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The tensor $epsilon_{ijk}$ is related to determinants?
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My textbook says the following in an appendix on tensor notation:
The tensor $epsilon_{rst} = begin{cases}
0 qquad & text{unless $r, s,$ and $t$ are distinct} \
+1 qquad & text{if $rst$ is an even permutation of $123$} \
-1 qquad & text{if $rst$ is an odd permutation of $123$}
end{cases}$
The tensor $epsilon_{ijk}$ is related to determinants: for three contravariant tensors $a^i$, $b^j$, and $c^k$, one verifies that $a^i b^j c^k epsilon_{ijk}$ is the determinant of the $3 times 3$ matrix with rows $a^i$, $b^j$, and $c^k$.
I don't understand what this excerpt is saying:
The tensor $epsilon_{ijk}$ is related to determinants: for three contravariant tensors $a^i$, $b^j$, and $c^k$, one verifies that $a^i b^j c^k epsilon_{ijk}$ is the determinant of the $3 times 3$ matrix with rows $a^i$, $b^j$, and $c^k$.
I would greatly appreciate it if people could please take the time to elaborate on this and demonstrate what is being referred to.
linear-algebra determinant tensors
asked 22 hours ago
The Pointer
2,73121232
add a comment |
up vote
0
down vote
favorite
My textbook says the following in an appendix on tensor notation:
The tensor $epsilon_{rst} = begin{cases}
0 qquad & text{unless $r, s,$ and $t$ are distinct} \
+1 qquad & text{if $rst$ is an even permutation of $123$} \
-1 qquad & text{if $rst$ is an odd permutation of $123$}
end{cases}$
The tensor $epsilon_{ijk}$ is related to determinants: for three contravariant tensors $a^i$, $b^j$, and $c^k$, one verifies that $a^i b^j c^k epsilon_{ijk}$ is the determinant of the $3 times 3$ matrix with rows $a^i$, $b^j$, and $c^k$.
I don't understand what this excerpt is saying:
The tensor $epsilon_{ijk}$ is related to determinants: for three contravariant tensors $a^i$, $b^j$, and $c^k$, one verifies that $a^i b^j c^k epsilon_{ijk}$ is the determinant of the $3 times 3$ matrix with rows $a^i$, $b^j$, and $c^k$.
I would greatly appreciate it if people could please take the time to elaborate on this and demonstrate what is being referred to.
linear-algebra determinant tensors
asked 22 hours ago
The Pointer
2,73121232
Are you familiar with Einsteins notation of sums?
– maxmilgram
22 hours ago
@maxmilgram I've encountered them before and scantily remember them. If you post an answer with them, as long as it's basic Einstein summation notation, I should be able to quickly review what I need to know to understand.
– The Pointer
22 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My textbook says the following in an appendix on tensor notation:
The tensor $epsilon_{rst} = begin{cases}
0 qquad & text{unless $r, s,$ and $t$ are distinct} \
+1 qquad & text{if $rst$ is an even permutation of $123$} \
-1 qquad & text{if $rst$ is an odd permutation of $123$}
end{cases}$
The tensor $epsilon_{ijk}$ is related to determinants: for three contravariant tensors $a^i$, $b^j$, and $c^k$, one verifies that $a^i b^j c^k epsilon_{ijk}$ is the determinant of the $3 times 3$ matrix with rows $a^i$, $b^j$, and $c^k$.
I don't understand what this excerpt is saying:
The tensor $epsilon_{ijk}$ is related to determinants: for three contravariant tensors $a^i$, $b^j$, and $c^k$, one verifies that $a^i b^j c^k epsilon_{ijk}$ is the determinant of the $3 times 3$ matrix with rows $a^i$, $b^j$, and $c^k$.
I would greatly appreciate it if people could please take the time to elaborate on this and demonstrate what is being referred to.
linear-algebra determinant tensors
asked 22 hours ago
The Pointer
2,73121232
My textbook says the following in an appendix on tensor notation:
The tensor $epsilon_{rst} = begin{cases}
0 qquad & text{unless $r, s,$ and $t$ are distinct} \
+1 qquad & text{if $rst$ is an even permutation of $123$} \
-1 qquad & text{if $rst$ is an odd permutation of $123$}
end{cases}$
The tensor $epsilon_{ijk}$ is related to determinants: for three contravariant tensors $a^i$, $b^j$, and $c^k$, one verifies that $a^i b^j c^k epsilon_{ijk}$ is the determinant of the $3 times 3$ matrix with rows $a^i$, $b^j$, and $c^k$.
I don't understand what this excerpt is saying:
The tensor $epsilon_{ijk}$ is related to determinants: for three contravariant tensors $a^i$, $b^j$, and $c^k$, one verifies that $a^i b^j c^k epsilon_{ijk}$ is the determinant of the $3 times 3$ matrix with rows $a^i$, $b^j$, and $c^k$.
I would greatly appreciate it if people could please take the time to elaborate on this and demonstrate what is being referred to.
linear-algebra determinant tensors
linear-algebra determinant tensors
asked 22 hours ago
The Pointer
2,73121232
asked 22 hours ago
The Pointer
2,73121232
asked 22 hours ago
The Pointer
2,73121232
asked 22 hours ago
The Pointer
2,73121232
asked 22 hours ago
The Pointer
2,73121232
2,73121232
Are you familiar with Einsteins notation of sums?
