# Irreducibility of determinant of symmetric matrix

It is quite known fact that the determinant of arbitrary symmetric matrix is an irreducible polynomial in algebra $\mathbb C [x_{ij}, 1\leq i,j\leq n]$ ($x_{ij}=x_{ji}$) (see this: atlas.mat.ub.es/personals/sombra/papers/cayley/cayley.ps ).

Is there any geometric proof of this statement like the proof of irreducibility of determinant (from the biduality theorem in Gelfand-Zelevinsky-Kapranov book)?

Upd: I have found an algebraic proof of this fact (But I still need geometric). Since $\det (A A^{T})=\det(A)^2$ our polynomial (if not irreducible) is a square of irreducible. Since $\det diag(a_1,\ldots ,a_n)=a_1\cdot\ldots\cdot a_n$ our polynomial cannot be a square of any polynomial.

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Could you please provide a reference to the exact place inside the book? –  David Corwin Dec 26 '10 at 1:43
Is it possible to decompose Newton polytope of the determinant of symmetric matrix into Minkowski sum of two polytopes? Negative answer implies irreducibility. –  Petya Dec 26 '10 at 3:22
I did not understand your algebraic proof. It is true for diagonal matrices (for example) that $\det (A A^{T})=det(A)^2$ and yet the determinant there is neither irreducible nor the square of an irreducible. What is the ingredient I am missing? –  Aaron Meyerowitz Dec 27 '10 at 4:37
@Davidac897: It follows from the irreducibility of resultant which follows from irreducibility of variety dual to irreducible (the first chapter). @Aaron: $A=(a_{i,j})$, $\det(A)\in\mathbb C[a_{i,j}, 1\leq i,j\leq n]$, $\det(A A^T)\in\mathbb C[a_{i,j}, 1\leq i,j\leq n]$ ($a_{i,j}\ne a_{j,i}$). Do you understand what I meant? @Petya: This is a still algebraic proof. –  zroslav Dec 27 '10 at 7:24
@Davidac897: the construction of resultants is given in the third chapter –  zroslav Dec 27 '10 at 7:31

I don't know a proof based on biduality, but here is a short geometric proof, which generalizes to other similar situations (all degenerate matrices, singular hypersurfaces and many other examples considered in the book by Gelfand, Kapranov and Zelevinsky).

In all these cases the discriminant variety admits a canonical'' resolution of singularities, which in the case of symmetric matrices is constructed as follows. Consider the space $X$ of couples (a degenerate symmetric $n\times n$ matrix, a 1-dimensional subspace in the kernel of the matrix). This projects both to the space of matrices and to $\mathbf{P}^n$. The second projection gives the structure of a vector bundle over $\mathbf{P}^n$ on $X$, so $X$ is irreducible. The image of $X$ under the first projection is the discriminant hypersurface, which is irreducible, so is given by an irreducible polynomial $f$. By the Nullstellensatz the determinant is a power of $f$ times a constant. Now consider the family of matrices with $t,1,1,\ldots, 1$ on the diagonal and zeroes elsewhere (here $t\in\mathbf{C}$). Restricting the determinant to this family we get $t$, so the determinant is precisely $f$.

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I'm maybe very silly now but why the first projection image is the discriminant variety? And where are we using the symmetricity of maytrix? –  zroslav Dec 26 '10 at 7:44
zroslav -- the image of the first projection is formed by all matrices that have a kernel, which is to say, all degenerate matrices. You are right, the symmetry condition is not very important, the same argument shows that the determinant considered as a polynomial in the coefficients of an arbitrary matrix is irreducible. –  algori Dec 26 '10 at 7:49
I was thinking that your proof is right but know I understand that this is not correct. For example, your proof must be correct if the set of matrices is diagonal but there is a obvious contradiction. Where are you using the simmetricity? –  zroslav Dec 27 '10 at 7:19
Ah, Ok. I understood that if the set of all possible kernels is irreducible then our set of matrices is irreducible too. –  zroslav Dec 27 '10 at 7:30

I think you must mean the characteristic polynomial (the determinant is a scalar, not a polynomial), but then what about the identity matrix---it has characteristic polynomial $(\lambda-1)^n$ and so is not irreducible unless $n = 1$.

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I think he actually means the determinant of the matrix whose entries are indeterminates x_{i,j}, as a polynomial in these indeterminates. –  Mattia Talpo Dec 25 '10 at 18:56
@Mattia: yes I meant exactly that what you've wrote –  zroslav Dec 25 '10 at 19:01
Ah, OK, I see. But then I think you probably should edit the question to make that clear (instead of arbitrary symmetric matrix say something like a symmetric matrix in indeterminates $X_{ij}$). Also. the use of the word "Hessian" in the title is confusing---this usually refers to the matrix of second partial derivatives of a function. –  Dick Palais Dec 25 '10 at 20:21
@Dick: Ok. Yes this is an equivalent problem for formal differential symbols –  zroslav Dec 25 '10 at 20:49

guess that if the characteristic polynomial is reducible you can specialize all variables to constants, but one, and get a contradiction of the kind some large class of matrices would always have a reducible characteristic polynomial...

luis $* * *$

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@Luis, there is no need to sign your name in answers, as your name is automatically added by the software. –  Mariano Suárez-Alvarez Dec 26 '10 at 1:12

The original question was for a GKZ-style proof of the irreducibility of the determinant of a symmetric matrix:

If you view symmetric matrices as quadratic polynomials, the determinant of the associated symmetric matrix is actually the discriminant of the quadratic form. The discriminant is also the equation of the dual variety to the quadratic Veronese variety, which is irreducible via the bi-duality theorem.

This is actually the first case in a large family. The next question is what about the determinant of symmetric tensors? First, what should you take as the determinant and second, is it irreducible?

I wrote a paper answering these questions, which I will shamelessly promote here: