Is there a way to simply this (vectorized matrix, derivative of a matrix)

Question

$vec^T((X^{-1})^2)^T\cdot|X^2|$

I got this as a part of

$\frac{d}{dX}|X^2|=vec^T((X^{-1})^2)^T\cdot|X^2|\cdot2X$

I'm afraid that the dimensions don't match so I might've done something wrong.

Answer 1

Let $Y=X^2$, then it appears that you wish to calculate the gradient of the Frobenius Norm of $Y$ (which I'll denote by $N$) with respect to $X$.

First, express the norm in terms of the Frobenius Product (denoted by a colon) and find the differential $$\eqalign{ N^2 &= Y:Y \cr 2N\,dN &= 2Y:dY \cr &= 2Y:(X\,dX + dX\,X) \cr &= 2(X^TY+YX^T):dX \cr\cr }$$ Next, find the gradient $$\eqalign{ \frac{\partial N}{\partial X} &= \frac{1}{N}(X^TY+YX^T) \cr &= \frac{1}{N}(X^TX^2+X^2X^T) \cr\cr }$$ If $X$ has any special properties, e.g. $X^T=X$, they can be used to simplify the result further.

Finally, you can (if you wish), vectorize both sides of the gradient. But the right-hand-side will be very messy, involving Kronecker products and a Kronecker permutation matrix (due to the presence of $X^T$).

Update

It occurs to me that your question might have been about the determinant, instead of the norm. In which case, the function, differential, and gradient are a bit simpler
$$\eqalign{ f &= \det(X^2) \cr &= \det(X)^2 \cr\cr df &= 2\,\det(X)\,\,d\det(X) \cr &= 2\,\det(X)\,X^{-T}:dX \cr\cr \frac{\partial f}{\partial X} &= 2\,\det(X)\,X^{-T} \cr &= 2\,{\rm adj}(X^T) \cr }$$ where ${\rm adj}(A)$ denotes the adjugate of the matrix $A$.