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Inner and Outer Product of 2 rank tensors
What does a zero tensor product imply?Is the tensor product of two torsion-free modules always non-zero?Universal Property of Tensor AlgebraTensor Product and Direct Sumcan quaternions be expressed in terms of tensor products?Product of Symmetric and Antisymmetric TensorsFunctor $R^2 otimes -$ for a ring $R$ is left-exact?Deduce that relation must hold for all unitary matricesTrace optimization of a function of a left stochastic matrixQuestion about creating a 3-rank tensor in numpy with python3
$begingroup$
I am trying to figure out whether or not the following identity is true:
$D_klA_ijA_kl=A_klD_klA_ij$
Or what I think is qually:
$D:(Aotimes A)= (A:D)A$
The matrices are both symmetric $A=A^T, B=B^T$.
I tried to "prove" it by simply doing the calculus with 2x2 matrices but I always end up wit another (wrong) result. I would very much appreciate it if you could help me or give me some hints whether this is right or wrong.
tensor-products matrix-calculus
asked Mar 4 at 18:54
Pablo JekenPablo Jeken
1084
$endgroup$
add a comment |
$begingroup$
I am trying to figure out whether or not the following identity is true:
$D_klA_ijA_kl=A_klD_klA_ij$
Or what I think is qually:
$D:(Aotimes A)= (A:D)A$
The matrices are both symmetric $A=A^T, B=B^T$.
I tried to "prove" it by simply doing the calculus with 2x2 matrices but I always end up wit another (wrong) result. I would very much appreciate it if you could help me or give me some hints whether this is right or wrong.
tensor-products matrix-calculus
asked Mar 4 at 18:54
Pablo JekenPablo Jeken
1084
$endgroup$
add a comment |
$begingroup$
I am trying to figure out whether or not the following identity is true:
$D_klA_ijA_kl=A_klD_klA_ij$
Or what I think is qually:
$D:(Aotimes A)= (A:D)A$
The matrices are both symmetric $A=A^T, B=B^T$.
I tried to "prove" it by simply doing the calculus with 2x2 matrices but I always end up wit another (wrong) result. I would very much appreciate it if you could help me or give me some hints whether this is right or wrong.
tensor-products matrix-calculus
asked Mar 4 at 18:54
Pablo JekenPablo Jeken
1084
$endgroup$
I am trying to figure out whether or not the following identity is true:
$D_klA_ijA_kl=A_klD_klA_ij$
Or what I think is qually:
$D:(Aotimes A)= (A:D)A$
The matrices are both symmetric $A=A^T, B=B^T$.
I tried to "prove" it by simply doing the calculus with 2x2 matrices but I always end up wit another (wrong) result. I would very much appreciate it if you could help me or give me some hints whether this is right or wrong.
tensor-products matrix-calculus
tensor-products matrix-calculus
asked Mar 4 at 18:54
Pablo JekenPablo Jeken
1084
asked Mar 4 at 18:54
Pablo JekenPablo Jeken
1084
asked Mar 4 at 18:54
Pablo JekenPablo Jeken
1084
asked Mar 4 at 18:54
Pablo JekenPablo Jeken
1084
asked Mar 4 at 18:54
Pablo JekenPablo Jeken
1084
1084
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
When employing index notation, you are effectively just summing over repeated indices, so if the resulting matrix is $M$, then
$$ M_ij = sum_k=1^n sum_l=1^n D_klA_ijA_kl= sum_k=1^n sum_l=1^nA_klD_klA_ij = sum_k=1^n sum_l=1^n A_ijD_klA_kl= cdots
$$
The terms being multiplied inside the summation can be taken in any order.
Thus, given definitions for "$otimes$" and "$:$" you can come up with different interpretations to an operation which is effectively just summing up the product of elements in two matrices/tensors in a certain order. All this provided the dimensions of the matrices allow for such an operation.
answered 2 days ago
TrawsTraws
6112
New contributor
Traws is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
When employing index notation, you are effectively just summing over repeated indices, so if the resulting matrix is $M$, then
$$ M_ij = sum_k=1^n sum_l=1^n D_klA_ijA_kl= sum_k=1^n sum_l=1^nA_klD_klA_ij = sum_k=1^n sum_l=1^n A_ijD_klA_kl= cdots
$$
The terms being multiplied inside the summation can be taken in any order.
Thus, given definitions for "$otimes$" and "$:$" you can come up with different interpretations to an operation which is effectively just summing up the product of elements in two matrices/tensors in a certain order. All this provided the dimensions of the matrices allow for such an operation.
answered 2 days ago
TrawsTraws
6112
New contributor
Traws is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
When employing index notation, you are effectively just summing over repeated indices, so if the resulting matrix is $M$, then
$$ M_ij = sum_k=1^n sum_l=1^n D_klA_ijA_kl= sum_k=1^n sum_l=1^nA_klD_klA_ij = sum_k=1^n sum_l=1^n A_ijD_klA_kl= cdots
$$
The terms being multiplied inside the summation can be taken in any order.
Thus, given definitions for "$otimes$" and "$:$" you can come up with different interpretations to an operation which is effectively just summing up the product of elements in two matrices/tensors in a certain order. All this provided the dimensions of the matrices allow for such an operation.
answered 2 days ago
TrawsTraws
6112
New contributor
Traws is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
When employing index notation, you are effectively just summing over repeated indices, so if the resulting matrix is $M$, then
$$ M_ij = sum_k=1^n sum_l=1^n D_klA_ijA_kl= sum_k=1^n sum_l=1^nA_klD_klA_ij = sum_k=1^n sum_l=1^n A_ijD_klA_kl= cdots
$$
The terms being multiplied inside the summation can be taken in any order.
Thus, given definitions for "$otimes$" and "$:$" you can come up with different interpretations to an operation which is effectively just summing up the product of elements in two matrices/tensors in a certain order. All this provided the dimensions of the matrices allow for such an operation.
answered 2 days ago
TrawsTraws
6112
New contributor
Traws is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
When employing index notation, you are effectively just summing over repeated indices, so if the resulting matrix is $M$, then
$$ M_ij = sum_k=1^n sum_l=1^n D_klA_ijA_kl= sum_k=1^n sum_l=1^nA_klD_klA_ij = sum_k=1^n sum_l=1^n A_ijD_klA_kl= cdots
$$
The terms being multiplied inside the summation can be taken in any order.
Thus, given definitions for "$otimes$" and "$:$" you can come up with different interpretations to an operation which is effectively just summing up the product of elements in two matrices/tensors in a certain order. All this provided the dimensions of the matrices allow for such an operation.
answered 2 days ago
TrawsTraws
6112
New contributor
Traws is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
TrawsTraws
6112
New contributor
Traws is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
TrawsTraws
6112
answered 2 days ago
TrawsTraws
6112
6112
New contributor
Traws is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Traws is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Traws is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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