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?



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   



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 $T$)

  • $\begingroup$ 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

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.


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