How to rewrite $M_1\otimes M_2$ isolating $M_2$?

Question

I have 2 matrices $M_1, M_2$.

Is there a way to rewrite $M_1\otimes M_2$ as $M \cdot M_2$? i.e. $M$ is a matrix that it's being multiplied by $M_2$.

My objective here is to isolate $M_2$ as a product of matrices.

Can we do something similar with $\operatorname{vec}(M_1\otimes M_2)$ as $M\cdot \operatorname{vec}(M_2)$?

For those who are voting to close. Please give me some feedback. The reasons to vote are too general... — An old man in the sea., Dec 21 '18 at 10:29

score 1 · Answer 1 · answered Dec 21 '18 at 10:22

1

Since $\operatorname{vec}(M_1 \otimes M_2)$ is linear in the entries of $M_2$, then letting $\mathbf{m} = \operatorname{vec}(M_2)$ you could write \begin{align} \operatorname{vec}(M_1 \otimes M_2) &= D \cdot \mathbf{m}_2 \end{align} where the columns of $D$ are given by $$ \begin{align} D_{:, i} &= \frac{\partial}{\partial m_i}\operatorname{vec}(M_1 \otimes M_2) \\ &= \operatorname{vec}\left(M_1 \otimes \frac{\partial}{\partial m_i}M_2\right) \end{align} $$

answered Dec 21 '18 at 10:22

Nadiels

2,648

Nadiels, Thanks for the answer. However, I don't understand where the derivative comes from... – An old man in the sea. Dec 21 '18 at 10:44
Let me reformulate my comment above... What's $m_i$? – An old man in the sea. Dec 21 '18 at 17:16
Can't elaborate 'cause phone, but someone else has given you a more complete answer. Basically once you vectorise everything you can write linear transformations as a matrix/vec product using their Jacobian. I was just too lazy to keep track of the change in indices moving to the vectorised form, hence my $\mathbf{m}$ – Nadiels Dec 21 '18 at 17:56
=OOOOO I would never have thought about something like that!!!!! WOW! =D thanks ;) – An old man in the sea. Dec 21 '18 at 17:58

greg · Accepted Answer · 2018-12-21T17:32:25.093

1

Assume the following dimensions for the matrices: $$\eqalign{ M_1\quad&is\quad(m\times n) \cr M_2\quad&is\quad(p\times q) \cr M\quad&is\quad(r\times p) \cr }$$ Then for their products: $$\eqalign{ M_1\otimes M_2\quad&is\quad(mp\times nq) \cr M\cdot M_2\quad&is\quad(r\times q) \cr }$$ Note that the final dimensions do not match, except in the trivial case $n=1$.

However, finding a matrix such that $${\rm vec}(M_1\otimes M_2)=M\cdot{\rm vec}(M_2)$$ is possible; something along the lines that Nadiels has suggested.

Let $c_k$ be the $k^{th}$ column of $M_1\,\,$ and let $\,(P,Q)\,$ be the $(p\times p)$ and $(q\times q)$ identity matrices, respectively. Then $$\eqalign{ M &= \pmatrix{Q\otimes c_1\cr Q\otimes c_2\cr \vdots\cr Q\otimes c_n}\otimes P\cr }$$

edited Dec 21 '18 at 17:32

answered Dec 21 '18 at 17:08

greg

35,825

You're completely right... I never thought about that. I'm a bit dumb... thanks for all the help. ;) – An old man in the sea. Dec 21 '18 at 17:15
One thing I didn't understand from the answer of Nadiels is what's $\frac{\partial}{\partial m_i}$... – An old man in the sea. Dec 21 '18 at 17:17
Nadiels's derivative zeros out all elements of $M_1$ and $m$ except for the one that corresponds to the $m_i$ component. So it transforms $m$ into a vector from the standard basis ($e_i$), and transforms $M_1$ into a single-entry matrix. – greg Dec 21 '18 at 17:40
Greg I always learn with you. Thank you! ;) – An old man in the sea. Dec 21 '18 at 18:40

score 1 · Answer 3 · answered Dec 26 '18 at 15:16

Another way to solve the problem is to take advantage of the block structure of the Kronecker product $$P = M_1\otimes M_2$$ Simply extract the $(1,1)$-block of the product and divide it by the $(1,1)$ element of the first matrix. Assuming the dimensions of the matrices are $$M_1\in{\mathbb R}^{m\times n},\quad M_2\in{\mathbb R}^{p\times q}$$ one can isolate the second matrix as follows $$\eqalign{ M_2 &= \frac{\pmatrix{I_{p} & 0}\Big(M_1\otimes M_2\Big)\pmatrix{I_{q}\cr 0}}{\pmatrix{1&0}M_1\pmatrix{1\cr 0}} \cr }$$

How to rewrite $M_1\otimes M_2$ isolating $M_2$?

3 Answers3

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