When the determinant of a 2x2 polynomial matrix is a square?

Consider a 2x2 matrix $A$ with entries from $\mathbb{C}[x,y]$. Assume that $\mathrm{det} A$ is a square. Is it true that then $A$ can be represented as a noncomuting product $A=A_1 A_2 … A_{2n}$, in which any given matrix $A_k$ occurs an even number of times up to transposition? (Formally, the latter means that for each k=1,…,2n there is an even number of indices j such that either $A_j=A_k$ or $A_j=A_k^T$).

If this is not true, is there a nice description of 2x2 polynomial matrices whose determinant is a square?

In fact the following particular case is of most interest for me:

Let $A=A(x,y)$ be a biquadratic polynomial with quaternionic coefficients. Assume that $|A|^2$ is a square of a real polynomial. Is it true that then $A$ can be represented as a noncomuting product $A=\lambda A_1 A_2 … A_{2n}$, in which any given quaternionic polynomial $A_k$ occurs an even number of times up to conjugation, and $\lambda$ is a real polynomial?

This question is a continuation of the following one: Pythagorean 5-tuples

• The matrix [[x,0],[0,x]] should be a counterexample to your first question. – ARupinski Oct 25 '11 at 11:12
• $\begin{pmatrix} x & 0 \\ 0 & x \end{pmatrix} = \begin{pmatrix} x & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ – David E Speyer Oct 25 '11 at 12:01
• @David Speyer: wow! Now rolled the factor $\lambda$ back in the first question. For the quaternions the trick does not work so well, thus $\lambda$ still remains there. – mikhail skopenkov Oct 25 '11 at 12:48
• Can skew-symmetric matrices be factored in this way? – M P Oct 25 '11 at 14:04
• @MP I had the same thought. In the $2 \times 2$ case, yes. I suspect that $\begin{pmatrix} 0 & u & v & w \\ -u & 0 & x & y \\ -v & -x & 0 & z \\ -w & -y & -z & 0 \end{pmatrix}$ cannot, but I can't prove it (yet). – David E Speyer Oct 25 '11 at 16:23

I can show that there is no universal formula. More precisely, let $k$ be an algebraically closed field and let $R = k[w,x,y,z,\Delta]/(\Delta^2-wz+xy)$. I claim that there do not exist $2 \times 2$ matrices $A_1$, $A_2$, ..., $A_{2t}$ with entries in $R$, such that every matrix occurs an even number of times in the list up to transpose, and $\left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right) = A_1 A_2 \cdots A_{2t}$. In fact, we are going to show something much stronger: It is impossible to write $\left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right)$ as $MN$ for two noninvertible $2 \times 2$ matrices with entries in $R$.

Since $R$ is a graded integral domain, $\det M$ and $\det N$ would have to be of pure degree. If $\det M$ has degree $0$ then $M$ is invertible. So we must have $\det M$ and $\det N$ of degree $1$.

Write $M=M_0 + M_1 + \cdots$ where $M_k$ is pure of degree $k$, and likewise for $N$. So $$\left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right) = M_0 N_1 + M_1 N_0.$$ Since $\det M$ has no constant term, we have $\det M_0 =0$ and, similarly, $\det N_0 =0$. Let $\left( \begin{smallmatrix} u_1 & u_2 \end{smallmatrix} \right) \in k^2$ be in the left kernel of $M_0$ and let $\left( \begin{smallmatrix} v_1 \\ v_2 \end{smallmatrix} \right) \in k^2$ be in the right kernel of $N_0$. So the above equation shows that $\left( \begin{smallmatrix} u_1 & u_2 \end{smallmatrix} \right) \left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right) \left( \begin{smallmatrix} v_1 \\ v_2 \end{smallmatrix} \right) =0$.

But there are no nonzero vectors in $k^2$ for which $\left( \begin{smallmatrix} u_1 & u_2 \end{smallmatrix} \right) \left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right) \left( \begin{smallmatrix} v_1 \\ v_2 \end{smallmatrix} \right) =0$, a contradiction.

