Understanding the Confusion: Tensor Product vs. Dyadic Product for Vectors

In summary, the first interpretation is consistent with the fundamental definition of the vector product, but the latter version is necessary for things like the rotation tensor definition to be true.
  • #1
wil3
179
1
Hello. I keep on encountering the need to find the Tensor or Kronecker product of two vectors. Based on the definition, If I found the product of two 2D vectors, I would get a 4-dimensional vector. Some authors claim this is the correct interpretation.

However the dyadic product, which many claim is just the 1st order case of the tensor product, would generate a second-order tensor. In other words, if I multiplied my two 2D vectors, I would get a 2x2 matrix. This is essentially finding the tensor product between the first vector and the transpose of the second vector.

This troubles me. The first interpretation is consistent with the fundamental definition of the vector product, but the latter version is necessary for things like the rotation tensor definition to be true:

[tex] R = I\cos\theta + \sin\theta[\mathbf u]_{\times} + (1-\cos\theta)\mathbf{u}\otimes\mathbf{u} [/tex]

Which version is correct, and how can it be made consistent with the other version? Thanks for any help.
 
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  • #2
I don't know much at all about tensors, but I was under the impression that a 2x2 matrix is a vector in a 4-dimensional space [itex]\mathcal{L}(V,V)[/itex] if [itex]V[/itex] is 2-dimensional.
 
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  • #3
That idea sounds promising, but I'm afraid I don't quite understand what you mean yet. Would you mind elucidating things a little? I'm new to tensors, and so there could be a critical spatial relationship that I am missing out on here.
 
  • #4
I made a little bit of a mistake but I'll clear it all up.

[itex]\mathcal{L}(V,W)[/itex] is the set of all linear transformations [itex]T:V\to W[/itex]. If [itex]V[/itex] has dimension [itex]m[/itex] and [itex]W[/itex] has dimension [itex]n[/itex], then [itex]\mathcal{L}(V,W)[/itex] can be shown to be a vector space with dimension [itex]mn[/itex].

The set [itex]M_{m,n}(\mathbb{R})[/itex] of [itex]m[/itex]-by-[itex]n[/itex] matrices with entries from [itex]\mathbb{R}[/itex] is isomorphic to [itex]\mathcal{L}(V,W)[/itex] and thus, is also a vector space of dimension [itex]mn[/itex]. If [itex]n=m=2[/itex], then the dimension of [itex]M_{2,2}(\mathbb{R})[/itex] is 4, and its elements are 2-by-2 matrices.

You can also find an isomorphism [itex]T:M_{2,2}(\mathbb{R})\to\mathbb{R}^{4}[/itex], such that [itex]T(\left[ {\begin{array}{cc}
x & y \\
z & w \\
\end{array} } \right])=(x,y,z,w)[/itex]

I hope that clears some things up.
 
  • #5
That actually very much does clear things up... I had never learned about the equivalency of, for example, spaces of 2x2 tensors and spaces of 1x4 tensors. Where could I find an example of a relationship that establishes the isomorphism? (ie, is there any sort of tensor A would make your relation
[tex]
T(\left[ {\begin{array}{cc}
x & y \\
z & w \\
\end{array} } \right])=(x,y,z,w)
[/tex]

true, or is it a more abstract sort of transformation?
 
  • #6
wil3 said:
That actually very much does clear things up... I had never learned about the equivalency of, for example, spaces of 2x2 tensors and spaces of 1x4 tensors. Where could I find an example of a relationship that establishes the isomorphism? (ie, is there any sort of tensor A would make your relation
[tex]
T(\left[ {\begin{array}{cc}
x & y \\
z & w \\
\end{array} } \right])=(x,y,z,w)
[/tex]

true, or is it a more abstract sort of transformation?

That is the isomorphism, you just need to prove it to be one.
 
  • #7
ah, I see. Thanks very much!
 

1. What is a tensor product for vectors?

A tensor product for vectors is a mathematical operation that combines two vectors to create a new vector. It is a way of multiplying vectors that takes into account both direction and magnitude.

2. How is a tensor product for vectors different from scalar multiplication?

The main difference between a tensor product for vectors and scalar multiplication is that a tensor product takes into account the direction of the vectors, while scalar multiplication only considers the magnitude. This means that the resulting vectors from a tensor product will have both magnitude and direction, while the result of scalar multiplication will only have magnitude.

3. What is the purpose of a tensor product for vectors in science?

A tensor product for vectors is commonly used in physics and engineering to represent and calculate various physical quantities, such as force, torque, and stress. It allows for a more accurate representation of these quantities, as it considers both magnitude and direction.

4. How do you perform a tensor product for vectors?

The most common way to perform a tensor product for vectors is to use the cross product or dot product. The cross product results in a vector that is perpendicular to both input vectors, while the dot product results in a scalar value. Both of these operations can be used to calculate the tensor product.

5. Can a tensor product be performed on more than two vectors?

Yes, a tensor product can be performed on any number of vectors. The resulting vector will have a magnitude and direction that is determined by the combination of all the input vectors. This is known as a multi-vector product or an outer product.

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