$A^TJA = J$ $ rightarrow det(A) = 1$ [closed]
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Why is this true? I cant seem to understand why the $det(A) = 1$ for this to hold?
ordinary-differential-equations pde determinant matrix-calculus
edited Dec 2 '18 at 20:42
Bernard
121k740116
asked Dec 2 '18 at 20:39
pablo_mathscobarpablo_mathscobar
996
$endgroup$
closed as unclear what you're asking by Did, Lord Shark the Unknown, KReiser, Shailesh, John B Dec 10 '18 at 11:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Why is this true? I cant seem to understand why the $det(A) = 1$ for this to hold?
ordinary-differential-equations pde determinant matrix-calculus
edited Dec 2 '18 at 20:42
Bernard
121k740116
asked Dec 2 '18 at 20:39
pablo_mathscobarpablo_mathscobar
996
$endgroup$
closed as unclear what you're asking by Did, Lord Shark the Unknown, KReiser, Shailesh, John B Dec 10 '18 at 11:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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What is $J$? If $J$ is a non-singular skew-symmetric matrix then one can indeed show that $det A=1$.
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– Fabian
Dec 2 '18 at 20:50
1
$begingroup$
Before dealing with matrices, try scalars. If $a^2 j = j$ then we can have $a = pm 1$. So it cannot be true for matrices.
$endgroup$
– copper.hat
Dec 2 '18 at 21:14
add a comment |
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Why is this true? I cant seem to understand why the $det(A) = 1$ for this to hold?
ordinary-differential-equations pde determinant matrix-calculus
edited Dec 2 '18 at 20:42
Bernard
121k740116
asked Dec 2 '18 at 20:39
pablo_mathscobarpablo_mathscobar
996
$endgroup$
Why is this true? I cant seem to understand why the $det(A) = 1$ for this to hold?
ordinary-differential-equations pde determinant matrix-calculus
ordinary-differential-equations pde determinant matrix-calculus
edited Dec 2 '18 at 20:42
Bernard
121k740116
asked Dec 2 '18 at 20:39
pablo_mathscobarpablo_mathscobar
996
edited Dec 2 '18 at 20:42
Bernard
121k740116
asked Dec 2 '18 at 20:39
pablo_mathscobarpablo_mathscobar
996
edited Dec 2 '18 at 20:42
Bernard
121k740116
edited Dec 2 '18 at 20:42
Bernard
121k740116
edited Dec 2 '18 at 20:42
Bernard
121k740116
121k740116
asked Dec 2 '18 at 20:39
pablo_mathscobarpablo_mathscobar
996
asked Dec 2 '18 at 20:39
pablo_mathscobarpablo_mathscobar
996
asked Dec 2 '18 at 20:39
pablo_mathscobarpablo_mathscobar
996
996
closed as unclear what you're asking by Did, Lord Shark the Unknown, KReiser, Shailesh, John B Dec 10 '18 at 11:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Did, Lord Shark the Unknown, KReiser, Shailesh, John B Dec 10 '18 at 11:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What is $J$? If $J$ is a non-singular skew-symmetric matrix then one can indeed show that $det A=1$.
$endgroup$
– Fabian
Dec 2 '18 at 20:50
1
$begingroup$
Before dealing with matrices, try scalars. If $a^2 j = j$ then we can have $a = pm 1$. So it cannot be true for matrices.
$endgroup$
– copper.hat
Dec 2 '18 at 21:14
add a comment |
$begingroup$
What is $J$? If $J$ is a non-singular skew-symmetric matrix then one can indeed show that $det A=1$.
$endgroup$
– Fabian
Dec 2 '18 at 20:50
1
$begingroup$
Before dealing with matrices, try scalars. If $a^2 j = j$ then we can have $a = pm 1$. So it cannot be true for matrices.
$endgroup$
– copper.hat
Dec 2 '18 at 21:14
$begingroup$
What is $J$? If $J$ is a non-singular skew-symmetric matrix then one can indeed show that $det A=1$.
