Dyad as a product of two vectors: what does ii or jj mean?

In summary: Not quite. If we have three, that is a triad ##u\otimes v \otimes w## and gives a cube: first the matrix and then weighted copies of it stacked. A general tensor of rank three is a linear combination of triads and so we can get any cube this way, or with linear combinations of dyads any matrix. A dyad itself is always a rank one matrix. Rank is ambiguous here, as the tensor has a rank which says how long the tensor products are, and whether they are composed of vectors or linear forms, which can be written as vectors, too: ##v^*\, = \,(\,x \mapsto \langle v
  • #1
Sorcerer
281
52
So I'm reading through some stuff trying to learn tensors, and I wanted to stop myself to grasp this concept before I go further.

Along the way it was mentioned in one of the sources I'm looking at that other than the dot product and cross product, you could multiply vectors in another way, and example given was UV forming a dyad. This was their definition:

UV = u1v1ii + u1v2ij + u1v3ik + u2v1ji + ...​

So my question is
, what does ii or ij even mean? This is multiplication of unit vectors, obviously, but what kind? Dot product? Cross product? That's what I don't understand here.

If i = (1, 0, 0) and j = (0, 1, 0), then i ⋅ j is obviously zero, and i X j = (0,0,1) = k.

But which type of multiplication are we talking about here with ii and ij and so on? Is it something else entirely? Or do we just leave it like that, and it just means each component has two directions? (not that I can't fathom of a vector doing that, but of course, a dyad isn't a vector, right?)Any insight is welcome. Thanks.
 
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  • #2
A dyad with ##u## and ##v## is the tensor product ##u \otimes v##. Now this tensor product is defined via it's algebraic properties, but we can also do it in coordinates. So let us assume we have column vectors ##u,v##. Then the tensor product is the matrix multiplication ##u \cdot v^\tau##, that is column times row. The result is a square matrix: first coordinate ##u_1## with the entire row ##v^\tau## for the first row of the matrix, and so on until the last coordinate ##u_n## again with the entire row ##v^\tau##to get the last row. This is what your formula say: here ##i \cdot i## means position ##(1,1)##, ##i\cdot j## position ##(1,2)## and so on.

Here's a short essay about tensors which I've written:
https://www.physicsforums.com/insights/what-is-a-tensor/
 
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  • #3
fresh_42 said:
A dyad with ##u## and ##v## is the tensor product ##u \otimes v##. Now this tensor product is defined via it's algebraic properties, but we can also do it in coordinates. So let us assume we have column vectors ##u,v##. Then the tensor product is the matrix multiplication ##u \cdot v^\tau##, that is column times row. The result is a square matrix: first coordinate ##u_1## with the entire row ##v^\tau## for the first row of the matrix, and so on until the last coordinate ##u_n## again with the entire row ##v^\tau##to get the last row. This is what your formula say: here ##i \cdot i## means position ##(1,1)##, ##i\cdot j## position ##(1,2)## and so on.

Here's a short essay about tensors which I've written:
https://www.physicsforums.com/insights/what-is-a-tensor/
Thanks. I'll read that asap.
 
  • #4
To be exact ##i \cdot j ## should better be written or at least thought of as
$$
i \otimes j = \begin{bmatrix}1\\0\\0\end{bmatrix} \cdot \begin{bmatrix}0&1&0\end{bmatrix}=\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}
$$
but ##i \,j## is simply faster and shorter.
 
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  • #5
fresh_42 said:
To be exact ##i \cdot j ## should better be written or at least thought of as
$$
i \otimes j = \begin{bmatrix}1\\0\\0\end{bmatrix} \cdot \begin{bmatrix}0&1&0\end{bmatrix}=\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}
$$
but ##i \,j## is simply faster and shorter.
That makes a lot of sense, since we're going up in, I guess, degree, by talking about dyads instead of vectors. Well I guess the term is rank, and the number of components for rank n would be 3n, right?
 
  • #6
Sorcerer said:
That makes a lot of sense, since we're going up in, I guess, degree, by talking about dyads instead of vectors. Well I guess the term is rank, and the number of components for rank n would be 3n, right?
Not quite. If we have three, that is a triad ##u\otimes v \otimes w## and gives a cube: first the matrix and then weighted copies of it stacked. A general tensor of rank three is a linear combination of triads and so we can get any cube this way, or with linear combinations of dyads any matrix. A dyad itself is always a rank one matrix. Rank is ambiguous here, as the tensor has a rank which says how long the tensor products are, and whether they are composed of vectors or linear forms, which can be written as vectors, too: ##v^*\, = \,(\,x \mapsto \langle v,x \rangle \,)## and so we can combine them to e.g. ##u^* \otimes v^* \otimes w## which is a rank ##(1,2)-##tensor: one vector, two linear forms. In coordinates it is still a cube, which brings us back to your ##3^n##. It is ##n^3##.