– maxmilgram
22 hours ago
@maxmilgram I've encountered them before and scantily remember them. If you post an answer with them, as long as it's basic Einstein summation notation, I should be able to quickly review what I need to know to understand.
– The Pointer
22 hours ago
add a comment |
Are you familiar with Einsteins notation of sums?
– maxmilgram
22 hours ago
@maxmilgram I've encountered them before and scantily remember them. If you post an answer with them, as long as it's basic Einstein summation notation, I should be able to quickly review what I need to know to understand.
– The Pointer
22 hours ago
Are you familiar with Einsteins notation of sums?
– maxmilgram
22 hours ago
Are you familiar with Einsteins notation of sums?
– maxmilgram
22 hours ago
@maxmilgram I've encountered them before and scantily remember them. If you post an answer with them, as long as it's basic Einstein summation notation, I should be able to quickly review what I need to know to understand.
– The Pointer
22 hours ago
@maxmilgram I've encountered them before and scantily remember them. If you post an answer with them, as long as it's basic Einstein summation notation, I should be able to quickly review what I need to know to understand.
– The Pointer
22 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
If you look at the rule of sarrus for $3times3$ matrices you find that the determinant is a sum of products of the matrix elements https://en.wikipedia.org/wiki/Rule_of_Sarrus.
The formula $a^ib^jc^kepsilon_{ijk}$ gives exactly that: It is a sum of products of three matrix elements together with the tensor which is $0,-1$ or $1$.
answered 22 hours ago
maxmilgram
4227
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Thanks for the answer. With regards to $a^ib^jc^kepsilon_{ijk}$, is this actually Einstein summation notation? I was under the impression that the convention requires indices repeated in the contravariant and covariant positions, as in, for instance, $a^i_j x^j_i$, which would sum over all $i$'s and $j$'s? $a^ib^jc^kepsilon_{ijk}$ doesn't have the same covariant/contravariant "alternating" pattern of indices?
– The Pointer
22 hours ago
2
Yes it is. Every index appears once as subscript and once as superscript. They do not need to be in "alternating" order.
– maxmilgram
22 hours ago
Ahh, yes, I see: "According to this convention, when an index variable appears twice in a single term and is not otherwise defined (see free and bound variables), it implies summation of that term over all the values of the index." en.wikipedia.org/wiki/Einstein_notation
– The Pointer
22 hours ago
Thank you for the clarification! :)
– The Pointer
22 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If you look at the rule of sarrus for $3times3$ matrices you find that the determinant is a sum of products of the matrix elements https://en.wikipedia.org/wiki/Rule_of_Sarrus.
The formula $a^ib^jc^kepsilon_{ijk}$ gives exactly that: It is a sum of products of three matrix elements together with the tensor which is $0,-1$ or $1$.
answered 22 hours ago
maxmilgram
4227
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Thanks for the answer. With regards to $a^ib^jc^kepsilon_{ijk}$, is this actually Einstein summation notation? I was under the impression that the convention requires indices repeated in the contravariant and covariant positions, as in, for instance, $a^i_j x^j_i$, which would sum over all $i$'s and $j$'s? $a^ib^jc^kepsilon_{ijk}$ doesn't have the same covariant/contravariant "alternating" pattern of indices?
– The Pointer
22 hours ago
2
Yes it is. Every index appears once as subscript and once as superscript. They do not need to be in "alternating" order.
– maxmilgram
22 hours ago
Ahh, yes, I see: "According to this convention, when an index variable appears twice in a single term and is not otherwise defined (see free and bound variables), it implies summation of that term over all the values of the index." en.wikipedia.org/wiki/Einstein_notation
– The Pointer
22 hours ago
Thank you for the clarification! :)
– The Pointer
22 hours ago
add a comment |
up vote
1
down vote
accepted
If you look at the rule of sarrus for $3times3$ matrices you find that the determinant is a sum of products of the matrix elements https://en.wikipedia.org/wiki/Rule_of_Sarrus.
The formula $a^ib^jc^kepsilon_{ijk}$ gives exactly that: It is a sum of products of three matrix elements together with the tensor which is $0,-1$ or $1$.
answered 22 hours ago
maxmilgram
4227
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Thanks for the answer. With regards to $a^ib^jc^kepsilon_{ijk}$, is this actually Einstein summation notation? I was under the impression that the convention requires indices repeated in the contravariant and covariant positions, as in, for instance, $a^i_j x^j_i$, which would sum over all $i$'s and $j$'s? $a^ib^jc^kepsilon_{ijk}$ doesn't have the same covariant/contravariant "alternating" pattern of indices?
– The Pointer
22 hours ago
2
Yes it is. Every index appears once as subscript and once as superscript. They do not need to be in "alternating" order.