$endgroup$
– Fabian
Dec 2 '18 at 20:50
$begingroup$
What is $J$? If $J$ is a non-singular skew-symmetric matrix then one can indeed show that $det A=1$.
$endgroup$
– Fabian
Dec 2 '18 at 20:50
1
1
$begingroup$
Before dealing with matrices, try scalars. If $a^2 j = j$ then we can have $a = pm 1$. So it cannot be true for matrices.
$endgroup$
– copper.hat
Dec 2 '18 at 21:14
$begingroup$
Before dealing with matrices, try scalars. If $a^2 j = j$ then we can have $a = pm 1$. So it cannot be true for matrices.
$endgroup$
– copper.hat
Dec 2 '18 at 21:14
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
It is not indeed the case. We might have $$det (A)=-1$$ since $$det(A^TJA)=det(J)$$yields to $$det(A^T)det(J)det(A)=det(J)$$and if $J$ is invertible (i.e. the determinant is non-zero) we conclude $$det(A)^2=1$$if $J$ is singuler ($det(J)=0$) we cannot determine $det (A)$
answered Dec 2 '18 at 20:46
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
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Thats what i thought, thank you
$endgroup$
– pablo_mathscobar
Dec 2 '18 at 20:47
$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 2 '18 at 20:48
add a comment |
$begingroup$
I assume the matrices are over the real numbers.
In general, $det J$ can be anything.
If $J$ is invertible, then one can show that $det A = pm 1$ (see for example the answer by MostafaAyaz)
If $J$ is invertible and antisymmetric with $J= -J^T$, then one can show that $det A =1$. The proof of this fact is not completely elementary as it involves the Pfaffian with $$operatorname{Pf}(A^T J A) = det(A) operatorname{Pf}( J) ,.$$
As $A^T J A =J$ and $J$ is not singular, we follow $det A=1$.
answered Dec 2 '18 at 20:54
FabianFabian
19.8k3674
$endgroup$
add a comment |
$begingroup$
It is not true in all cases. The determinant is multiplicative, , and $det vphantom{A}^{mathrm tmkern -5mu}A=det A$, so
$$det(vphantom{A}^{mathrm tmkern -5mu}AJA)=(det A)^2det J$$
and if it is equal to $det J$ and $det Jne 0$, you can deduce $det A=pm1$.
answered Dec 2 '18 at 20:48
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
We assume $det A >0$
The multiplicative property of determinant implies $$ det A^T JA =(det A)^2 det J =det J$$
You may cancel $det J $ and the result follows. We are also-assuming that $det J ne 0$
answered Dec 2 '18 at 20:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
It is not true in full generality. For instance, if you are dealing with $1×1$ matrices, then matrix multiplication boils down to ordinary multiplication and the equation would be $Acdot Acdot J = J$, so assuming we are working over the rationals (though this can be generalized), we have $J=0$ or $A=1$ or $A=-1$.
In fact, this example can be generalized by applying the determinant to both sides of the equation and noting that $det (A^T)=det(A)$. This gives $det(A)^2 times det(J) = det(J)$, so if $det(J)$ is not zero, then $det(A)=1$ or $det(A)=-1$.