Correction, you are right. I thought of dimension ##n## and three vectors, and you probably of dimension ##3## and ##n## vectors. Sorry, for the misunderstanding. There is another remark to be made. Whether as in my example it is called ##(1,2)-##tensor or ##(2,1)-##tensor can vary from author to author.
 
  • #7
fresh_42 said:
Not quite. If we have three, that is a triad ##u\otimes v \otimes w## and gives a cube: first the matrix and then weighted copies of it stacked. A general tensor of rank three is a linear combination of triads and so we can get any cube this way, or with linear combinations of dyads any matrix. A dyad itself is always a rank one matrix. Rank is ambiguous here, as the tensor has a rank which says how long the tensor products are, and whether they are composed of vectors or linear forms, which can be written as vectors, too: ##v^*\, = \,(\,x \mapsto \langle v,x \rangle \,)## and so we can combine them to e.g. ##u^* \otimes v^* \otimes w## which is a rank ##(1,2)-##tensor: one vector, two linear forms. In coordinates it is still a cube, which brings us back to your ##3^n##. It is ##n^3##.

Correction, you are right. I thought of dimension ##n## and three vectors, and you probably of dimension ##3## and ##n## vectors. Sorry, for the misunderstanding. There is another remark to be made. Whether as in my example it is called ##(1,2)-##tensor or ##(2,1)-##tensor can vary from author to author.
Yeah I was thinking of dimension 3. With 4-vectors would it be 4n? I recall something about a tensor in the Einstein Equation having 16 components, 4 squared.

Anyway it looks like I’m about to open Pandora’s Box here, based on what you’ve posted here. Or, I guess Kansas is going bye bye. Or whatever metaphor, because this looks like a whole other level of complexity.
 
  • #8
Sorcerer said:

Yeah I was thinking of dimension 3. With 4-vectors would it be 4n?
In four dimensions with ##n## vectors, i.e. ##u_1\otimes u_2 \otimes \ldots \otimes u_n##, then yes, this makes ##4^n## coordinates.
I recall something about a tensor in the Einstein Equation having 16 components, 4 squared.

Anyway it looks like I’m about to open Pandora’s Box here, based on what you’ve posted here. Or, I guess Kansas is going bye bye. Or whatever metaphor, because this looks like a whole other level of complexity.
It's not that complicated. It's linear after all, or better, multilinear. E.g. think of a linear mapping ##\varphi \, : \, V \longrightarrow W## with a matrix ##A##. Then we write ##\varphi (u) = Au = a_1w_1 + \ldots +a_nw_n##. Now we can also write ##A=\sum v^*_i \otimes w_j## and ##Au=\sum v_i^*(u)w_j = \sum \langle u,v_i \rangle \,w_j## and the ##a_i = \langle u,v_i \rangle##.
 

Related to Dyad as a product of two vectors: what does ii or jj mean?

1. What is a dyad?

A dyad is a mathematical term that refers to a product of two vectors, also known as a dyadic product. It is a way of multiplying two vectors to create a new vector that represents the combination of the two original vectors.

2. How is a dyad represented?

A dyad is typically represented using the notation ii or jj, where the first letter represents the first vector and the second letter represents the second vector. For example, the dyad of vectors a and b would be represented as ab.

3. What does ii or jj mean in the context of a dyad?

In the context of a dyad, ii or jj represents the direction and magnitude of the resulting vector created by multiplying the two original vectors. It is a way of representing the combination of the two vectors in a concise and consistent manner.

4. How is a dyad calculated?

A dyad is calculated by multiplying the first vector by the second vector, and then multiplying each component of the first vector by each component of the second vector. The resulting vector is the dyad of the two original vectors.

5. What are the applications of a dyad in science?

Dyads have various applications in different fields of science, such as physics, engineering, and mathematics. They are used to represent the relationships between different physical quantities, such as force and displacement, and to calculate moments of inertia and angular momentum. They also have applications in tensor analysis and quantum mechanics.

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