UPDATE: Pushing this argument further, I can show that $\begin{pmatrix} 0 & u & v & w \\ -u & 0 & x & y \\ -v & -x & 0 & z \\ -w & -y & -z & 0 \end{pmatrix}$ cannot be expressed as $MN$ with $\det M$ and $\det N$ nonconstant. Proof: Abbreviate the above matrix to $S$. As before, deduce that we can write $M_0 N_0 =0$ and $M_0 N_1 + M_1 N_0 = S$ with $\det M_0 = \det N_0 = 0$. So $\mathrm{rank} M_0 + \mathrm{rank} N_0 \leq 4$. Without loss of generality, suppose that $M_0$ has nullity at least $2$ and $N_0$ has nullity at least $1$. So, as before, we can find linearly independent constant vectors $p$ and $q$ with $p^T M_0=q^T M_0 =0$ and a nonzero vector $r$ with $N_0 r=0$. We have $p^T S r = q^T S r = 0$ as before.

Let $p=(p_1, p_2, p_3, p_4)$ and $r=(r_1, r_2, r_3, r_4)$. Looking at the coefficient of $u$ in $p^T S r$, we get $p_1 r_2 = p_2 r_1$. Looking at $v$, $w$, $x$, $y$ and $z$, we see that $p_i r_ = p_j r_i$ for all $(i,j)$, so $p$ and $r$ are proportional. Similarly, $q$ and $r$ are proportional, so $p$ and $q$ are proportional, contradicting that we choose them linearly independent. QED.

I have no $2 \times 2$ counter-example.

• Thanks, that's really amazing! Could anyone correct a bit misleading misprint: $\det(M)\cdot \det(N)=\Delta^2$ should stand instead of $\det(M)=\det(N)=\Delta^2$? – mikhail skopenkov Mar 6 '12 at 11:59
• Statement removed altogether. On reflection, I'm not sure whether R is a UFD, and the rest of the paragraph is correct without it. – David E Speyer Mar 6 '12 at 13:07
• @David : I know the statement that $k[x_1,\ldots,x_n]/(x_1^2+\ldots+x_n^2)$ is a UFD for $n \geq 5$, thus $R$ is a UFD (at least in characteristic not 2...). – François Brunault Mar 6 '12 at 14:26

EDIT: does not work as stated here, see in the comments. Although maybe it is possible to do some kind of reduction similar to the SNF.

We can at least reduce to the case of unimodular matrices. Take the Smith Normal Form of your polynomial, which will look like

$$E \begin{bmatrix} p & 0\\0& pq^2 \end{bmatrix} F,$$

where we can assume that $\det E=\det F=1$. The thesis is equivalent to showing that each of the factors

$$E,\, \begin{bmatrix} p & 0\\0& p \end{bmatrix},\, \begin{bmatrix} 1 & 0\\0& q^2 \end{bmatrix},\, F$$

admits a factorization of the kind you are interested in. David Speyer's trick in the comments works for the second factor, and the third is trivial. So the only missing step is the following lemma.

Lemma If $\det F=1$, then $F$ can be factored as per your request.

I am not so familiar with matrices with entries in a ring. Is it true, for instance, that in this case the $PLU$ factorization of $F$ can be done entirely in the ring? This would be enough to prove the lemma.

• Thank you very much for your answer. Your Lemma seems a right start point to attack the first question. However, I do not understand the reduction: the article on Smith normal form (en.wikipedia.org/wiki/Smith_normal_form) says that the form can be always found for matrices over a principle ideal domain (en.wikipedia.org/wiki/Principal_ideal_domain) while C[x,y] is not a principle ideal domain. – mikhail skopenkov Mar 6 '12 at 11:52
• You're right, sorry, I thought it worked on all polynomial rings. Although maybe in the bi-variate case it is possible to do something similar to the proof on Wikipedia, with successive triangular reduction steps? – Federico Poloni Mar 6 '12 at 11:57