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Every matrix $A\in M_4(\mathbb{R})$ can be represented in the form of $$A=\sum_{i=1}^{n(A)} B_i\otimes C_i$$ for $B_i,C_i\in M_2(\mathbb{R})$.

What is the least uniform upper bound $M$ for such $n(A)$? In other words, what is the least integer $M$ such that every $A$ admit such a representation with $n(A)\leq M$?

Is this least upper bound equal to the corresponding least upper bound for all matrices $A$ which are a matrix representation of quaternions $a+bi+cj+dk$?

As another question about tensor product representation: What is a sufficient condition for a $4\times 4$ matrix $A$ to be represented in the form of $A=B\otimes C -C\otimes B$?

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Because $M_4(\mathbb R) = M_2(\mathbb R) \otimes M_2(\mathbb R)$ as vector spaces (and as algebras, but we won't use this), we can replace $M_2(\mathbb R)$ by an arbitrary $4$-dimensional vector space $V$ and $M_4(\mathbb R)$ by $V \otimes V$. We can represent elements of $V\otimes V$ conveniently as $4\times 4$ matrices, where simple tensors are rank one matrices. The question is then equivalent to asking the minimum number of rank $1$ matrices it takes to write a $4 \times 4$ matrix as a sum of rank $1$ matrices. The answer is obviously $4$.

For quaternions acting by left multiplication by quaternions in the standard basis:

$$1= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix} \otimes\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}$$

$$i= \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix} \otimes\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}$$

$$j= \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \otimes\begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix}$$

$$k= \begin{pmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \otimes\begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix}$$

Because only two terms appear on the right side, the matrices clearly have rank at most $2$ at tensor product, and some obtain rank exactly $2$, so the answer is two.

For the last question, the answer is isomorphic to the analogous answer for writing an element of $V \otimes V$ as $v_1 \otimes v_2 - v_2 \otimes v_1$ for $V$ a four-dimensional vector space. The condition is given by a skew-symmetry condition (i.e. 10 linear conditions) plus a Pfaffian condition (a quadratic condition). More precisely the general such matrix can be written as

$$ \begin{pmatrix} 0 & a & -a & 0 \\ b & c & d & e \\ -b & -d & -c & -e \\ 0 & f & -f & 0 \\ \end{pmatrix}$$

such that $af-be+cd=0$

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    $\begingroup$ "this is equivalent to asking the minimum number of rank 1 matrices it takes to write a 4×4 matrix as a sum of rank 1 matrices. The answer is obviously 4." @WillSavin I do not understand this sentence. $\endgroup$
    – Paul Broussous
    Commented May 15, 2019 at 9:01
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    $\begingroup$ @PaulBroussous We are asking what are the minimum $n$ such that all elements of $V \otimes W$ can be written as $\sum_{i=1}^{n} v_i \otimes w_i$ for $v_i \in V, w_i \in W$, where $V$ and $W$ are four-dimensional vector spaces. We can represent elements of $V \otimes W$ as $4 \times 4$ matrices, in which case the elements of the form $v_i \otimes w_i$ are the rank $1$ matrices. So we are asking for the minimum $n$ to write any $4 \times 4$ matrix as a sum of $n$ rank one matrix. The answer is $4$ - e.g. we can take the columns or the rows. $\endgroup$
    – Will Sawin
    Commented May 15, 2019 at 13:32
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    $\begingroup$ OK, I understand. Here you use the canonical isomorphism $V\otimes V^* \simeq {\rm End}(V)$, which has nothing to do with the natural isomorphism $M_4 ({\mathbb R})\simeq M_2 ({\mathbb R})\otimes M_2 ({\mathbb R})$. Indeed in that last isomorphim, $A\otimes B$, $A$, $B\not= 0$, is not necessarily a rank $1$ matrix ! $\endgroup$
    – Paul Broussous
    Commented May 15, 2019 at 13:57
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    $\begingroup$ @PaulBroussous Yes, the structure of $M_4(\mathbb R)$ is irrelevant for this problem, so we discard it, viewing $M_4(\mathbb R)$ only as $M_2(\mathbb R)\otimes M_2(\mathbb R)$, and then the structure of $M_2(\mathbb R)$ is irrelevant as well. I had already forgotten about this after the first line and so didn't realize that bringing in $4 \times 4$ matrices again would be potentially confusing. $\endgroup$
    – Will Sawin
    Commented May 15, 2019 at 15:39
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    $\begingroup$ @ZachTeitler It's skew symmetric when written in the standard matrix notation for elements of $V \otimes V$, $V$ a four-dimensional vector space, and the Pfaffian is the standard Pfaffian. But if we specialize to $V= M_2(\mathbb R)$ we have another way to write elements of the same vector space as matrices, using $M_2(\mathbb R) \otimes M_2(\mathbb R) = M_4(\mathbb R)$. I have written it in that way, so it no longer appears skew-symmetric. Concretely this just means we permute the entries. It might be fun to work out what this permutation does and why this formula is right - it was for me. $\endgroup$
    – Will Sawin
    Commented May 16, 2019 at 0:40

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