# 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?

## Answers and Replies

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BvU
Homework Helper
2019 Award
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
Mentor
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
Gold Member
2019 Award
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)