I Confused by this result for the tensor product of two vectors

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

Science Advisor
Homework Helper
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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.
 
Last edited:

fresh_42

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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

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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/threads/find-the-basis-of-a-vector-subspace-of-r-2-2.929266/#post-5868072

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

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