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Specific, Index-Notation Formula of the determinant of 2-dimensional matrices
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Expressing the determinant of a sum of two matrices?Formula for determinant of sum of matricesIs there a formula for the determinant of the wedge product of two matrices?determinant of the Fubini-Study metricDeterminant with Levi-Civita Symbol?Do we have $det=e^1wedgecdotswedge e^n$?How to prove the following identity?On Selberg's Proof of the formula for the Selberg IntegralProduct of two epsilon tensors with n indices.Proving Equation for Timelike Geodesics of a Conformally Flat MetricConfused by use of Cauchy Binet formulaA way of writing the Leibniz formula for determinants
$begingroup$
I was practicing some old tensor calculus exams and had to answer this question (paraphrased, it has multiple subquestions):
If $eta_ij$ is the component of the metric $H$ on a 2D Riemannian manifold (i.e. $H=eta_ij dx^i otimes dx^j$) and $w_i$ is the componont of a smooth covector field $Omega=w_i dx^i$, construct a symmetric tensor $Gamma=gamma_ijdx^i otimes dx^j=(eta_ij-w_i w_j)dx^i otimes dx^j$.
Show that $det (gamma_ij)$ is equal to $det(eta_ij) (1-eta^ijw_i w_j)$.
In the solutions the answer is just "A direct computation shows this". I have verified the correctness of the final formula by just bruteforcing the summations in 2D, but I want to be able to verify this in index notation. So far I've found two threads which contain (a variant) of this formula: Formula for determinant of sum of matrices and Expressing the determinant of a sum of two matrices?. However, one proves it using standard matrix notation and the other only proves it in the n-dimensional cases which I found a quite hard to follow since this is my first actual mathematics course (physicist). (The formula is, for an invertible matrix A, $det(A+B)=det(A)+det(B)+trace(A^-1 B)$, this is just a special case of that formula)
So far I've got this:
$det(gamma_ij)=frac12epsilon^i_1 i_2epsilon^j_1 j_2gamma_i_1 j_1gamma_i_2 j_2=frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1-w_i_1w_j_1)(eta_i_2 j_2-w_i_2w_j_2)=frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1eta_i_2 j_2+w_i_1w_j_1w_i_2w_j_2-eta_i_1 j_1w_i_2w_j_2-eta_i_2 j_2w_i_1w_j_1)=$
$=det(eta_ij)+det(w_i w_j)-frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1w_i_2w_j_2+eta_i_2 j_2w_i_1w_j_1)$
The determinant of the ww matrix is obviously zero, so now all that's left to show is
$frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1w_i_2w_j_2+eta_i_2 j_2w_i_1w_j_1)=det(eta_ab)eta^ijw_i w_j=frac12epsilon^i_1 i_2epsilon^j_1 j_2eta_i_1 j_1eta_i_2 j_2eta^alpha betaw_alphaw_beta$
Now, in my understanding the terms in the brackets on the l.h.s are equal to eachother after summing with the antisymmetric symbols, so all that's left to do is show, using only valid index juggling tricks, that
$eta_i_1 j_1w_i_2w_j_2$ can be written as $frac12eta_i_1 j_1eta_i_2 j_2eta^alpha betaw_alphaw_beta$. This is where I'm stuck, however of course I could have made (multiple) mistakes before arriving there but I can't seem to find it and any help would be greatly appreciated.
Edit: I got it now and it rests upon some property of the antisymmetric symbol; $epsilon^a b c_a i c _b k= epsilon_i k det (c) $ but I'm too lazy too type it and I'm not allowed to post pictures yet.
determinant index-notation
$endgroup$
add a comment |
$begingroup$
I was practicing some old tensor calculus exams and had to answer this question (paraphrased, it has multiple subquestions):
If $eta_ij$ is the component of the metric $H$ on a 2D Riemannian manifold (i.e. $H=eta_ij dx^i otimes dx^j$) and $w_i$ is the componont of a smooth covector field $Omega=w_i dx^i$, construct a symmetric tensor $Gamma=gamma_ijdx^i otimes dx^j=(eta_ij-w_i w_j)dx^i otimes dx^j$.
Show that $det (gamma_ij)$ is equal to $det(eta_ij) (1-eta^ijw_i w_j)$.
