Let $n$ be a positive integer, and let $M = (m_{ij})$ be a skew $2n \times 2n$ matrix. That is, we have $m_{ij} = -m_{ji}$ for $1 \leq i, j \leq 2n$. Then it is well-known that

$$\det M = p(M)^2,$$

where $p$ is a polynomial in the entries $m_{ij}$. The polynomial $p(M)$ is called the Pfaffian of $M$.

Is there a generalization of this? That is, is there a natural family of $kn \times kn$ matrices whose determinants are perfect $k$-th powers of polynomials in the entries?

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    $\begingroup$ Singular matrices. :-) $\endgroup$ – LSpice Jan 9 at 0:57
  • $\begingroup$ see : mathoverflow.net/questions/278981/… $\endgroup$ – Libli Jan 9 at 6:31
  • $\begingroup$ @Libli The $n = 4m$ case seems closely related to what I need... if you could expand on this and give a reference I will accept that as the answer $\endgroup$ – Stanley Yao Xiao Jan 9 at 7:51
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    $\begingroup$ Block matrices with an $n \times n$ block repeated $k$ times :-) $\endgroup$ – Zach Teitler Jan 9 at 21:01

A good class of examples of this is given by Clifford algebras: If $V$ is a real vector space with endowed with a quadratic form $q:V\to\mathbb{R}$, the algebra $Cl(q)$ is the algebra generated by the elements of $V$ subject to the multiplication rule $x^2 = -q(x)$. If $M$ is a $Cl(q)$-module, say $M\simeq\mathbb{R}^m$, then we have an inclusion $V\hookrightarrow\mathrm{End}(M)$ and the characteristic polynomial of $x\in V\subseteq\mathrm{End}(M)$ is easily seen to be $(t^2+q(x))^{m/2}$, so we have $$ \det(x) = q(x)^{m/2} $$ for all $x\in V$.

For example, if $V$ is $\mathbb{R}^8$ with its standard Euclidean quadratic form $q$, then $Cl(q)$ is isomorphic to $\mathrm{End}_{\mathbb{R}}(\mathbb{R}^{16})$, so we can take $M=\mathbb{R}^{16}$ (and every $Cl(q)$-module is $\mathbb{R}^{16k}$ for some integer $k$). Thus, in this case, we have $\det(x) = p(x)^8$ where $p(x) = |x|^2$ for all $x\in V$.

In general, when $V\simeq\mathbb{R}^n$ and $q_n:V\to\mathbb{R}$ is nondegenerate, the dimension of a minimal nontrivial $Cl(q_n)$-module grows (roughly) exponentially with $n$, so the minimal $m$ grows exponentially with $n$. This shows that there are nontrivial 'irreducible' examples with $\det(x) = p(x)^k$ for $k$ arbitrarily large and that there is no bound on the possible dimension $n$ of the subspace $V\subset\mathrm{End}(M)$.

Remark: Given a linear subspace $V\subset\mathrm{End}(\mathbb{R}^{m})$ such that there exists a polynomial $p:V\to\mathbb{R}$ and an integer $k = m/\deg(p)>1$ such that $\det(x) = p(x)^k$, we say that the pair $(V,\mathbb{R}^m)$ is irreducible if there is no nontrivial subspace $M\subset\mathbb{R}^m$ such that $x(M)\subset M$ for all $x\in V$ and $\det(x_{|M}) = p(x)^j$ for all $x\in V$, where, necessarily, $j = (\dim M)/\deg(p)$.

The interesting problem for linear subspaces $V\subset\mathrm{End}(\mathbb{R}^m)$ on which the $\det$-function is a higher power of a polynomial on $V$ is to classify the irreducible ones of maximal dimension for a given $m$.


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