# Representation of $4\times4$ matrices in the form of $\sum B_i\otimes C_i$

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 $$A=B\otimes C -C\otimes B$$?

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 $$af-be+cd=0$$

• "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. – Paul Broussous May 15 at 9:01
• @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. – Will Sawin May 15 at 13:32
• 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 ! – Paul Broussous May 15 at 13:57
• @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. – Will Sawin May 15 at 15:39
• @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. – Will Sawin May 16 at 0:40