For an skew-symmetric matrix $A$ (meaning $A^T=-A$), the Pfaffian is defined by the equation $(\text{Pf}\,A)^2=\det A$. It is my understanding that this is defined for anti-symmetric matrices because it is known that the determinant of an anti-symmetric matrix is always a square of a polynomial in the entries of the matrix.

Now, skew-symmetry is sufficient to prove that the determinant is a square of a polynomial, but it is not necessary. The simplest example is the $2n\times 2n$ matrix $A=a I_{2n}$ with $a\in\mathbb{C}$ and $I_k$ the $k\times k$ identity matrix. The determinant is $\det A = a^{2n} = (a^n)^2$. Of course, for $a\neq 0$, $A$ is not skew-symmetric.

I have a few questions about this.

  1. Is there a generalization of a Pfaffian for any matrix whose determinant is a square of a polynomial?
  2. Is there a characterization (or some known set of properties) of matrices whose determinants are squares of polynomials?
  3. (Edit) Are there any known necessary and sufficient conditions for a matrix to have its determinant be the square of a polynomial (aside from skew-symmetry being sufficient)?

(Edit 2) For those who are curious, these questions arise from a problem from physics I am working on. I have a certain class of matrices whose characteristic polynomials (which arise as the determinant of a non-skew-symmetric matrix) appear to be the squares of Chebyshev polynomials. If I could prove that these characteristic polynomials must be squares of polynomials (using properties of the matrix) then I may be able to use some of the properties attributed to Pfaffians (or the proper generalization to non-skew-symmetric matrices) to confirm that they are indeed squared Chebyshev polynomials.

(Edit 3) To be as concrete as possible, I am looking for any information (e.g., answers to questions 1-3) on the set $$\{A\in\mathcal{M}_n(\mathbb{C}): \det A = p(\{a_{ij}\})^2\text{ with }p\text{ a polynomial} \}$$ where $\mathcal{M}_n(\mathbb{C})$ is the set of $n\times n$ complex matrices and $a_{ij}$ is the $i,j$'th entry of $A$.

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    $\begingroup$ Determinants can be squares for various random reasons. Are you asking about certain families of matrices whose determinants are squares? Otherwise I'd say it's a rather vague question. $\endgroup$ – darij grinberg Nov 27 '18 at 19:44
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    $\begingroup$ I don't know if there exist an extension of Pfaffians to matrices other than the antisymmetric ones. But if you are interested into these polynomials, here is a paper that enlarges the point of view, in particular by using exterior algebra : kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/1302-14.pdf $\endgroup$ – Jean Marie Nov 27 '18 at 19:59
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    $\begingroup$ @JeanMarie Thanks for the response. I have actually read quite a bit on Pfaffians in the context of Soliton theory (in particular the book "The Direct Method in Soliton Theory" by Hirota - same author as that paper you linked). They have many nice properties that could be very useful, but are defined only for anti-symmetric matrices (and rely on this fact quite heavily it seems). $\endgroup$ – UglyMousanova19 Nov 27 '18 at 23:21
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    $\begingroup$ Maybe a way to rephrase your question to appeal to those who deem it vague: Let $R$ be the ring of polynomials in matrix entries, so $\det\in R$. Let us call an ideal $I\subseteq R$ Pfaffian if $\det$ becomes a square in $R/I$ and denote by $J$ the intersection of all Pfaffian ideals. What is $J$? It characterizes the largest subvariety of our matrix space where we can define something like a Pfaffian. And also a question: Can you name a Pfaffian ideal that does not contain the vanishing ideal of all skew-symmetric matrices? $\endgroup$ – Jesko Hüttenhain Nov 27 '18 at 23:25
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    $\begingroup$ @JeskoHüttenhain: What about the ideal defining $\operatorname{SL}_n\left(K\right)$? That would be Pfaffian, too. $\endgroup$ – darij grinberg Nov 28 '18 at 3:34

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