How to find the general linear transformation that preserves a certain scalar product?












1












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So I have got the scalar product of two vectors defined in this way:
$$(x,y)_c = x_1y_1+cx_2y_2$$ where $x = (x1,x2)$ and $y=(y1,y2)$. Now I need to find the set of tranformation matrices that preserves the scalar product defined above. I know that for the standard scalar product between two vectors the set of orthogonal matrices ${R_n}$ s.t. $R^T R = I$ preserves the scalar products:
$$(Rp, Rq) = (p,q) forall p,q$$



How do I find the equivalent transformation matrices for the scalar product above?










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    1












    $begingroup$


    So I have got the scalar product of two vectors defined in this way:
    $$(x,y)_c = x_1y_1+cx_2y_2$$ where $x = (x1,x2)$ and $y=(y1,y2)$. Now I need to find the set of tranformation matrices that preserves the scalar product defined above. I know that for the standard scalar product between two vectors the set of orthogonal matrices ${R_n}$ s.t. $R^T R = I$ preserves the scalar products:
    $$(Rp, Rq) = (p,q) forall p,q$$



    How do I find the equivalent transformation matrices for the scalar product above?










    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      So I have got the scalar product of two vectors defined in this way:
      $$(x,y)_c = x_1y_1+cx_2y_2$$ where $x = (x1,x2)$ and $y=(y1,y2)$. Now I need to find the set of tranformation matrices that preserves the scalar product defined above. I know that for the standard scalar product between two vectors the set of orthogonal matrices ${R_n}$ s.t. $R^T R = I$ preserves the scalar products:
      $$(Rp, Rq) = (p,q) forall p,q$$



      How do I find the equivalent transformation matrices for the scalar product above?










      share|cite|improve this question











      $endgroup$




      So I have got the scalar product of two vectors defined in this way:
      $$(x,y)_c = x_1y_1+cx_2y_2$$ where $x = (x1,x2)$ and $y=(y1,y2)$. Now I need to find the set of tranformation matrices that preserves the scalar product defined above. I know that for the standard scalar product between two vectors the set of orthogonal matrices ${R_n}$ s.t. $R^T R = I$ preserves the scalar products:
      $$(Rp, Rq) = (p,q) forall p,q$$



      How do I find the equivalent transformation matrices for the scalar product above?







      linear-algebra matrices linear-transformations






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 14 at 16:53







      daljit97

















      asked Jan 14 at 16:37









      daljit97daljit97

      178111




      178111






















          1 Answer
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          $begingroup$

          Let $C : c_{1,1} = 1, c_{2,2} = c, c_{1,2}=c_{2,1} = 0$. You want $(ACx)^TAy = (Cx)^Ty$, that is: $x^TCA^TAy = (Cx)^Ty$, then once again $A^TA=I$.



          Edit:
          Note that you can extend that to arbitrary diagonal matrices $C$. So you can literally have $x cdot_c y = sum_{i=0}^{n}{c_{i,i}x_iy_i}$ in arbitrary dimensions and the requirement will still be $A^TA=I$ as seen above.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What do you mean by the notation $C : c_{1,1}$?
            $endgroup$
            – daljit97
            Jan 14 at 18:19










          • $begingroup$
            A matrix $C$, $c_{i,j}$ is the element in row $i$ and column $j$.
            $endgroup$
            – lightxbulb
            Jan 14 at 18:28











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          0












          $begingroup$

          Let $C : c_{1,1} = 1, c_{2,2} = c, c_{1,2}=c_{2,1} = 0$. You want $(ACx)^TAy = (Cx)^Ty$, that is: $x^TCA^TAy = (Cx)^Ty$, then once again $A^TA=I$.



