Suppose A is a rank-n tensor.
Question 1: What is the name given to the embedded rank (n-1) tensor
Z found along the diagonal of A? (assume that dim(A) = dim(Z)^2 )

Question 2: What is the symbology/formula used to extract this
diagonal (n-1) tensor?

Example:
Suppose A was the rank-2 tensor:

A = [1, 2, 3]
[4, 5, 6]
[7, 8, 9]

The "embedded diagonal rank-1 tensor" here is [1, 5, 9]. What is the
proper name for this entity?

And for the other question:
If,
A = B x C
where
x denotes tensor product
B is an n/2 rank tensor with dimension SQRT(dim(A))
C is an n/2 rank tensor with dimension SQRT(dim(A))

How can I extract this "embedded diagonal rank (n-1) tensor" from A?

Any help appreciated.

Schoenfeld wrote:
Suppose A was the rank-2 tensor:

A = [1, 2, 3]
[4, 5, 6]
[7, 8, 9]

The "embedded diagonal rank-1 tensor" here is [1, 5, 9]. What is the
proper name for this entity?

'Not a tensor', as another poster already told you. A rank 1 tensor
has to be either a vector or a covector, and this is neither. It
isn't a geometric object of any sort, since knowing the numbers
[1,5,9] isn't enough information to specify the diagonal
of A under a change of basis---other components of A get mixed
in with them.

Robert Low wrote:
Schoenfeld wrote:
Suppose A was the rank-2 tensor:

A = [1, 2, 3]
[4, 5, 6]
[7, 8, 9]

The "embedded diagonal rank-1 tensor" here is [1, 5, 9]. What is the
proper name for this entity?

'Not a tensor', as another poster already told you. A rank 1 tensor
has to be either a vector or a covector, and this is neither.

It is something relevant, but obviously there is no point continuing
discussion. Thanks for you help, please avoid any more in future.

It
isn't a geometric object of any sort, since knowing the numbers
[1,5,9] isn't enough information to specify the diagonal
of A under a change of basis---other components of A get mixed
in with them.

Schoenfeld wrote:
Robert Low wrote:
Schoenfeld wrote:
'Not a tensor', as another poster already told you. A rank 1 tensor
has to be either a vector or a covector, and this is neither.
It is something relevant, but obviously there is no point continuing
discussion.

It may well be something relevant, but that something isn't a
tensor. There's no point in getting indignant because of that.
Maybe if you explained what you want it for it would be easier
to provide you with more useful information.

Robert Low wrote:
Schoenfeld wrote:

Robert Low wrote:

Schoenfeld wrote:
'Not a tensor', as another poster already told you. A rank 1 tensor
has to be either a vector or a covector, and this is neither.

It is something relevant, but obviously there is no point continuing
discussion.


It may well be something relevant, but that something isn't a
tensor. There's no point in getting indignant because of that.
Maybe if you explained what you want it for it would be easier
to provide you with more useful information.

Weren't you just told "Thanks for you help, please avoid any more in
future."?

I understand that school's back in session, but what happened to manners?

Schoenfeld wrote:
Suppose A is a rank-n tensor.
Question 1: What is the name given to the embedded rank (n-1) tensor
Z found along the diagonal of A? (assume that dim(A) = dim(Z)^2 )

Question 2: What is the symbology/formula used to extract this
diagonal (n-1) tensor?

Example:
Suppose A was the rank-2 tensor:

A = [1, 2, 3]
[4, 5, 6]
[7, 8, 9]

The "embedded diagonal rank-1 tensor" here is [1, 5, 9]. What is the
proper name for this entity?

As others have said, this isn't a rank 1 tensor. But I would call it
the "diagonal entries".

Incidently, if A is a mixed rank 2 tensor, then 1+5+9 is a rank 0
tensor. This is called the "trace."

Does this help?

Stephen

TMG wrote:
Weren't you just told "Thanks for you help, please avoid any more in
future."?
I understand that school's back in session, but what happened to manners?

Anybody whose response to being told they're wrong is to tell you
to be quiet needs all the help you can throw at them :-)

Robert Low wrote:
TMG wrote:
Weren't you just told "Thanks for you help, please avoid any more in
future."?
I understand that school's back in session, but what happened to manners?

Anybody whose response to being told they're wrong is to tell you
to be quiet needs all the help you can throw at them :-)

Robert, I understand these components can be used to form such n-1
tensor, this is why i called it "embedded". In the same way the
components of a vector can be used to form a scalar. This is clearly
what I implied. That you persist with your objection implies this
conversation is useless and thread unfortunately corrupted.

Stephen Montgomery-Smith wrote:
Schoenfeld wrote:
Suppose A is a rank-n tensor.
Question 1: What is the name given to the embedded rank (n-1) tensor
Z found along the diagonal of A? (assume that dim(A) = dim(Z)^2 )

Question 2: What is the symbology/formula used to extract this
diagonal (n-1) tensor?

Example:
Suppose A was the rank-2 tensor:

A = [1, 2, 3]
[4, 5, 6]
[7, 8, 9]

The "embedded diagonal rank-1 tensor" here is [1, 5, 9]. What is the
proper name for this entity?

As others have said, this isn't a rank 1 tensor. But I would call it
the "diagonal entries".

I clearly said it was "embedded" implying that the resultant tensor can
be easily constructed from the collection of components found along
this diagonal.

In this instance, the vector here is simply Z = SUM(i=1)^3 A_ii Bi {Bi
= i'th basis vector).


Incidently, if A is a mixed rank 2 tensor, then 1+5+9 is a rank 0
tensor. This is called the "trace."

Does this help?

No, because trace(A) = Z . SUM{i=1}^(SQRT(dim(A))) Bi). I needed Z not
the trace(A). I found the solution anyway, it's related to the
hardamard product. Thanks anyway.


Stephen

Schoenfeld wrote:
Stephen Montgomery-Smith wrote:
Schoenfeld wrote:
Suppose A is a rank-n tensor.
Question 1: What is the name given to the embedded rank (n-1) tensor
Z found along the diagonal of A? (assume that dim(A) = dim(Z)^2 )

Question 2: What is the symbology/formula used to extract this
diagonal (n-1) tensor?

Example:
Suppose A was the rank-2 tensor:

A = [1, 2, 3]
[4, 5, 6]
[7, 8, 9]

The "embedded diagonal rank-1 tensor" here is [1, 5, 9]. What is the
proper name for this entity?

As others have said, this isn't a rank 1 tensor. But I would call it
the "diagonal entries".

I clearly said it was "embedded" implying that the resultant tensor can
be easily constructed from the collection of components found along
this diagonal.

In this instance, the vector here is simply Z = SUM(i=1)^3 A_ii Bi {Bi
= i'th basis vector).


Incidently, if A is a mixed rank 2 tensor, then 1+5+9 is a rank 0
tensor. This is called the "trace."

Does this help?

No, because trace(A) = Z . SUM{i=1}^(SQRT(dim(A))) Bi). I needed Z not
the trace(A). I found the solution anyway, it's related to the
hardamard product. Thanks anyway.

typo: hadamard product


Stephen