# A SPECIAL Derivative of Matrix Determinant (tensor involved)

1. Oct 13, 2009

### roger1318

There is a surface defined by setting implicit function g(x)=0, where x is a 3 by 1 column vector, denoting a point on the surface;

3X1 vector $$\nabla$$g(x) is the Gradient(surface normal at point x;

3X3 matrix H(g(x)) = $$\nabla^2$$(g(x)) is the Hessian Matrix;

3X3X3 tensor $$\nabla^3g(\bold{x})$$ is $$\frac{\partial \bold{H}}{\partial \bold{x}}$$

The goal is to find $${\color{red}\frac{\partial (det(\bold{H}))}{\partial \bold{x}}}$$ ; which should be a 3X1 vector since the determinant of H is a scalar.

I found the formula for calculating the derivative of the determinant of a square matrix with respect to itself at (http://en.wikipedia.org/wiki/Matrix_calculus" [Broken]), which in my case here is a 3X3 matrix.

$$\frac{\partial (det(\bold{H}))}{\partial \bold{H}}=det(\bold{H})\cdot\bold{H}^-^1}$$

But what I want is the derivative with respect to that point x, not with respect to the matrix itself in the conventional sense.

So I attempted to use chain rule as we do in most cases:

$$\frac{\partial (det(\bold{H}))}{\partial \bold{x}}=\frac{\partial (det(\bold{H}))}{\partial \bold{H}}\cdot\frac{\partial \bold{H}}{\partial \bold{x}}=det(\bold{H})\cdot \underline{\bold{H}^-^1}}\cdot \underline{\underline{\nabla^3 g(\bold{x})}}$$

Now here comes the problem:
$${\color{blue}\bold{H}^-^1}}$$ is a 3X3 matrix and $${\color{blue}\nabla^3g(\bold{x})}$$ is 3X3X3 tensor; their multiplication product is still a 3X3X3 tensor but not the 3X1 vector as expected.

I am pretty sure something must've gone wrong; anybody could tell me where? And how am I supposed to deal with this determinant derivative issue? If for some reason the chain rule doesn't apply here, what rule should I use to get the 3X1 vector?