Equivalent Conditions of Nondegenerate Bilinear Forms and the Gram Matrix












1












$begingroup$


One can often find the following theorem describing equivalent conditions for non degenerate bilinear forms.



$textbf{Theorem 1}$: Let $V$ be a vector space over the field $mathbb{F}$ equipped with bilinear form $beta : V times V to mathbb{F} $. The following are equivalent:



(1) Let ${ e_i } $ be a basis of $V$. The matrix $B = || beta(e_i, e_j) || $ is invertible



(2) $forall v in V / { 0 }, exists u in V $ such that that $beta(v,u) neq 0 $.



We then say a bilinear form is nondegenerate if the above conditions hold for $beta$. Examples of such theorem are provided here in $textbf{Proposition} 3.11$ and here in $textbf{Theorem} 3.1 $.



It is my understanding the matrix $B := || beta ( e_i, e_j)||$ in the above theorem is by definition the Gram Matrix. The Gram matrix then satisfies the following theorem.



$textbf{Theorem 2}:$ If $V$ is an vector space equipped with an inner product $ langle cdot, cdot rangle $. The set of vectors ${ v_1, ldots, v_n } in V$ is linearly independent iff $det(B_{ij}) neq 0$.



The proof of this theorem is shown in this question.



It appears to me there is a contradiction between these theorems. In $textbf{Theorem 1}$ since ${ e_i}$ is a basis it's also linearly ind. by definition and therefore by $textbf{Theorem 2}$ (and the invertible matrix theorem) the matrix $B:= || beta ( e_i, e_j)||$ is invertible which then would imply every bilinear form is nondegenerate which can't be true. I am thus failing to recognize some important assumptions. Can someone point out to me what information I am failing to recognize? Thank you for any help.






linear-algebra inner-product-space bilinear-form







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  • 1




    $begingroup$
    your Theorem 2 has an inner product. Over the real field, this is defined positive definite, therefore nondegenerate. There is an analogous version for complexes as well. In brief, an inner product is a very special case of a bilinear form
    $endgroup$
    – Will Jagy
    Dec 27 '18 at 17:40










  • $begingroup$
    @WillJagy Right! I just realized this after I posted it.
    $endgroup$
    – MaTheoPhys1994
    Dec 27 '18 at 18:15
















1












$begingroup$


One can often find the following theorem describing equivalent conditions for non degenerate bilinear forms.



$textbf{Theorem 1}$: Let $V$ be a vector space over the field $mathbb{F}$ equipped with bilinear form $beta : V times V to mathbb{F} $. The following are equivalent:



(1) Let ${ e_i } $ be a basis of $V$. The matrix $B = || beta(e_i, e_j) || $ is invertible



(2) $forall v in V / { 0 }, exists u in V $ such that that $beta(v,u) neq 0 $.



We then say a bilinear form is nondegenerate if the above conditions hold for $beta$. Examples of such theorem are provided here in $textbf{Proposition} 3.11$ and here in $textbf{Theorem} 3.1 $.



It is my understanding the matrix $B := || beta ( e_i, e_j)||$ in the above theorem is by definition the Gram Matrix. The Gram matrix then satisfies the following theorem.



$textbf{Theorem 2}:$ If $V$ is an vector space equipped with an inner product $ langle cdot, cdot rangle $. The set of vectors ${ v_1, ldots, v_n } in V$ is linearly independent iff $det(B_{ij}) neq 0$.



The proof of this theorem is shown in this question.



It appears to me there is a contradiction between these theorems. In $textbf{Theorem 1}$ since ${ e_i}$ is a basis it's also linearly ind. by definition and therefore by $textbf{Theorem 2}$ (and the invertible matrix theorem) the matrix $B:= || beta ( e_i, e_j)||$ is invertible which then would imply every bilinear form is nondegenerate which can't be true. I am thus failing to recognize some important assumptions. Can someone point out to me what information I am failing to recognize? Thank you for any help.






linear-algebra inner-product-space bilinear-form







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    your Theorem 2 has an inner product. Over the real field, this is defined positive definite, therefore nondegenerate. There is an analogous version for complexes as well. In brief, an inner product is a very special case of a bilinear form
    $endgroup$
    – Will Jagy
    Dec 27 '18 at 17:40










  • $begingroup$
    @WillJagy Right! I just realized this after I posted it.
    $endgroup$
    – MaTheoPhys1994
    Dec 27 '18 at 18:15














1












1








1





$begingroup$


One can often find the following theorem describing equivalent conditions for non degenerate bilinear forms.



