Schoenfeld wrote:
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.

Man, what *is* your beef? Almost everything you say is incorrect yet
you keep offending people who thought they could help you.
Specifically:

I understand these components can be used to form such n-1
tensor,

They can't. (n-2)-rank tensor - yes.

this is why i called it "embedded". In the same way the
components of a vector can be used to form a scalar.

They can't.

This is clearly
what I implied.

Not only implied, you plainly stated it. The point is, it's wrong. No
big deal. Get over it.

That you persist with your objection implies this
conversation is useless and thread unfortunately corrupted.

Whatever.

--
Jan Bielawski

JanPB wrote:
Schoenfeld wrote:
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.

Man, what *is* your beef? Almost everything you say is incorrect yet
you keep offending people who thought they could help you.
Specifically:

I understand these components can be used to form such n-1
tensor,

They can't. (n-2)-rank tensor - yes.

*sigh* Listen very carefully:

Consider the matrix A:
[1, 2, 3
4, 5, 6
7, 8, 9]

The diagonal of this matrix is the set S = [1,5,9].

Now consider the following series:

Z = SUM(i=1)^3 ( S_i * B_i ) here B_i denotes i'th basis vector.
= SUM(i=1)^3 ( A_ii * B_i)

Z IS A VECTOR NOT A SCALAR! Do you understand yet?

The trace of A is not what I was referring to, the trace of A is
related to Z in the following way.

trace(A) = Z . SUM(i=1)^3 ( B_i ) { here . denotes vector dot product
}

Knowing the trace(A) is INSUFFICIENT to determine Z, since there are
(partition(trace(A)) - 1) solutions to Z.

Now in the same way that vector Z can be constructed for matrix A, you
can construct (n-1) tensor for n tensor. Get it yet?

this is why i called it "embedded". In the same way the
components of a vector can be used to form a scalar.

They can't.

This is clearly
what I implied.

Not only implied, you plainly stated it. The point is, it's wrong. No
big deal. Get over it.

Repeat:

Z = SUM(i=1)^3 ( A_ii * B_i)

That you persist with your objection implies this
conversation is useless and thread unfortunately corrupted.

Whatever.

hint: Math is more that what you rote learned.

--
Jan Bielawski

Schoenfeld wrote:

*sigh* Listen very carefully:

Consider the matrix A:
[1, 2, 3
4, 5, 6
7, 8, 9]

The diagonal of this matrix is the set S = [1,5,9].

Now consider the following series:

Z = SUM(i=1)^3 ( S_i * B_i ) here B_i denotes i'th basis vector.
= SUM(i=1)^3 ( A_ii * B_i)

Z IS A VECTOR NOT A SCALAR! Do you understand yet?

It is a vector defined in terms of a fixed basis. If you allow this
kind of vectors, then everything under the sun defines a vector: for
example the three numbers separated by the chess knight move: [1,8,3].
Or the triple consisting of my height, the temperature today, and the
population of San Francisco. Such "vectors" are coordinate-dependent
objects which is a totally different thing than an "(n-1)-tensor" you
asked for originally.

The trace of A is not what I was referring to, the trace of A is
related to Z in the following way.

trace(A) = Z . SUM(i=1)^3 ( B_i ) { here . denotes vector dot product
}

Sure.

Knowing the trace(A) is INSUFFICIENT to determine Z, since there are
(partition(trace(A)) - 1) solutions to Z.

Why not infinitely many? Are your matrices over positive integers only?
The plot thickens... :-)

Now in the same way that vector Z can be constructed for matrix A, you
can construct (n-1) tensor for n tensor. Get it yet?

No, you cannot. You can define an (n-1)-tensor for *n-tensor AND a
fixed basis*. But that's a sort of vacuous "tensor", just like saying
that my height, room temperature, and the population of San Francisco
is a vector in 3-dimensional space.

--
Jan Bielawski

JanPB wrote:
Schoenfeld wrote:

*sigh* Listen very carefully:

Consider the matrix A:
[1, 2, 3
4, 5, 6
7, 8, 9]

The diagonal of this matrix is the set S = [1,5,9].

Now consider the following series:

Z = SUM(i=1)^3 ( S_i * B_i ) here B_i denotes i'th basis vector.
= SUM(i=1)^3 ( A_ii * B_i)

Z IS A VECTOR NOT A SCALAR! Do you understand yet?

It is a vector defined in terms of a fixed basis. If you allow this
kind of vectors, then everything under the sun defines a vector: for
example the three numbers separated by the chess knight move: [1,8,3].
Or the triple consisting of my height, the temperature today, and the
population of San Francisco. Such "vectors" are coordinate-dependent
objects which is a totally different thing than an "(n-1)-tensor" you
asked for originally.

*sigh*

Someone says:
"(a+b)(a+b) = (a+b)^2"
JanPB responds:
"no that's all wrong, that doesn't make any sense, a,b could be
anticommutative, why are you insulting me?"