          Edit:
          Note that you can extend that to arbitrary diagonal matrices $C$. So you can literally have $x cdot_c y = sum_{i=0}^{n}{c_{i,i}x_iy_i}$ in arbitrary dimensions and the requirement will still be $A^TA=I$ as seen above.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What do you mean by the notation $C : c_{1,1}$?
            $endgroup$
            – daljit97
            Jan 14 at 18:19










          • $begingroup$
            A matrix $C$, $c_{i,j}$ is the element in row $i$ and column $j$.
            $endgroup$
            – lightxbulb
            Jan 14 at 18:28
















          0












          $begingroup$

          Let $C : c_{1,1} = 1, c_{2,2} = c, c_{1,2}=c_{2,1} = 0$. You want $(ACx)^TAy = (Cx)^Ty$, that is: $x^TCA^TAy = (Cx)^Ty$, then once again $A^TA=I$.



          Edit:
          Note that you can extend that to arbitrary diagonal matrices $C$. So you can literally have $x cdot_c y = sum_{i=0}^{n}{c_{i,i}x_iy_i}$ in arbitrary dimensions and the requirement will still be $A^TA=I$ as seen above.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What do you mean by the notation $C : c_{1,1}$?
            $endgroup$
            – daljit97
            Jan 14 at 18:19










          • $begingroup$
            A matrix $C$, $c_{i,j}$ is the element in row $i$ and column $j$.
            $endgroup$
            – lightxbulb
            Jan 14 at 18:28














          0












          0








          0





          $begingroup$

          Let $C : c_{1,1} = 1, c_{2,2} = c, c_{1,2}=c_{2,1} = 0$. You want $(ACx)^TAy = (Cx)^Ty$, that is: $x^TCA^TAy = (Cx)^Ty$, then once again $A^TA=I$.



          Edit:
          Note that you can extend that to arbitrary diagonal matrices $C$. So you can literally have $x cdot_c y = sum_{i=0}^{n}{c_{i,i}x_iy_i}$ in arbitrary dimensions and the requirement will still be $A^TA=I$ as seen above.






          share|cite|improve this answer











          $endgroup$



          Let $C : c_{1,1} = 1, c_{2,2} = c, c_{1,2}=c_{2,1} = 0$. You want $(ACx)^TAy = (Cx)^Ty$, that is: $x^TCA^TAy = (Cx)^Ty$, then once again $A^TA=I$.



          Edit:
          Note that you can extend that to arbitrary diagonal matrices $C$. So you can literally have $x cdot_c y = sum_{i=0}^{n}{c_{i,i}x_iy_i}$ in arbitrary dimensions and the requirement will still be $A^TA=I$ as seen above.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 14 at 18:30

























          answered Jan 14 at 17:19









          lightxbulblightxbulb

          900211




          900211












          • $begingroup$
            What do you mean by the notation $C : c_{1,1}$?
            $endgroup$
            – daljit97
            Jan 14 at 18:19










          • $begingroup$
            A matrix $C$, $c_{i,j}$ is the element in row $i$ and column $j$.
            $endgroup$
            – lightxbulb
            Jan 14 at 18:28


















          • $begingroup$
            What do you mean by the notation $C : c_{1,1}$?
            $endgroup$
            – daljit97
            Jan 14 at 18:19










          • $begingroup$
            A matrix $C$, $c_{i,j}$ is the element in row $i$ and column $j$.
            $endgroup$
            – lightxbulb
            Jan 14 at 18:28
















          $begingroup$
          What do you mean by the notation $C : c_{1,1}$?
          $endgroup$
          – daljit97
          Jan 14 at 18:19




          $begingroup$
          What do you mean by the notation $C : c_{1,1}$?
          $endgroup$
          – daljit97
          Jan 14 at 18:19












          $begingroup$
          A matrix $C$, $c_{i,j}$ is the element in row $i$ and column $j$.
          $endgroup$
          – lightxbulb
          Jan 14 at 18:28




          $begingroup$
          A matrix $C$, $c_{i,j}$ is the element in row $i$ and column $j$.
          $endgroup$
          – lightxbulb
          Jan 14 at 18:28


















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