1
$\begingroup$

This is the origin of my problem: I have a set of data which expresses which user ($U$ set) applies what tag ($T$ set) to which item ($I$ set). So it is actually a $U×I×T$ tensor $A$ (or 3-dimensional matrix). Now I unfold this tensor A to be two 2-dimensional matrices. Matrix $A_{(u)}$ is $U×IT$, and matrix $A_{(i)}$ is $I×UT$. So these two matrices actually come from the same initial tensor, with differently arranged elements.

The two matrices are the unfolding matrices of the same tensor by different mode.

I am wondering if these two matrices have the same singular values, or the same rank? Or any other relations?

E.g.

$A_{(u)}$=

u_1,i_1,t_1 u_1,i_2,t_1 u_1,i_1,t_2 u_1,i_2,t_2
u_2,i_1,t_1 u_2,i_2,t_1 u_2,i_1,t_2 u_2,i_2,t_2   

and

$A_{(i)}$=

u_1,i_1,t_1 u_2,i_1,t_1 u_1,i_1,t_2 u_2,i_1,t_2 
u_1,i_2,t_1 u_2,i_2,t_1 u_1,i_2,t_2 u_2,i_2,t_2 

(Note e.g.,(u_1,i_2,t_1) or $(u_1,i_2,t_1)$ is a numeral element with 1 as index of $u$ in $U$, 2 as index of $i$ in $I$ and 1 as index of $t$ in $T$)

$\endgroup$
  • $\begingroup$ E.g., if $b_2=0$, $A_{(u)}$ can be (by a suitable choice of the other parameters) made to have rank 2, while $A_{(i)}$ would have rank 1. $\endgroup$ – Algebraic Pavel Aug 17 '14 at 1:18
0
+50
$\begingroup$

If you have a single user, $|U|=1$, then $A_{(u)}$ is rank 1, while $A_{(i)}$ is an arbitrary matrix, perhaps of full rank. So neither the rank nor the singular values need to match. In this case $A_{(u)}$ is essentially the vector you get by unfolding $A_{(i)}$.

If $\;|U|=u$, then $A_{(u)}$ is essentially the $u$ rows you get from "unfolding" $A_{(i)}^\top$ so that each block of $u$ rows moves as a block. The unfolding process acts on blocks, where each block corresponds to a value in $T$. We can see that $A_{(u)}$ has the blocks in a row, while $A_{(i)}^\top$ has the blocks in a column.

Suppose $A_{(u)}$ has rank $r$. Then each block has rank at most $r$. Since $A_{(i)}^\top$ is composed of $|T|=t$ of these blocks, the rank of $A_{(i)}^\top$ (and thus of $A_{(i)}$) can be at most $t \cdot r$. If $r<u$ then this gives a nontrivial restriction on the rank of $A_{(i)}$. So you are right that the ranks (and therefore the singular values) of the two matrices are not completely independent, specifically, the ranks must be within a factor of $t$ of each other.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.