Confused by this result for the tensor product of two vectors

[Prez Cannady]

Given two probability distributions $p \in R^{m}_{+}$ and $q \in R^{n}_{+}$ (the "+" subscript simply indicates non-negative elements), this paper (page 4) writes down the tensor product as

$$p \otimes q := \begin{pmatrix}
p(1)q(1) \\
p(1)q(2) \\
\vdots \\
p(1)q(n) \\
\vdots \\
p(m)q(n)
\end{pmatrix}, $$

yet I expected to see

$$p \otimes q := \begin{pmatrix}
p(1)q(1) && p(1)q(2) && \cdots && p(1)q(n-1) && p(1)q(n) \\
p(2)q(1) && \ddots && && && p(2)q(n) \\
\vdots && && \ddots && && \vdots \\
p(m-1)q(1) && && && \ddots && p(m-1)q(n) \\
p(m)q(1) && p(m)q(2) && \cdots && p(m)q(n-1) && p(m)q(n)
\end{pmatrix}, $$

Am I missing something?

 

[BvU]

Well, they define, so they call the shots. At least they explain clearly what they mean:

Above, we have introduced the notation ⊗ to denote the tensor product, which in general maps a pair of vectors with dimensions $m, n$ to a single vector with dimension $mn$

so all the elements you expected to see are present.

 

[fresh_42]

I'd also prefer the matrix notation, because the tensor product then becomes an ordinary matrix multiplication: column times row. But who says you can't write a matrix as a column? I think it's more convenient for type setting than it is mathematically, but in the end it depends on what you want to do with it.

 

[StoneTemplePython]

Given two probability distributions $p \in R^{m}_{+}$ and $q \in R^{n}_{+}$ (the "+" subscript simply indicates non-negative elements), this paper (page 4) writes down the tensor product as

$$p \otimes q := \begin{pmatrix}
p(1)q(1) \\
p(1)q(2) \\
\vdots \\
p(1)q(n) \\
\vdots \\
p(m)q(n)
\end{pmatrix}, $$

yet I expected to see

$$p \otimes q := \begin{pmatrix}
p(1)q(1) && p(1)q(2) && \cdots && p(1)q(n-1) && p(1)q(n) \\
p(2)q(1) && \ddots && && && p(2)q(n) \\
\vdots && && \ddots && && \vdots \\
p(m-1)q(1) && && && \ddots && p(m-1)q(n) \\
p(m)q(1) && p(m)q(2) && \cdots && p(m)q(n-1) && p(m)q(n)
\end{pmatrix}, $$

Am I missing something?

For the Kronecker product this is actually a very common definition. In fact if you use the standard definition for the Kronecker product of

$\mathbf X \otimes \mathbf Y = \begin{bmatrix} x_{1,1}\mathbf Y & \cdots & x_{1,n}\mathbf Y\\ \vdots & \ddots & \vdots \\ x_{m,1}\mathbf Y & \cdots & x_{m,n}\mathbf Y \end{bmatrix}$

where $\mathbf X$ is $\text{m x n}$
- - - -
and you then constrain $\mathbf X$ and $\mathbf Y$ to be column vectors, you really have no choice but to have

$\mathbf {xy}^* = \mathbf x \otimes \mathbf y^*$

and
$\mathbf x^* \otimes \mathbf y = \mathbf y \mathbf x^*$

but

$\mathbf x \otimes \mathbf y$

is a column vector.

Notation and definitions can be tweaked slightly to get very different results, which is unfortunately, confusing.

- - - -
There's a nice free 12 page sample chapter and walkthrough of Kronecker products in Laub's "Matrix Analysis for Scientists & Engineers" -- I gave an indirect link here:

https://www.physicsforums.com/threa...vector-subspace-of-r-2-2.929266/#post-5868072

(page 2 of said sample chapter addresses your question directly)