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? (always assume that dim(A) =
SQRT(dim(Z)) )

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 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? (always assume that dim(A) =
SQRT(dim(Z)) )

Typo: that should be
" Assume that dim(A) = dim(z)^2 "

Your question seems a bit premature to me - it quietly assumes things
that are not true.

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?

Unless I misunderstand your description, this embedded "tensor" is not
a tensor. You can easily cook up an example, say a rank-2 tensor T on
R^2:

T = [1 1]
[0 2]

The "embedded tensor" would be - well, here is your first problem -
would it be [1 2] or [1 2]-transpose?

It's easy to see neither of them transforms correctly under coordinate
change. Say you change coordinates by a matrix A:

A = [1 1]
[0 1]

Then T in the new coordinates is:

A^(-1)TA = [1 0]
[0 2]

....whose "embedded tensor" would be [1 2] or [1 2]-transpose,

neither of which corresponds to your original [1 2] or [1 2]-transpose
under none of A, A-transpose, A^(-1), (A^(-1))-transpose.

Excuse the tedium, there is probably a faster way to show this.

(always assume that dim(A) =
SQRT(dim(Z)) )

What do you need this for? The dimensionality of the space doesn't
matter.

So the answer to your remaining questions is "N/A".

--
Jan Bielawski

JanPB wrote:
Your question seems a bit premature to me - it quietly assumes things
that are not true.

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?

Unless I misunderstand your description, this embedded "tensor" is not
a tensor. You can easily cook up an example, say a rank-2 tensor T on
R^2:

T = [1 1]
[0 2]

The "embedded tensor" would be - well, here is your first problem -
would it be [1 2] or [1 2]-transpose?

Well it is a collection of components which can form an n-1 tensor just
like the components of a rank-1 tensor are rank-0 tensors. This is very
obvious. I am interested in a way to extract such components from
tensor product B x C = A and expressing this collection as an n-1
tensor. Raising red herrings will not achieve much, but as usual, one
needs to solve problems themselves to get anything done.

It's easy to see neither of them transforms correctly under coordinate
change. Say you change coordinates by a matrix A:

A = [1 1]
[0 1]

Then T in the new coordinates is:

A^(-1)TA = [1 0]
[0 2]

...whose "embedded tensor" would be [1 2] or [1 2]-transpose,

neither of which corresponds to your original [1 2] or [1 2]-transpose
under none of A, A-transpose, A^(-1), (A^(-1))-transpose.

Excuse the tedium, there is probably a faster way to show this.

(always assume that dim(A) =
SQRT(dim(Z)) )

What do you need this for? The dimensionality of the space doesn't
matter.

So the answer to your remaining questions is "N/A".

--
Jan Bielawski

Schoenfeld wrote:

Well it is a collection of components which can form an n-1 tensor just
like the components of a rank-1 tensor are rank-0 tensors.

But the components of a rank-1 tensor are not rank-0 tensors.

You might have a use for the diagonal of a matrix, (or
any other tensor), but it isn't a tensor. So you'll have
to be careful with it, because whatever property you
are interested in might be very different in a different
basis.

In article .com,
Schoenfeld wrote:

JanPB wrote:
Your question seems a bit premature to me - it quietly assumes things
that are not true.

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?

Unless I misunderstand your description, this embedded "tensor" is not
a tensor. You can easily cook up an example, say a rank-2 tensor T on
R^2:

T = [1 1]
[0 2]

The "embedded tensor" would be - well, here is your first problem -
would it be [1 2] or [1 2]-transpose?

Well it is a collection of components which can form an n-1 tensor just
like the components of a rank-1 tensor are rank-0 tensors. This is very
obvious. I am interested in a way to extract such components from
tensor product B x C = A and expressing this collection as an n-1
tensor. Raising red herrings will not achieve much, but as usual, one
needs to solve problems themselves to get anything done.

