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Matricial product not commutative unlike tensor 2D
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I would like to know why,as a general rule and which under conditions, one says that tensor product is commutative unlike matricial product.
For example, for a product of 2 matrices, we get the element (i,j) by writing (with Einstein notations) :
$$C_ij = A_ik B_kj$$
and
$$D_ij = B_ik A_kj$$
So $C_ij neq D_ij$
But with tensors, it seems that it doesn't matter to multiply the first tensor by the second or the contrary :
$$C_ik= A_ilB_lk = B_lkA_il=B_klA_li=D_ki$$
You can notice that I have supposed A and B as symetric tensors.
What are the rules for a tensor to respect for not consider the order in multiplications and get commutative product (unlike matricial product) ?
Taking symetric tensors is sufficient ? I don't think so since there are also antsymetric tensors (like Maxwell tensor).
If someone could give me a simple example highlighting this issue ?
Any clarifications is welcome, regards.
tensor-products tensors matrix-calculus
asked Jul 29 at 0:21
youpilat13
6911
 |Â
show 1 more comment
up vote
0
down vote
favorite
I would like to know why,as a general rule and which under conditions, one says that tensor product is commutative unlike matricial product.
For example, for a product of 2 matrices, we get the element (i,j) by writing (with Einstein notations) :
$$C_ij = A_ik B_kj$$
and
$$D_ij = B_ik A_kj$$
So $C_ij neq D_ij$
But with tensors, it seems that it doesn't matter to multiply the first tensor by the second or the contrary :
$$C_ik= A_ilB_lk = B_lkA_il=B_klA_li=D_ki$$
You can notice that I have supposed A and B as symetric tensors.
What are the rules for a tensor to respect for not consider the order in multiplications and get commutative product (unlike matricial product) ?
Taking symetric tensors is sufficient ? I don't think so since there are also antsymetric tensors (like Maxwell tensor).
If someone could give me a simple example highlighting this issue ?
Any clarifications is welcome, regards.
tensor-products tensors matrix-calculus
asked Jul 29 at 0:21
youpilat13
6911
Your "tensor product" is the same as the matrix product. The actual tensor product would have 4 indices for $C$ (so the result is not a matrix): $$C_ijkl = A_ijB_kl$$
– mr_e_man
Jul 29 at 23:54
-@mr_e_man it may be a contraction between $j$ and $k$, such that we could write : $C_ijkl = C_il=A_inB_nl$, couldn't we ?
– youpilat13
Jul 29 at 23:59
$C$ is not equal to its contraction! en.wikipedia.org/wiki/Tensor#Contraction
– mr_e_man
Jul 30 at 0:01
The first contraction of $C$ is the matrix product, and the second contraction is the trace $texttr(AB)$.
– mr_e_man
Jul 30 at 0:03
-@mr_e_man ok, you're right. I can't find an example showing the cases where order of factors matters (for matrices) and those when order doesn't matter (like for tensors), could you illustrate please this issue ?
– youpilat13
Jul 30 at 0:26
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I would like to know why,as a general rule and which under conditions, one says that tensor product is commutative unlike matricial product.
For example, for a product of 2 matrices, we get the element (i,j) by writing (with Einstein notations) :
$$C_ij = A_ik B_kj$$
and
$$D_ij = B_ik A_kj$$
So $C_ij neq D_ij$
But with tensors, it seems that it doesn't matter to multiply the first tensor by the second or the contrary :
$$C_ik= A_ilB_lk = B_lkA_il=B_klA_li=D_ki$$
You can notice that I have supposed A and B as symetric tensors.
What are the rules for a tensor to respect for not consider the order in multiplications and get commutative product (unlike matricial product) ?
Taking symetric tensors is sufficient ? I don't think so since there are also antsymetric tensors (like Maxwell tensor).
If someone could give me a simple example highlighting this issue ?
Any clarifications is welcome, regards.
tensor-products tensors matrix-calculus
asked Jul 29 at 0:21
youpilat13
6911
I would like to know why,as a general rule and which under conditions, one says that tensor product is commutative unlike matricial product.
For example, for a product of 2 matrices, we get the element (i,j) by writing (with Einstein notations) :
$$C_ij = A_ik B_kj$$
and
$$D_ij = B_ik A_kj$$
So $C_ij neq D_ij$
But with tensors, it seems that it doesn't matter to multiply the first tensor by the second or the contrary :
$$C_ik= A_ilB_lk = B_lkA_il=B_klA_li=D_ki$$
You can notice that I have supposed A and B as symetric tensors.
What are the rules for a tensor to respect for not consider the order in multiplications and get commutative product (unlike matricial product) ?
Taking symetric tensors is sufficient ? I don't think so since there are also antsymetric tensors (like Maxwell tensor).
If someone could give me a simple example highlighting this issue ?