edited Dec 3 '18 at 4:55
Tianlalu
3,08621038
answered Dec 2 '18 at 21:09
Rik BosRik Bos
1
$endgroup$
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is not indeed the case. We might have $$det (A)=-1$$ since $$det(A^TJA)=det(J)$$yields to $$det(A^T)det(J)det(A)=det(J)$$and if $J$ is invertible (i.e. the determinant is non-zero) we conclude $$det(A)^2=1$$if $J$ is singuler ($det(J)=0$) we cannot determine $det (A)$
answered Dec 2 '18 at 20:46
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
Thats what i thought, thank you
$endgroup$
– pablo_mathscobar
Dec 2 '18 at 20:47
$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 2 '18 at 20:48
add a comment |
$begingroup$
It is not indeed the case. We might have $$det (A)=-1$$ since $$det(A^TJA)=det(J)$$yields to $$det(A^T)det(J)det(A)=det(J)$$and if $J$ is invertible (i.e. the determinant is non-zero) we conclude $$det(A)^2=1$$if $J$ is singuler ($det(J)=0$) we cannot determine $det (A)$
answered Dec 2 '18 at 20:46
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
Thats what i thought, thank you
$endgroup$
– pablo_mathscobar
Dec 2 '18 at 20:47
$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 2 '18 at 20:48
add a comment |
$begingroup$
It is not indeed the case. We might have $$det (A)=-1$$ since $$det(A^TJA)=det(J)$$yields to $$det(A^T)det(J)det(A)=det(J)$$and if $J$ is invertible (i.e. the determinant is non-zero) we conclude $$det(A)^2=1$$if $J$ is singuler ($det(J)=0$) we cannot determine $det (A)$
answered Dec 2 '18 at 20:46
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
It is not indeed the case. We might have $$det (A)=-1$$ since $$det(A^TJA)=det(J)$$yields to $$det(A^T)det(J)det(A)=det(J)$$and if $J$ is invertible (i.e. the determinant is non-zero) we conclude $$det(A)^2=1$$if $J$ is singuler ($det(J)=0$) we cannot determine $det (A)$
answered Dec 2 '18 at 20:46
Mostafa AyazMostafa Ayaz
15.6k3939
answered Dec 2 '18 at 20:46
Mostafa AyazMostafa Ayaz
15.6k3939
answered Dec 2 '18 at 20:46
Mostafa AyazMostafa Ayaz
15.6k3939
answered Dec 2 '18 at 20:46
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
$begingroup$
Thats what i thought, thank you
$endgroup$
– pablo_mathscobar
Dec 2 '18 at 20:47
$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 2 '18 at 20:48
add a comment |
$begingroup$
Thats what i thought, thank you
$endgroup$
– pablo_mathscobar
Dec 2 '18 at 20:47
$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 2 '18 at 20:48
$begingroup$
Thats what i thought, thank you
$endgroup$
– pablo_mathscobar
Dec 2 '18 at 20:47
$begingroup$
Thats what i thought, thank you
$endgroup$
– pablo_mathscobar
Dec 2 '18 at 20:47
$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 2 '18 at 20:48
$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 2 '18 at 20:48
add a comment |
$begingroup$
I assume the matrices are over the real numbers.
In general, $det J$ can be anything.
If $J$ is invertible, then one can show that $det A = pm 1$ (see for example the answer by MostafaAyaz)
If $J$ is invertible and antisymmetric with $J= -J^T$, then one can show that $det A =1$. The proof of this fact is not completely elementary as it involves the Pfaffian with $$operatorname{Pf}(A^T J A) = det(A) operatorname{Pf}( J) ,.$$
As $A^T J A =J$ and $J$ is not singular, we follow $det A=1$.
answered Dec 2 '18 at 20:54
FabianFabian
19.8k3674
$endgroup$
add a comment |
$begingroup$
I assume the matrices are over the real numbers.
In general, $det J$ can be anything.
If $J$ is invertible, then one can show that $det A = pm 1$ (see for example the answer by MostafaAyaz)
If $J$ is invertible and antisymmetric with $J= -J^T$, then one can show that $det A =1$. The proof of this fact is not completely elementary as it involves the Pfaffian with $$operatorname{Pf}(A^T J A) = det(A) operatorname{Pf}( J) ,.$$
As $A^T J A =J$ and $J$ is not singular, we follow $det A=1$.
answered Dec 2 '18 at 20:54
FabianFabian
19.8k3674
$endgroup$
add a comment |
$begingroup$
I assume the matrices are over the real numbers.
In general, $det J$ can be anything.
