The Determinant is a Hyperdeterminant of a Matrix

September 27, 2008

In the previous post we defined the hyperdeterminant. To illustrate how the definition works and to justify using the same notation as we do for determinants, we need to work out the hyperdeterminant for an N x M matrix A (which may not be square) and show that it is the equivalent to the definition of the determinant for this case.

We start with the bilinear form of the matrix A over two vector spaces of dimension N and M which can be written is matrix notation

F(x,y) = xTAy

F has a stationary point at (x,y) when

∂F/∂x = Ay = 0      and       ∂F/∂y = xTA = 0

The point is a singularity if x0 and y0. From the well known theory of matrices we know that this is equivalent to saying that A has both a left and right eigenvector with zero eigenvalue. When the matrix A is square (M = N), this happens if and only if A is singular, i.e. det(A) = 0. This confirms that the definition of a singular hypermatrix when applied to square matrices is equivalent to the usual definition of a singular matrix. It also tells us that the hyperdeterminant for a square matrix only exists when the matrix is square and that it must be the determinant times some factor. To confirm that the factor is one we must see that the coefficient of the first term in the determinant is one when the terms are ordered as required in the definition of the hyperdeterminant. This follows from the formula for the determinant which is

det(A) = Σ sign(σ)A1 σ(1) … AN σ(N)

Where the sum is over all permutations σ which permute the values of the indices and sign(σ) = ±1 is the signature of the permutation. The first term under the given ordering of terms is the diagonal term (+1) A11…ANN which has coeficient one as required.

In the case of a non-square matrix with N < M we can still get singular points, but the singular matrices are determined by more than one polynomial equation (M-N+1 of them in fact) these polynomials are determinants of N x N submatrices of the N x M matrix. Because the set of singular matrices in this case is determined by more than one polynomial equation there is no hyperdeterminant.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: