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.
 A: 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 $f$.
A: 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:
Adv.Math.(2012)
arXiv
Slides
A: 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 $n = 1$.
A: 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
$* * *$