A = B x C can be thought of as a vector, an axis of rotation, in three
dimensions. In four or more dimensions there are an infinite number of
axes that are mutually perpendicular to B and C, so you can't reduce a
resulting rotation to an axis. You would express it as a plane of
rotation. E.g. not e1 x e2 = e3, but e1^e2.
--
"Then they placed the ark of the Lord on the cart; along with the box
containing the golden mice and the images of the hemorrhoids."
-- 1 Samuel 6:11

Schoenfeld wrote:
JanPB wrote:

The "embedded tensor" would be - well, here is your first problem -
would it be [1 2] or [1 2]-transpose?

Well it is a collection of components which can form an n-1 tensor just
like the components of a rank-1 tensor are rank-0 tensors.

But they aren't! A rank-0 tensor is a number that doesn't change under
choordinate changes while selected components of a vector (say) do
change their values.

--
Jan Bielawski

Gregory L. Hansen wrote:
In article .com,
Schoenfeld wrote:

JanPB wrote:
Your question seems a bit premature to me - it quietly assumes things
that are not true.

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?

Unless I misunderstand your description, this embedded "tensor" is not
a tensor. You can easily cook up an example, say a rank-2 tensor T on
R^2:

T = [1 1]
[0 2]

The "embedded tensor" would be - well, here is your first problem -
would it be [1 2] or [1 2]-transpose?

Well it is a collection of components which can form an n-1 tensor just
like the components of a rank-1 tensor are rank-0 tensors. This is very
obvious. I am interested in a way to extract such components from
tensor product B x C = A and expressing this collection as an n-1
tensor. Raising red herrings will not achieve much, but as usual, one
needs to solve problems themselves to get anything done.

A = B x C can be thought of as a vector, an axis of rotation, in three
dimensions. In four or more dimensions there are an infinite number of
axes that are mutually perpendicular to B and C, so you can't reduce a
resulting rotation to an axis. You would express it as a plane of
rotation. E.g. not e1 x e2 = e3, but e1^e2.

Thanks Greg. But that wasn't related to my question. The question, for
rank-2 A, was finding Z = SUM(i=1)^3 A_ii B_i {where B_x denotes x'th
basis vector} as an expression of C D where since A = C x D. I have
found the solution for rank-2 A and it is related to the entrywise
product of C and D. I suspect it can be extended to higher rank A by
extending to kronecker product (related to tensor product).



--
"Then they placed the ark of the Lord on the cart; along with the box
containing the golden mice and the images of the hemorrhoids."
-- 1 Samuel 6:11

Schoenfeld wrote:

Thanks Greg. But that wasn't related to my question. The question, for
rank-2 A, was finding Z = SUM(i=1)^3 A_ii B_i {where B_x denotes x'th
basis vector} as an expression of C D where since A = C x D. I have
found the solution for rank-2 A and it is related to the entrywise
product of C and D. I suspect it can be extended to higher rank A by
extending to kronecker product (related to tensor product).

Your original question bore no resemblance to what you write here. (In
case you are wondering...)

--
Jan Bielawski

JanPB wrote:
Schoenfeld wrote:

Thanks Greg. But that wasn't related to my question. The question, for
rank-2 A, was finding Z = SUM(i=1)^3 A_ii B_i {where B_x denotes x'th
basis vector} as an expression of C D where since A = C x D. I have
found the solution for rank-2 A and it is related to the entrywise
product of C and D. I suspect it can be extended to higher rank A by
extending to kronecker product (related to tensor product).

Your original question bore no resemblance to what you write here. (In
case you are wondering...)

Actually it's not. I just wanted to know what the collection of
components found along the "diagonal" of a tensor is called and related
properties of this entity (formulas, etc). Under certain conditions
(conditions stated in original post) you can construct an n-1 tensor
using those components. I called that tensor "the embedded tensor" -
loose terminology perhaps, but I thought that it wouldn't be an issue.
But since my choice of terminology cannot be found in the textbooks
rote learned by the responders, it must wrong on that basis alone,
according to the responders. What you call "different" is the same
question for special case rank(A) = 2 but more refined.





--
Jan Bielawski