The trace of A is not what I was referring to, the trace of A is
related to Z in the following way.

trace(A) = Z . SUM(i=1)^3 ( B_i ) { here . denotes vector dot product
}

Sure.

So it only took 5 posts and two threads for you to acknowledge this
elementary statement.

Knowing the trace(A) is INSUFFICIENT to determine Z, since there are
(partition(trace(A)) - 1) solutions to Z.

Why not infinitely many? Are your matrices over positive integers only?
The plot thickens... :-)

In my scenario I was considering only all positive components. In this
case there are partition(trace(A))-1 possible solutions to Z. In other
cases, there are more. This just makes my point stronger - knowing
trace(A) is USELESS for determining Z.

Now in the same way that vector Z can be constructed for matrix A, you
can construct (n-1) tensor for n tensor. Get it yet?

No, you cannot. You can define an (n-1)-tensor for *n-tensor AND a
fixed basis*.

Okay, so because I didn't explicitly state "under some fixed basis" you
decided to waste all this time?

Someone says:
"(a+b)(a+b) = (a+b)^2"
JanPB responds:
"no that's all wrong, that doesn't make any sense, a,b could be
anticommutative, why are you insulting me?"

But that's a sort of vacuous "tensor", just like saying
that my height, room temperature, and the population of San Francisco
is a vector in 3-dimensional space.

*Complete* nonsense. I must admit, you're as useless as rubber lips on
a woodpecker.

--
Jan Bielawski

JanPB wrote:
Schoenfeld wrote:
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.
Man, what *is* your beef?

I think his problem is that there's something he wants
to do with matrices, and he's using the wrong terminology,
then getting all offended when we keep saying 'but that
isn't a tensor'. It's exacerbated by the plain fact
that he hasn't a clue what a tensor is (but interprets
that as us being a bunch of recitation automata).

In any case, I'm tempted to agree with him that the
conversation is useless.

"Schoenfeld" wrote in message oups.com...

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.

That is one of the most ungrateful replies I have ever
seen and a perfect example of how not to get help
from a newsgroup.

I note that you continue in the same manner below.

Martin Hogbin

Robert Low wrote:
JanPB wrote:
Schoenfeld wrote:
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.
Man, what *is* your beef?

I think his problem is that there's something he wants
to do with matrices, and he's using the wrong terminology,
then getting all offended when we keep saying 'but that
isn't a tensor'. It's exacerbated by the plain fact
that he hasn't a clue what a tensor is (but interprets
that as us being a bunch of recitation automata).

In any case, I'm tempted to agree with him that the
conversation is useless.

A review of your post reveals nothing quantative, just emotional
opinion. I am treating a tensor as a multidimensional array of scalars.
What I called the "embedded n-1 tensor" from a rank-2 tensor is
equivalent to the vector formed from the diagonal of square matrix
using a diagonal basis.

If you have a rank-3 tensor represented as a 3 dimensional array of
scalars - a cube - the "embedded rank-2 tensor along the diagonal"
refers to the square made by a top edge of the cube and the bottom edge
of a cube. This square consists of a set of values that can form rank-2
tensor. Treating tensors as multidimensional arrays of scalars is
perfectly valid.

If ignorance were a disability, you'd be on the full pension.

Schoenfeld:

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).

A_ii is the sum of its diagonal elements, otherwise known as
the trace. A_ii B_i doesn't mean anything, except possibly
Tr(A) B_i. Z_i = A_ij B_j is a vector. The vectors of A are the
columns.


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.

Yeah, right....

Bilge wrote:
Schoenfeld:

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).

A_ii is the sum of its diagonal elements, otherwise known as
the trace. A_ii B_i doesn't mean anything, except possibly
Tr(A) B_i. Z_i = A_ij B_j is a vector. The vectors of A are the
columns.

No, A_ii is the scalar component at row i, col i. If I were using
Einstein notation I would indicate so. SUM(i=1)^3 A_ii Bi refers to a
special vector which I was interested in (each term of the trace sum is
given a basis vector). This can be extended to tensors of higher rank.


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.

Yeah, right....

Are you getting emotional again?

Schoenfeld wrote:
Robert Low wrote:
It's exacerbated by the plain fact
that he hasn't a clue what a tensor is (but interprets
that as us being a bunch of recitation automata).

In any case, I'm tempted to agree with him that the
conversation is useless.
A review of your post reveals nothing quantative, just emotional
opinion. I am treating a tensor as a multidimensional array of scalars.

Ah, that's your problem then. You see, a tensor is not
an array of scalars. Or then again, perhaps you don't.
But hey, that's just me emoting.


If ignorance were a disability, you'd be on the full pension.

Gee, I'm glad only one of us is rude and ignorant. Just
think how things would degenerate if you were too...