In the solutions the answer is just "A direct computation shows this". I have verified the correctness of the final formula by just bruteforcing the summations in 2D, but I want to be able to verify this in index notation. So far I've found two threads which contain (a variant) of this formula: Formula for determinant of sum of matrices and Expressing the determinant of a sum of two matrices?. However, one proves it using standard matrix notation and the other only proves it in the n-dimensional cases which I found a quite hard to follow since this is my first actual mathematics course (physicist). (The formula is, for an invertible matrix A, $det(A+B)=det(A)+det(B)+trace(A^-1 B)$, this is just a special case of that formula)
So far I've got this:
$det(gamma_ij)=frac12epsilon^i_1 i_2epsilon^j_1 j_2gamma_i_1 j_1gamma_i_2 j_2=frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1-w_i_1w_j_1)(eta_i_2 j_2-w_i_2w_j_2)=frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1eta_i_2 j_2+w_i_1w_j_1w_i_2w_j_2-eta_i_1 j_1w_i_2w_j_2-eta_i_2 j_2w_i_1w_j_1)=$
$=det(eta_ij)+det(w_i w_j)-frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1w_i_2w_j_2+eta_i_2 j_2w_i_1w_j_1)$
The determinant of the ww matrix is obviously zero, so now all that's left to show is
$frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1w_i_2w_j_2+eta_i_2 j_2w_i_1w_j_1)=det(eta_ab)eta^ijw_i w_j=frac12epsilon^i_1 i_2epsilon^j_1 j_2eta_i_1 j_1eta_i_2 j_2eta^alpha betaw_alphaw_beta$
Now, in my understanding the terms in the brackets on the l.h.s are equal to eachother after summing with the antisymmetric symbols, so all that's left to do is show, using only valid index juggling tricks, that
$eta_i_1 j_1w_i_2w_j_2$ can be written as $frac12eta_i_1 j_1eta_i_2 j_2eta^alpha betaw_alphaw_beta$. This is where I'm stuck, however of course I could have made (multiple) mistakes before arriving there but I can't seem to find it and any help would be greatly appreciated.
Edit: I got it now and it rests upon some property of the antisymmetric symbol; $epsilon^a b c_a i c _b k= epsilon_i k det (c) $ but I'm too lazy too type it and I'm not allowed to post pictures yet.
determinant index-notation
$endgroup$
add a comment |
$begingroup$
I was practicing some old tensor calculus exams and had to answer this question (paraphrased, it has multiple subquestions):
If $eta_ij$ is the component of the metric $H$ on a 2D Riemannian manifold (i.e. $H=eta_ij dx^i otimes dx^j$) and $w_i$ is the componont of a smooth covector field $Omega=w_i dx^i$, construct a symmetric tensor $Gamma=gamma_ijdx^i otimes dx^j=(eta_ij-w_i w_j)dx^i otimes dx^j$.
Show that $det (gamma_ij)$ is equal to $det(eta_ij) (1-eta^ijw_i w_j)$.
In the solutions the answer is just "A direct computation shows this". I have verified the correctness of the final formula by just bruteforcing the summations in 2D, but I want to be able to verify this in index notation. So far I've found two threads which contain (a variant) of this formula: Formula for determinant of sum of matrices and Expressing the determinant of a sum of two matrices?. However, one proves it using standard matrix notation and the other only proves it in the n-dimensional cases which I found a quite hard to follow since this is my first actual mathematics course (physicist). (The formula is, for an invertible matrix A, $det(A+B)=det(A)+det(B)+trace(A^-1 B)$, this is just a special case of that formula)
So far I've got this:
$det(gamma_ij)=frac12epsilon^i_1 i_2epsilon^j_1 j_2gamma_i_1 j_1gamma_i_2 j_2=frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1-w_i_1w_j_1)(eta_i_2 j_2-w_i_2w_j_2)=frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1eta_i_2 j_2+w_i_1w_j_1w_i_2w_j_2-eta_i_1 j_1w_i_2w_j_2-eta_i_2 j_2w_i_1w_j_1)=$
$=det(eta_ij)+det(w_i w_j)-frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1w_i_2w_j_2+eta_i_2 j_2w_i_1w_j_1)$
The determinant of the ww matrix is obviously zero, so now all that's left to show is
$frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1w_i_2w_j_2+eta_i_2 j_2w_i_1w_j_1)=det(eta_ab)eta^ijw_i w_j=frac12epsilon^i_1 i_2epsilon^j_1 j_2eta_i_1 j_1eta_i_2 j_2eta^alpha betaw_alphaw_beta$
Now, in my understanding the terms in the brackets on the l.h.s are equal to eachother after summing with the antisymmetric symbols, so all that's left to do is show, using only valid index juggling tricks, that
$eta_i_1 j_1w_i_2w_j_2$ can be written as $frac12eta_i_1 j_1eta_i_2 j_2eta^alpha betaw_alphaw_beta$. This is where I'm stuck, however of course I could have made (multiple) mistakes before arriving there but I can't seem to find it and any help would be greatly appreciated.
