    This is an homogeneous system of equations and it has either 1 solution (the trivial solution $X=0$) or infinite solutions. It's easy to prove that the only solution is the trivial solution. Then $X^TCY neq Y^TCX$ if $Yneq0$ and $Xneq0$






    share|cite|improve this answer











    $endgroup$

















      1












      $begingroup$

      I hope this can help:



      Let X and Y be vectors in $R^3$



      begineqnarray
      X&=&beginpmatrix x_1 \ x_2 \ x_3 endpmatrix\
      Y&=&beginpmatrix y_1 \ y_2 \ y_3 endpmatrix
      endeqnarray



      beginequation
      X^TCY = beginpmatrix x_1 & x_2 & x_3 endpmatrixbeginpmatrix 1 & 2 & 3\ -1 & 1 & 1\ 1 & 0 & 0endpmatrixbeginpmatrix y_1 \ y_2 \ y_3 endpmatrix
      endequation



      beginequation
      X^TCY = (x_1-x_2+x_3)y_1+(2x_1+x_2)y_2+(3x_1+x_2)y_3
      endequation



      NOTE THAT $X^TCY$ IS AN SCALAR. Similarly:
      beginequation
      Y^TCX = beginpmatrix y_1 & y_2 & y_3 endpmatrixbeginpmatrix 1 & 2 & 3\ -1 & 1 & 1\ 1 & 0 & 0endpmatrixbeginpmatrix x_1 \ x_2 \ x_3 endpmatrix
      endequation



      beginequation
      Y^TCX = (x_1+2x_2+3x_3)y_1+(-x_1+x_2+x_3)y_2+x_1y_3
      endequation



      Now suppose that we want to know when $X^TCY = Y^TCX$, then:



      begineqnarray
      X^TCY &=& Y^TCX\
      (x_1-x_2+x_3)y_1+(2x_1&+&x_2)y_2+(3x_1+x_2)y_3=\(x_1+2x_2+3x_3)y_1+(-x_1+x_2&+&x_3)y_2+x_1y_3
      endeqnarray



      begincases x_1-x_2+x_3= x_1+2x_2+3x_3 \ 2x_1+x_2=-x_1+x_2+x_3\ 3x_1+x_2=x_1 endcases



      begincases 3x_2+2x_3= 0\ 3x_1-x_3=0\ 2x_1+x_2=0 endcases



      This is an homogeneous system of equations and it has either 1 solution (the trivial solution $X=0$) or infinite solutions. It's easy to prove that the only solution is the trivial solution. Then $X^TCY neq Y^TCX$ if $Yneq0$ and $Xneq0$






      share|cite|improve this answer











      $endgroup$















        1












        1








        1





        $begingroup$

        I hope this can help:



        Let X and Y be vectors in $R^3$



        begineqnarray
        X&=&beginpmatrix x_1 \ x_2 \ x_3 endpmatrix\
        Y&=&beginpmatrix y_1 \ y_2 \ y_3 endpmatrix
        endeqnarray



        beginequation
        X^TCY = beginpmatrix x_1 & x_2 & x_3 endpmatrixbeginpmatrix 1 & 2 & 3\ -1 & 1 & 1\ 1 & 0 & 0endpmatrixbeginpmatrix y_1 \ y_2 \ y_3 endpmatrix
        endequation



        beginequation
        X^TCY = (x_1-x_2+x_3)y_1+(2x_1+x_2)y_2+(3x_1+x_2)y_3
        endequation



        NOTE THAT $X^TCY$ IS AN SCALAR. Similarly:
        beginequation
        Y^TCX = beginpmatrix y_1 & y_2 & y_3 endpmatrixbeginpmatrix 1 & 2 & 3\ -1 & 1 & 1\ 1 & 0 & 0endpmatrixbeginpmatrix x_1 \ x_2 \ x_3 endpmatrix
        endequation



        beginequation
        Y^TCX = (x_1+2x_2+3x_3)y_1+(-x_1+x_2+x_3)y_2+x_1y_3
        endequation



        Now suppose that we want to know when $X^TCY = Y^TCX$, then:



        begineqnarray
        X^TCY &=& Y^TCX\
        (x_1-x_2+x_3)y_1+(2x_1&+&x_2)y_2+(3x_1+x_2)y_3=\(x_1+2x_2+3x_3)y_1+(-x_1+x_2&+&x_3)y_2+x_1y_3
        endeqnarray



        begincases x_1-x_2+x_3= x_1+2x_2+3x_3 \ 2x_1+x_2=-x_1+x_2+x_3\ 3x_1+x_2=x_1 endcases



        begincases 3x_2+2x_3= 0\ 3x_1-x_3=0\ 2x_1+x_2=0 endcases



        This is an homogeneous system of equations and it has either 1 solution (the trivial solution $X=0$) or infinite solutions. It's easy to prove that the only solution is the trivial solution. Then $X^TCY neq Y^TCX$ if $Yneq0$ and $Xneq0$






        share|cite|improve this answer











        $endgroup$



        I hope this can help:



        Let X and Y be vectors in $R^3$



        begineqnarray
        X&=&beginpmatrix x_1 \ x_2 \ x_3 endpmatrix\
        Y&=&beginpmatrix y_1 \ y_2 \ y_3 endpmatrix
        endeqnarray



        beginequation
        X^TCY = beginpmatrix x_1 & x_2 & x_3 endpmatrixbeginpmatrix 1 & 2 & 3\ -1 & 1 & 1\ 1 & 0 & 0endpmatrixbeginpmatrix y_1 \ y_2 \ y_3 endpmatrix
        endequation



        beginequation
        X^TCY = (x_1-x_2+x_3)y_1+(2x_1+x_2)y_2+(3x_1+x_2)y_3
        endequation



        NOTE THAT $X^TCY$ IS AN SCALAR. Similarly:
        beginequation
        Y^TCX = beginpmatrix y_1 & y_2 & y_3 endpmatrixbeginpmatrix 1 & 2 & 3\ -1 & 1 & 1\ 1 & 0 & 0endpmatrixbeginpmatrix x_1 \ x_2 \ x_3 endpmatrix
        endequation



        beginequation
        Y^TCX = (x_1+2x_2+3x_3)y_1+(-x_1+x_2+x_3)y_2+x_1y_3
        endequation



        Now suppose that we want to know when $X^TCY = Y^TCX$, then:



        begineqnarray
        X^TCY &=& Y^TCX\
        (x_1-x_2+x_3)y_1+(2x_1&+&x_2)y_2+(3x_1+x_2)y_3=\(x_1+2x_2+3x_3)y_1+(-x_1+x_2&+&x_3)y_2+x_1y_3
        endeqnarray



        begincases x_1-x_2+x_3= x_1+2x_2+3x_3 \ 2x_1+x_2=-x_1+x_2+x_3\ 3x_1+x_2=x_1 endcases



        begincases 3x_2+2x_3= 0\ 3x_1-x_3=0\ 2x_1+x_2=0 endcases



        This is an homogeneous system of equations and it has either 1 solution (the trivial solution $X=0$) or infinite solutions. It's easy to prove that the only solution is the trivial solution. Then $X^TCY neq Y^TCX$ if $Yneq0$ and $Xneq0$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 19 at 22:58

























        answered Mar 19 at 20:56









        Carlos E. González C.Carlos E. González C.

        535




        535













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