If $J$ is invertible, then one can show that $det A = pm 1$ (see for example the answer by MostafaAyaz)
If $J$ is invertible and antisymmetric with $J= -J^T$, then one can show that $det A =1$. The proof of this fact is not completely elementary as it involves the Pfaffian with $$operatorname{Pf}(A^T J A) = det(A) operatorname{Pf}( J) ,.$$
As $A^T J A =J$ and $J$ is not singular, we follow $det A=1$.
answered Dec 2 '18 at 20:54
FabianFabian
19.8k3674
$endgroup$
I assume the matrices are over the real numbers.
In general, $det J$ can be anything.
If $J$ is invertible, then one can show that $det A = pm 1$ (see for example the answer by MostafaAyaz)
If $J$ is invertible and antisymmetric with $J= -J^T$, then one can show that $det A =1$. The proof of this fact is not completely elementary as it involves the Pfaffian with $$operatorname{Pf}(A^T J A) = det(A) operatorname{Pf}( J) ,.$$
As $A^T J A =J$ and $J$ is not singular, we follow $det A=1$.
answered Dec 2 '18 at 20:54
FabianFabian
19.8k3674
answered Dec 2 '18 at 20:54
FabianFabian
19.8k3674
answered Dec 2 '18 at 20:54
FabianFabian
19.8k3674
answered Dec 2 '18 at 20:54
FabianFabian
19.8k3674
19.8k3674
add a comment |
add a comment |
$begingroup$
It is not true in all cases. The determinant is multiplicative, , and $det vphantom{A}^{mathrm tmkern -5mu}A=det A$, so
$$det(vphantom{A}^{mathrm tmkern -5mu}AJA)=(det A)^2det J$$
and if it is equal to $det J$ and $det Jne 0$, you can deduce $det A=pm1$.
answered Dec 2 '18 at 20:48
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
It is not true in all cases. The determinant is multiplicative, , and $det vphantom{A}^{mathrm tmkern -5mu}A=det A$, so
$$det(vphantom{A}^{mathrm tmkern -5mu}AJA)=(det A)^2det J$$
and if it is equal to $det J$ and $det Jne 0$, you can deduce $det A=pm1$.
answered Dec 2 '18 at 20:48
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
It is not true in all cases. The determinant is multiplicative, , and $det vphantom{A}^{mathrm tmkern -5mu}A=det A$, so
$$det(vphantom{A}^{mathrm tmkern -5mu}AJA)=(det A)^2det J$$
and if it is equal to $det J$ and $det Jne 0$, you can deduce $det A=pm1$.
answered Dec 2 '18 at 20:48
BernardBernard
121k740116
$endgroup$
It is not true in all cases. The determinant is multiplicative, , and $det vphantom{A}^{mathrm tmkern -5mu}A=det A$, so
$$det(vphantom{A}^{mathrm tmkern -5mu}AJA)=(det A)^2det J$$
and if it is equal to $det J$ and $det Jne 0$, you can deduce $det A=pm1$.
answered Dec 2 '18 at 20:48
BernardBernard
121k740116
answered Dec 2 '18 at 20:48
BernardBernard
121k740116
answered Dec 2 '18 at 20:48
BernardBernard
121k740116
answered Dec 2 '18 at 20:48
BernardBernard
121k740116
121k740116
add a comment |
add a comment |
$begingroup$
We assume $det A >0$
The multiplicative property of determinant implies $$ det A^T JA =(det A)^2 det J =det J$$
You may cancel $det J $ and the result follows. We are also-assuming that $det J ne 0$
answered Dec 2 '18 at 20:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
We assume $det A >0$
The multiplicative property of determinant implies $$ det A^T JA =(det A)^2 det J =det J$$
You may cancel $det J $ and the result follows. We are also-assuming that $det J ne 0$
answered Dec 2 '18 at 20:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
We assume $det A >0$
The multiplicative property of determinant implies $$ det A^T JA =(det A)^2 det J =det J$$
You may cancel $det J $ and the result follows. We are also-assuming that $det J ne 0$
answered Dec 2 '18 at 20:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
We assume $det A >0$
The multiplicative property of determinant implies $$ det A^T JA =(det A)^2 det J =det J$$
You may cancel $det J $ and the result follows. We are also-assuming that $det J ne 0$
answered Dec 2 '18 at 20:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 2 '18 at 20:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 2 '18 at 20:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 2 '18 at 20:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
$begingroup$
It is not true in full generality. For instance, if you are dealing with $1×1$ matrices, then matrix multiplication boils down to ordinary multiplication and the equation would be $Acdot Acdot J = J$, so assuming we are working over the rationals (though this can be generalized), we have $J=0$ or $A=1$ or $A=-1$.