Edit: I got it now and it rests upon some property of the antisymmetric symbol; $epsilon^a b c_a i c _b k= epsilon_i k det (c) $ but I'm too lazy too type it and I'm not allowed to post pictures yet.
determinant index-notation
$endgroup$
I was practicing some old tensor calculus exams and had to answer this question (paraphrased, it has multiple subquestions):
If $eta_ij$ is the component of the metric $H$ on a 2D Riemannian manifold (i.e. $H=eta_ij dx^i otimes dx^j$) and $w_i$ is the componont of a smooth covector field $Omega=w_i dx^i$, construct a symmetric tensor $Gamma=gamma_ijdx^i otimes dx^j=(eta_ij-w_i w_j)dx^i otimes dx^j$.
Show that $det (gamma_ij)$ is equal to $det(eta_ij) (1-eta^ijw_i w_j)$.
In the solutions the answer is just "A direct computation shows this". I have verified the correctness of the final formula by just bruteforcing the summations in 2D, but I want to be able to verify this in index notation. So far I've found two threads which contain (a variant) of this formula: Formula for determinant of sum of matrices and Expressing the determinant of a sum of two matrices?. However, one proves it using standard matrix notation and the other only proves it in the n-dimensional cases which I found a quite hard to follow since this is my first actual mathematics course (physicist). (The formula is, for an invertible matrix A, $det(A+B)=det(A)+det(B)+trace(A^-1 B)$, this is just a special case of that formula)
So far I've got this:
$det(gamma_ij)=frac12epsilon^i_1 i_2epsilon^j_1 j_2gamma_i_1 j_1gamma_i_2 j_2=frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1-w_i_1w_j_1)(eta_i_2 j_2-w_i_2w_j_2)=frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1eta_i_2 j_2+w_i_1w_j_1w_i_2w_j_2-eta_i_1 j_1w_i_2w_j_2-eta_i_2 j_2w_i_1w_j_1)=$
$=det(eta_ij)+det(w_i w_j)-frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1w_i_2w_j_2+eta_i_2 j_2w_i_1w_j_1)$
The determinant of the ww matrix is obviously zero, so now all that's left to show is
$frac12epsilon^i_1 i_2epsilon^j_1 j_2(eta_i_1 j_1w_i_2w_j_2+eta_i_2 j_2w_i_1w_j_1)=det(eta_ab)eta^ijw_i w_j=frac12epsilon^i_1 i_2epsilon^j_1 j_2eta_i_1 j_1eta_i_2 j_2eta^alpha betaw_alphaw_beta$
Now, in my understanding the terms in the brackets on the l.h.s are equal to eachother after summing with the antisymmetric symbols, so all that's left to do is show, using only valid index juggling tricks, that
$eta_i_1 j_1w_i_2w_j_2$ can be written as $frac12eta_i_1 j_1eta_i_2 j_2eta^alpha betaw_alphaw_beta$. This is where I'm stuck, however of course I could have made (multiple) mistakes before arriving there but I can't seem to find it and any help would be greatly appreciated.
Edit: I got it now and it rests upon some property of the antisymmetric symbol; $epsilon^a b c_a i c _b k= epsilon_i k det (c) $ but I'm too lazy too type it and I'm not allowed to post pictures yet.
determinant index-notation
determinant index-notation
edited Apr 1 at 22:06
Cipier
asked Apr 1 at 18:56
CipierCipier
62
62
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