What is the Tensor Product of Vectors and How Does It Differ Across Contexts?

[Harel]

Hey it might be a stupid question but I saw that the tensor product of 2 vectors with dim m and n gives another vector with dimension mn and in another context I saw that the tensor product of vector gives a metrix. For example from sean carroll's book: "If T is a (k,l) tensor and S is a (m, n) tensor, we define a (k + m, l + n) tensor T ⊗ S"
so the tensor product of two type 1 tensors,k=1,vectors, is a metrix
and in the context of quantum mechanic I saw
(1,0)⊗(1,0)↦(1,0,0,0) when those our basis vectors.
I'm sure I'm just getting something wrong but I am hopefull that you can explain me what.

 

[Daeho Ro]

so the tensor product of two type 1 tensors,k=1,vectors, is a metrix

Why it would be? If one of them is dual vector, then it might be. I will give you some examples:
\begin{pmatrix} 1 \\ 2 \end{pmatrix} \otimes \begin{pmatrix} 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 \times \begin{pmatrix} 1 & 2 \end{pmatrix}\\ 2 \times \begin{pmatrix} 1 & 2 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix},
\begin{pmatrix} 1 & 2 \end{pmatrix} \otimes \begin{pmatrix} 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 \times \begin{pmatrix} 1 & 2 \end{pmatrix} & 2 \times \begin{pmatrix} 1 & 2 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 1 & 2 & 2 & 4 \end{pmatrix}

 

[Harel]

Why it would be? If one of them is dual vector, then it might be. I will give you some examples:
\begin{pmatrix} 1 \\ 2 \end{pmatrix} \otimes \begin{pmatrix} 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 \times \begin{pmatrix} 1 & 2 \end{pmatrix}\\ 2 \times \begin{pmatrix} 1 & 2 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix},
\begin{pmatrix} 1 & 2 \end{pmatrix} \otimes \begin{pmatrix} 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 \times \begin{pmatrix} 1 & 2 \end{pmatrix} & 2 \times \begin{pmatrix} 1 & 2 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 1 & 2 & 2 & 4 \end{pmatrix}

Because by the defenition of sean, let's take a type (1,0) tensor which is a vector and another (1,0) tensor which is also a vector and the product will be a (2,0) tensor, which is a metrix.

 

[WWGD]

Hey it might be a stupid question but I saw that the tensor product of 2 vectors with dim m and n gives another vector with dimension mn and in another context I saw that the tensor product of vector gives a metrix. For example from sean carroll's book: "If T is a (k,l) tensor and S is a (m, n) tensor, we define a (k + m, l + n) tensor T ⊗ S"
so the tensor product of two type 1 tensors,k=1,vectors, is a metrix
and in the context of quantum mechanic I saw
(1,0)⊗(1,0)↦(1,0,0,0) when those our basis vectors.
I'm sure I'm just getting something wrong but I am hopefull that you can explain me what.

A vector can be seen as a $1 \times n$ matrix.

 

[Daeho Ro]

Because by the defenition of sean, let's take a type (1,0) tensor which is a vector and another (1,0) tensor which is also a vector and the product will be a (2,0) tensor.

That is true. But, the results of mine are matrices with the size 2 \times 2 and 1 \times 4.

 

[WWGD]

Hey it might be a stupid question but I saw that the tensor product of 2 vectors with dim m and n gives another vector with dimension mn and in another context I saw that the tensor product of vector gives a metrix. For example from sean carroll's book: "If T is a (k,l) tensor and S is a (m, n) tensor, we define a (k + m, l + n) tensor T ⊗ S"
so the tensor product of two type 1 tensors,k=1,vectors, is a metrix
<Snip>.

Actually, if your map is k-linear ( in any " coordinate") for k>2 (where you may have quadratic forms), it is not representable as a matrix anymore. That is the actual point of tensors: to represent k - , or j- ( k,j pos. integers) linear maps in many variables, which is not feasible with matrices alone whenever you have an index >2.
Only linear and bilinear maps may be represented using matrices.

 

[HallsofIvy]

I feel compelled to point out that a tensor can be represented by a matrix in a given coordinate system, but, strictly speaking, a tensor is NOT a matrix.

 

[WWGD]

I feel compelled to point out that a tensor can be represented by a matrix in a given coordinate system, but, strictly speaking, a tensor is NOT a matrix.

How do you represent a higher-order tensor as a matrix? e.g., a 3-linear map .

 

[HallsofIvy]

In the same sense that a "vector" is a 1 by 3 matrix, so a higher order tensor can be represented by a "3 by 3 by 3" or higher matrix. I admit that is stretching the concept of "matrix" a bit far. My point was simply that a matrix is NOT a tensor.