$textbf{Theorem 1}$: Let $V$ be a vector space over the field $mathbb{F}$ equipped with bilinear form $beta : V times V to mathbb{F} $. The following are equivalent:



(1) Let ${ e_i } $ be a basis of $V$. The matrix $B = || beta(e_i, e_j) || $ is invertible



(2) $forall v in V / { 0 }, exists u in V $ such that that $beta(v,u) neq 0 $.



We then say a bilinear form is nondegenerate if the above conditions hold for $beta$. Examples of such theorem are provided here in $textbf{Proposition} 3.11$ and here in $textbf{Theorem} 3.1 $.



It is my understanding the matrix $B := || beta ( e_i, e_j)||$ in the above theorem is by definition the Gram Matrix. The Gram matrix then satisfies the following theorem.



$textbf{Theorem 2}:$ If $V$ is an vector space equipped with an inner product $ langle cdot, cdot rangle $. The set of vectors ${ v_1, ldots, v_n } in V$ is linearly independent iff $det(B_{ij}) neq 0$.



The proof of this theorem is shown in this question.



It appears to me there is a contradiction between these theorems. In $textbf{Theorem 1}$ since ${ e_i}$ is a basis it's also linearly ind. by definition and therefore by $textbf{Theorem 2}$ (and the invertible matrix theorem) the matrix $B:= || beta ( e_i, e_j)||$ is invertible which then would imply every bilinear form is nondegenerate which can't be true. I am thus failing to recognize some important assumptions. Can someone point out to me what information I am failing to recognize? Thank you for any help.






linear-algebra inner-product-space bilinear-form







share|cite|improve this question









$endgroup$




One can often find the following theorem describing equivalent conditions for non degenerate bilinear forms.



$textbf{Theorem 1}$: Let $V$ be a vector space over the field $mathbb{F}$ equipped with bilinear form $beta : V times V to mathbb{F} $. The following are equivalent:



(1) Let ${ e_i } $ be a basis of $V$. The matrix $B = || beta(e_i, e_j) || $ is invertible



(2) $forall v in V / { 0 }, exists u in V $ such that that $beta(v,u) neq 0 $.



We then say a bilinear form is nondegenerate if the above conditions hold for $beta$. Examples of such theorem are provided here in $textbf{Proposition} 3.11$ and here in $textbf{Theorem} 3.1 $.



It is my understanding the matrix $B := || beta ( e_i, e_j)||$ in the above theorem is by definition the Gram Matrix. The Gram matrix then satisfies the following theorem.



$textbf{Theorem 2}:$ If $V$ is an vector space equipped with an inner product $ langle cdot, cdot rangle $. The set of vectors ${ v_1, ldots, v_n } in V$ is linearly independent iff $det(B_{ij}) neq 0$.



The proof of this theorem is shown in this question.



It appears to me there is a contradiction between these theorems. In $textbf{Theorem 1}$ since ${ e_i}$ is a basis it's also linearly ind. by definition and therefore by $textbf{Theorem 2}$ (and the invertible matrix theorem) the matrix $B:= || beta ( e_i, e_j)||$ is invertible which then would imply every bilinear form is nondegenerate which can't be true. I am thus failing to recognize some important assumptions. Can someone point out to me what information I am failing to recognize? Thank you for any help.






linear-algebra inner-product-space bilinear-form




linear-algebra inner-product-space bilinear-form






share|cite|improve this question













share|cite|improve this question











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asked Dec 27 '18 at 17:24









MaTheoPhys1994MaTheoPhys1994

62




62








  • 1




    $begingroup$
    your Theorem 2 has an inner product. Over the real field, this is defined positive definite, therefore nondegenerate. There is an analogous version for complexes as well. In brief, an inner product is a very special case of a bilinear form
    $endgroup$
    – Will Jagy
    Dec 27 '18 at 17:40










  • $begingroup$
    @WillJagy Right! I just realized this after I posted it.
    $endgroup$
    – MaTheoPhys1994
    Dec 27 '18 at 18:15














  • 1




    $begingroup$
    your Theorem 2 has an inner product. Over the real field, this is defined positive definite, therefore nondegenerate. There is an analogous version for complexes as well. In brief, an inner product is a very special case of a bilinear form
    $endgroup$
    – Will Jagy
    Dec 27 '18 at 17:40










  • $begingroup$
    @WillJagy Right! I just realized this after I posted it.
    $endgroup$
    – MaTheoPhys1994
    Dec 27 '18 at 18:15