In fact, this example can be generalized by applying the determinant to both sides of the equation and noting that $det (A^T)=det(A)$. This gives $det(A)^2 times det(J) = det(J)$, so if $det(J)$ is not zero, then $det(A)=1$ or $det(A)=-1$.
edited Dec 3 '18 at 4:55
Tianlalu
3,08621038
answered Dec 2 '18 at 21:09
Rik BosRik Bos
1
$endgroup$
add a comment |
$begingroup$
It is not true in full generality. For instance, if you are dealing with $1×1$ matrices, then matrix multiplication boils down to ordinary multiplication and the equation would be $Acdot Acdot J = J$, so assuming we are working over the rationals (though this can be generalized), we have $J=0$ or $A=1$ or $A=-1$.
In fact, this example can be generalized by applying the determinant to both sides of the equation and noting that $det (A^T)=det(A)$. This gives $det(A)^2 times det(J) = det(J)$, so if $det(J)$ is not zero, then $det(A)=1$ or $det(A)=-1$.
edited Dec 3 '18 at 4:55
Tianlalu
3,08621038
answered Dec 2 '18 at 21:09
Rik BosRik Bos
1
$endgroup$
add a comment |
$begingroup$
It is not true in full generality. For instance, if you are dealing with $1×1$ matrices, then matrix multiplication boils down to ordinary multiplication and the equation would be $Acdot Acdot J = J$, so assuming we are working over the rationals (though this can be generalized), we have $J=0$ or $A=1$ or $A=-1$.
In fact, this example can be generalized by applying the determinant to both sides of the equation and noting that $det (A^T)=det(A)$. This gives $det(A)^2 times det(J) = det(J)$, so if $det(J)$ is not zero, then $det(A)=1$ or $det(A)=-1$.
edited Dec 3 '18 at 4:55
Tianlalu
3,08621038
answered Dec 2 '18 at 21:09
Rik BosRik Bos
1
$endgroup$
It is not true in full generality. For instance, if you are dealing with $1×1$ matrices, then matrix multiplication boils down to ordinary multiplication and the equation would be $Acdot Acdot J = J$, so assuming we are working over the rationals (though this can be generalized), we have $J=0$ or $A=1$ or $A=-1$.
In fact, this example can be generalized by applying the determinant to both sides of the equation and noting that $det (A^T)=det(A)$. This gives $det(A)^2 times det(J) = det(J)$, so if $det(J)$ is not zero, then $det(A)=1$ or $det(A)=-1$.
edited Dec 3 '18 at 4:55
Tianlalu
3,08621038
answered Dec 2 '18 at 21:09
Rik BosRik Bos
1
edited Dec 3 '18 at 4:55
Tianlalu
3,08621038
edited Dec 3 '18 at 4:55
Tianlalu
3,08621038
edited Dec 3 '18 at 4:55
Tianlalu
3,08621038
3,08621038
answered Dec 2 '18 at 21:09
Rik BosRik Bos
1
answered Dec 2 '18 at 21:09
Rik BosRik Bos
1
answered Dec 2 '18 at 21:09
Rik BosRik Bos
1
1
add a comment |
add a comment |
$begingroup$
What is $J$? If $J$ is a non-singular skew-symmetric matrix then one can indeed show that $det A=1$.
$endgroup$
– Fabian
Dec 2 '18 at 20:50
1
$begingroup$
Before dealing with matrices, try scalars. If $a^2 j = j$ then we can have $a = pm 1$. So it cannot be true for matrices.
$endgroup$
– copper.hat
Dec 2 '18 at 21:14