Equivalent Conditions of Nondegenerate Bilinear Forms and the Gram Matrix
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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
asked Dec 27 '18 at 17:24
MaTheoPhys1994MaTheoPhys1994
62
$endgroup$
add a comment |
$begingroup$
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
asked Dec 27 '18 at 17:24
MaTheoPhys1994MaTheoPhys1994
62
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 27 '18 at 17:24
MaTheoPhys1994MaTheoPhys1994
62
$endgroup$
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
asked Dec 27 '18 at 17:24
MaTheoPhys1994MaTheoPhys1994
62
asked Dec 27 '18 at 17:24
MaTheoPhys1994MaTheoPhys1994
62
asked Dec 27 '18 at 17:24
MaTheoPhys1994MaTheoPhys1994
62
asked Dec 27 '18 at 17:24
MaTheoPhys1994MaTheoPhys1994
62
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
add a comment |
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
add a comment |
1 Answer
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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.
answered Dec 27 '18 at 18:13
MaTheoPhys1994MaTheoPhys1994
62
$endgroup$
add a comment |
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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.
answered Dec 27 '18 at 18:13
MaTheoPhys1994MaTheoPhys1994
62
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 27 '18 at 18:13
MaTheoPhys1994MaTheoPhys1994
62
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 27 '18 at 18:13
MaTheoPhys1994MaTheoPhys1994
62
$endgroup$
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.
answered Dec 27 '18 at 18:13
MaTheoPhys1994MaTheoPhys1994
62
answered Dec 27 '18 at 18:13
MaTheoPhys1994MaTheoPhys1994
62
answered Dec 27 '18 at 18:13
MaTheoPhys1994MaTheoPhys1994
62
answered Dec 27 '18 at 18:13
MaTheoPhys1994MaTheoPhys1994
62
62
add a comment |
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$begingroup$
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