1




1




$begingroup$
your Theorem 2 has an inner product. Over the real field, this is defined positive definite, therefore nondegenerate. There is an analogous version for complexes as well. In brief, an inner product is a very special case of a bilinear form
$endgroup$
– Will Jagy
Dec 27 '18 at 17:40




$begingroup$
your Theorem 2 has an inner product. Over the real field, this is defined positive definite, therefore nondegenerate. There is an analogous version for complexes as well. In brief, an inner product is a very special case of a bilinear form
$endgroup$
– Will Jagy
Dec 27 '18 at 17:40












$begingroup$
@WillJagy Right! I just realized this after I posted it.
$endgroup$
– MaTheoPhys1994
Dec 27 '18 at 18:15




$begingroup$
@WillJagy Right! I just realized this after I posted it.
$endgroup$
– MaTheoPhys1994
Dec 27 '18 at 18:15










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In the immediate aftermath of writing this question I believe I have uncovered the error in my reasoning. In $textbf{Theorem 2}$ it is already assumed the bilinear form $langle cdot, cdot rangle $ is an inner product which implies it is already nondegenerate by definition and therefore satisfies $textbf{Theorem 1}$. Therefore, there is no contradiction.



Furthermore, for an arbitrary bilinear form $beta$, $B:= || beta ( e_i, e_j)||$ is not invertible in general because not every bilinear is an inner product which means we cannot apply $textbf{Theorem 2}$ to the forms in $textbf{Theorem 1}$. This was my main source of error.



In summary, every inner product is a bilinear form but every bilinear form is not an inner product as a result of their respective definitions.






share|cite|improve this answer









$endgroup$













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    $begingroup$

    In the immediate aftermath of writing this question I believe I have uncovered the error in my reasoning. In $textbf{Theorem 2}$ it is already assumed the bilinear form $langle cdot, cdot rangle $ is an inner product which implies it is already nondegenerate by definition and therefore satisfies $textbf{Theorem 1}$. Therefore, there is no contradiction.



    Furthermore, for an arbitrary bilinear form $beta$, $B:= || beta ( e_i, e_j)||$ is not invertible in general because not every bilinear is an inner product which means we cannot apply $textbf{Theorem 2}$ to the forms in $textbf{Theorem 1}$. This was my main source of error.



    In summary, every inner product is a bilinear form but every bilinear form is not an inner product as a result of their respective definitions.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      In the immediate aftermath of writing this question I believe I have uncovered the error in my reasoning. In $textbf{Theorem 2}$ it is already assumed the bilinear form $langle cdot, cdot rangle $ is an inner product which implies it is already nondegenerate by definition and therefore satisfies $textbf{Theorem 1}$. Therefore, there is no contradiction.



      Furthermore, for an arbitrary bilinear form $beta$, $B:= || beta ( e_i, e_j)||$ is not invertible in general because not every bilinear is an inner product which means we cannot apply $textbf{Theorem 2}$ to the forms in $textbf{Theorem 1}$. This was my main source of error.



      In summary, every inner product is a bilinear form but every bilinear form is not an inner product as a result of their respective definitions.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        In the immediate aftermath of writing this question I believe I have uncovered the error in my reasoning. In $textbf{Theorem 2}$ it is already assumed the bilinear form $langle cdot, cdot rangle $ is an inner product which implies it is already nondegenerate by definition and therefore satisfies $textbf{Theorem 1}$. Therefore, there is no contradiction.



        Furthermore, for an arbitrary bilinear form $beta$, $B:= || beta ( e_i, e_j)||$ is not invertible in general because not every bilinear is an inner product which means we cannot apply $textbf{Theorem 2}$ to the forms in $textbf{Theorem 1}$. This was my main source of error.



        In summary, every inner product is a bilinear form but every bilinear form is not an inner product as a result of their respective definitions.






        share|cite|improve this answer









        $endgroup$



        In the immediate aftermath of writing this question I believe I have uncovered the error in my reasoning. In $textbf{Theorem 2}$ it is already assumed the bilinear form $langle cdot, cdot rangle $ is an inner product which implies it is already nondegenerate by definition and therefore satisfies $textbf{Theorem 1}$. Therefore, there is no contradiction.



        Furthermore, for an arbitrary bilinear form $beta$, $B:= || beta ( e_i, e_j)||$ is not invertible in general because not every bilinear is an inner product which means we cannot apply $textbf{Theorem 2}$ to the forms in $textbf{Theorem 1}$. This was my main source of error.



        In summary, every inner product is a bilinear form but every bilinear form is not an inner product as a result of their respective definitions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 18:13









        MaTheoPhys1994MaTheoPhys1994

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