– maxmilgram
22 hours ago
Ahh, yes, I see: "According to this convention, when an index variable appears twice in a single term and is not otherwise defined (see free and bound variables), it implies summation of that term over all the values of the index." en.wikipedia.org/wiki/Einstein_notation
– The Pointer
22 hours ago
Thank you for the clarification! :)
– The Pointer
22 hours ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If you look at the rule of sarrus for $3times3$ matrices you find that the determinant is a sum of products of the matrix elements https://en.wikipedia.org/wiki/Rule_of_Sarrus.
The formula $a^ib^jc^kepsilon_{ijk}$ gives exactly that: It is a sum of products of three matrix elements together with the tensor which is $0,-1$ or $1$.
answered 22 hours ago
maxmilgram
4227
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If you look at the rule of sarrus for $3times3$ matrices you find that the determinant is a sum of products of the matrix elements https://en.wikipedia.org/wiki/Rule_of_Sarrus.
The formula $a^ib^jc^kepsilon_{ijk}$ gives exactly that: It is a sum of products of three matrix elements together with the tensor which is $0,-1$ or $1$.
answered 22 hours ago
maxmilgram
4227
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 22 hours ago
maxmilgram
4227
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 22 hours ago
maxmilgram
4227
answered 22 hours ago
maxmilgram
4227
4227
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Thanks for the answer. With regards to $a^ib^jc^kepsilon_{ijk}$, is this actually Einstein summation notation? I was under the impression that the convention requires indices repeated in the contravariant and covariant positions, as in, for instance, $a^i_j x^j_i$, which would sum over all $i$'s and $j$'s? $a^ib^jc^kepsilon_{ijk}$ doesn't have the same covariant/contravariant "alternating" pattern of indices?
– The Pointer
22 hours ago
2
Yes it is. Every index appears once as subscript and once as superscript. They do not need to be in "alternating" order.
– maxmilgram
22 hours ago
Ahh, yes, I see: "According to this convention, when an index variable appears twice in a single term and is not otherwise defined (see free and bound variables), it implies summation of that term over all the values of the index." en.wikipedia.org/wiki/Einstein_notation
– The Pointer
22 hours ago
Thank you for the clarification! :)
– The Pointer
22 hours ago
add a comment |
1
Thanks for the answer. With regards to $a^ib^jc^kepsilon_{ijk}$, is this actually Einstein summation notation? I was under the impression that the convention requires indices repeated in the contravariant and covariant positions, as in, for instance, $a^i_j x^j_i$, which would sum over all $i$'s and $j$'s? $a^ib^jc^kepsilon_{ijk}$ doesn't have the same covariant/contravariant "alternating" pattern of indices?
– The Pointer
22 hours ago
2
Yes it is. Every index appears once as subscript and once as superscript. They do not need to be in "alternating" order.
– maxmilgram
22 hours ago
Ahh, yes, I see: "According to this convention, when an index variable appears twice in a single term and is not otherwise defined (see free and bound variables), it implies summation of that term over all the values of the index." en.wikipedia.org/wiki/Einstein_notation
– The Pointer
22 hours ago
Thank you for the clarification! :)
– The Pointer
22 hours ago
1
1
Thanks for the answer. With regards to $a^ib^jc^kepsilon_{ijk}$, is this actually Einstein summation notation? I was under the impression that the convention requires indices repeated in the contravariant and covariant positions, as in, for instance, $a^i_j x^j_i$, which would sum over all $i$'s and $j$'s? $a^ib^jc^kepsilon_{ijk}$ doesn't have the same covariant/contravariant "alternating" pattern of indices?
– The Pointer
22 hours ago
Thanks for the answer. With regards to $a^ib^jc^kepsilon_{ijk}$, is this actually Einstein summation notation? I was under the impression that the convention requires indices repeated in the contravariant and covariant positions, as in, for instance, $a^i_j x^j_i$, which would sum over all $i$'s and $j$'s? $a^ib^jc^kepsilon_{ijk}$ doesn't have the same covariant/contravariant "alternating" pattern of indices?
– The Pointer
22 hours ago
2
2
Yes it is. Every index appears once as subscript and once as superscript. They do not need to be in "alternating" order.
– maxmilgram
22 hours ago
Yes it is. Every index appears once as subscript and once as superscript. They do not need to be in "alternating" order.
– maxmilgram
22 hours ago
Ahh, yes, I see: "According to this convention, when an index variable appears twice in a single term and is not otherwise defined (see free and bound variables), it implies summation of that term over all the values of the index." en.wikipedia.org/wiki/Einstein_notation
– The Pointer
22 hours ago
Ahh, yes, I see: "According to this convention, when an index variable appears twice in a single term and is not otherwise defined (see free and bound variables), it implies summation of that term over all the values of the index." en.wikipedia.org/wiki/Einstein_notation
– The Pointer
22 hours ago
Thank you for the clarification! :)
– The Pointer
22 hours ago
Thank you for the clarification! :)
– The Pointer
22 hours ago
add a comment |
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Are you familiar with Einsteins notation of sums?
– maxmilgram
22 hours ago
@maxmilgram I've encountered them before and scantily remember them. If you post an answer with them, as long as it's basic Einstein summation notation, I should be able to quickly review what I need to know to understand.
– The Pointer
22 hours ago