Bilinear form in finite dimension space












0












$begingroup$


How to prove that there exist a unique operator $T:Vto V$ such that



$$f(x,y)=g(Tx,y)$$



where $f$ is any bilinear form and $g$ is a non-degenerate bilinear form, and $V$ is a finite dimensional space.



I proved the uniqueness but I dont know how to prove the existence.






linear-algebra







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
    $endgroup$
    – Omnomnomnom
    Dec 21 '18 at 17:37










  • $begingroup$
    yes, g is a non-degenerate bilinear form.
    $endgroup$
    – Bilal Jafar Karaki
    Dec 21 '18 at 17:38
















0












$begingroup$


How to prove that there exist a unique operator $T:Vto V$ such that



$$f(x,y)=g(Tx,y)$$



where $f$ is any bilinear form and $g$ is a non-degenerate bilinear form, and $V$ is a finite dimensional space.



I proved the uniqueness but I dont know how to prove the existence.






linear-algebra







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
    $endgroup$
    – Omnomnomnom
    Dec 21 '18 at 17:37










  • $begingroup$
    yes, g is a non-degenerate bilinear form.
    $endgroup$
    – Bilal Jafar Karaki
    Dec 21 '18 at 17:38














0












0








0





$begingroup$


How to prove that there exist a unique operator $T:Vto V$ such that



$$f(x,y)=g(Tx,y)$$



where $f$ is any bilinear form and $g$ is a non-degenerate bilinear form, and $V$ is a finite dimensional space.



I proved the uniqueness but I dont know how to prove the existence.






linear-algebra







share|cite|improve this question











$endgroup$




How to prove that there exist a unique operator $T:Vto V$ such that



$$f(x,y)=g(Tx,y)$$



where $f$ is any bilinear form and $g$ is a non-degenerate bilinear form, and $V$ is a finite dimensional space.



I proved the uniqueness but I dont know how to prove the existence.






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 17:39







Bilal Jafar Karaki

















asked Dec 21 '18 at 17:31









Bilal Jafar KarakiBilal Jafar Karaki

392113




392113








  • 2




    $begingroup$
    We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
    $endgroup$
    – Omnomnomnom
    Dec 21 '18 at 17:37










  • $begingroup$
    yes, g is a non-degenerate bilinear form.
    $endgroup$
    – Bilal Jafar Karaki
    Dec 21 '18 at 17:38














  • 2




    $begingroup$
    We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
    $endgroup$
    – Omnomnomnom
    Dec 21 '18 at 17:37










  • $begingroup$
    yes, g is a non-degenerate bilinear form.
    $endgroup$
    – Bilal Jafar Karaki
    Dec 21 '18 at 17:38








2




2




$begingroup$
We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
$endgroup$
– Omnomnomnom
Dec 21 '18 at 17:37




$begingroup$
We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
$endgroup$
– Omnomnomnom
Dec 21 '18 at 17:37












$begingroup$
yes, g is a non-degenerate bilinear form.
$endgroup$
– Bilal Jafar Karaki
Dec 21 '18 at 17:38




$begingroup$
yes, g is a non-degenerate bilinear form.
$endgroup$
– Bilal Jafar Karaki
Dec 21 '18 at 17:38










2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
$$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
Now show there exists a matrix $T$ such that $T^top B = A$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
    $$
    hat{g}(x)colon ymapsto g(x,y)
    $$
    Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.



    Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
    $$
    hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
    $$
    and therefore, for every $yin V$,
    $$
    hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
    $$
    which is the same as saying that $f(x,y)=g(Tx,y)$.




    How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
      $$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
      Now show there exists a matrix $T$ such that $T^top B = A$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
        $$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
        Now show there exists a matrix $T$ such that $T^top B = A$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
          $$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
          Now show there exists a matrix $T$ such that $T^top B = A$.






          share|cite|improve this answer









          $endgroup$



          Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
          $$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
          Now show there exists a matrix $T$ such that $T^top B = A$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 21 '18 at 17:40









          SvanNSvanN

          2,0661422




          2,0661422























              1












              $begingroup$

              The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
              $$
              hat{g}(x)colon ymapsto g(x,y)
              $$
              Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.



              Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
              $$
              hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
              $$
              and therefore, for every $yin V$,
              $$
              hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
              $$
              which is the same as saying that $f(x,y)=g(Tx,y)$.




              How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
                $$
                hat{g}(x)colon ymapsto g(x,y)
                $$
                Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.



                Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
                $$
                hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
                $$
                and therefore, for every $yin V$,
                $$
                hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
                $$
                which is the same as saying that $f(x,y)=g(Tx,y)$.




                How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
                  $$
                  hat{g}(x)colon ymapsto g(x,y)
                  $$
                  Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.



                  Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
                  $$
                  hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
                  $$
                  and therefore, for every $yin V$,
                  $$
                  hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
                  $$
                  which is the same as saying that $f(x,y)=g(Tx,y)$.




                  How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.






                  share|cite|improve this answer









                  $endgroup$



                  The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
                  $$
                  hat{g}(x)colon ymapsto g(x,y)
                  $$
                  Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.



                  Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
                  $$
                  hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
                  $$
                  and therefore, for every $yin V$,
                  $$
                  hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
                  $$
                  which is the same as saying that $f(x,y)=g(Tx,y)$.




                  How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 21 '18 at 18:47









                  egregegreg

                  185k1486206




                  185k1486206






























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