Any clarifications is welcome, regards.
tensor-products tensors matrix-calculus
asked Jul 29 at 0:21
youpilat13
6911
edited Jul 29 at 23:40
asked Jul 29 at 0:21
youpilat13
6911
asked Jul 29 at 0:21
youpilat13
6911
asked Jul 29 at 0:21
youpilat13
6911
6911
Your "tensor product" is the same as the matrix product. The actual tensor product would have 4 indices for $C$ (so the result is not a matrix): $$C_ijkl = A_ijB_kl$$
– mr_e_man
Jul 29 at 23:54
-@mr_e_man it may be a contraction between $j$ and $k$, such that we could write : $C_ijkl = C_il=A_inB_nl$, couldn't we ?
– youpilat13
Jul 29 at 23:59
$C$ is not equal to its contraction! en.wikipedia.org/wiki/Tensor#Contraction
– mr_e_man
Jul 30 at 0:01
The first contraction of $C$ is the matrix product, and the second contraction is the trace $texttr(AB)$.
– mr_e_man
Jul 30 at 0:03
-@mr_e_man ok, you're right. I can't find an example showing the cases where order of factors matters (for matrices) and those when order doesn't matter (like for tensors), could you illustrate please this issue ?
– youpilat13
Jul 30 at 0:26
 |Â
show 1 more comment
Your "tensor product" is the same as the matrix product. The actual tensor product would have 4 indices for $C$ (so the result is not a matrix): $$C_ijkl = A_ijB_kl$$
– mr_e_man
Jul 29 at 23:54
-@mr_e_man it may be a contraction between $j$ and $k$, such that we could write : $C_ijkl = C_il=A_inB_nl$, couldn't we ?
– youpilat13
Jul 29 at 23:59
$C$ is not equal to its contraction! en.wikipedia.org/wiki/Tensor#Contraction
– mr_e_man
Jul 30 at 0:01
The first contraction of $C$ is the matrix product, and the second contraction is the trace $texttr(AB)$.
– mr_e_man
Jul 30 at 0:03
-@mr_e_man ok, you're right. I can't find an example showing the cases where order of factors matters (for matrices) and those when order doesn't matter (like for tensors), could you illustrate please this issue ?
– youpilat13
Jul 30 at 0:26
Your "tensor product" is the same as the matrix product. The actual tensor product would have 4 indices for $C$ (so the result is not a matrix): $$C_ijkl = A_ijB_kl$$
– mr_e_man
Jul 29 at 23:54
Your "tensor product" is the same as the matrix product. The actual tensor product would have 4 indices for $C$ (so the result is not a matrix): $$C_ijkl = A_ijB_kl$$
– mr_e_man
Jul 29 at 23:54
-@mr_e_man it may be a contraction between $j$ and $k$, such that we could write : $C_ijkl = C_il=A_inB_nl$, couldn't we ?
– youpilat13
Jul 29 at 23:59
-@mr_e_man it may be a contraction between $j$ and $k$, such that we could write : $C_ijkl = C_il=A_inB_nl$, couldn't we ?
– youpilat13
Jul 29 at 23:59
$C$ is not equal to its contraction! en.wikipedia.org/wiki/Tensor#Contraction
– mr_e_man
Jul 30 at 0:01
$C$ is not equal to its contraction! en.wikipedia.org/wiki/Tensor#Contraction
– mr_e_man
Jul 30 at 0:01
The first contraction of $C$ is the matrix product, and the second contraction is the trace $texttr(AB)$.
– mr_e_man
Jul 30 at 0:03
The first contraction of $C$ is the matrix product, and the second contraction is the trace $texttr(AB)$.
– mr_e_man
Jul 30 at 0:03
-@mr_e_man ok, you're right. I can't find an example showing the cases where order of factors matters (for matrices) and those when order doesn't matter (like for tensors), could you illustrate please this issue ?
– youpilat13
Jul 30 at 0:26
-@mr_e_man ok, you're right. I can't find an example showing the cases where order of factors matters (for matrices) and those when order doesn't matter (like for tensors), could you illustrate please this issue ?
– youpilat13
Jul 30 at 0:26
 |Â
show 1 more comment
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Your "tensor product" is the same as the matrix product. The actual tensor product would have 4 indices for $C$ (so the result is not a matrix): $$C_ijkl = A_ijB_kl$$
– mr_e_man
Jul 29 at 23:54
-@mr_e_man it may be a contraction between $j$ and $k$, such that we could write : $C_ijkl = C_il=A_inB_nl$, couldn't we ?
– youpilat13
Jul 29 at 23:59
$C$ is not equal to its contraction! en.wikipedia.org/wiki/Tensor#Contraction
– mr_e_man
Jul 30 at 0:01
The first contraction of $C$ is the matrix product, and the second contraction is the trace $texttr(AB)$.
– mr_e_man
Jul 30 at 0:03
-@mr_e_man ok, you're right. I can't find an example showing the cases where order of factors matters (for matrices) and those when order doesn't matter (like for tensors), could you illustrate please this issue ?
– youpilat13
Jul 30 at 0:26