Matrix derivative of $|I-2iT\Sigma|^{-\frac{n}{2}}$

Question

I just started learning matrix derivatives and I'm trying to find derivative of $$f(T) = |I-2iT\Sigma|^{-\frac{n}{2}},$$ where $T$ and $\Sigma$ are symmetric matrices. This is what I already have: $$\frac{df(T)}{dT} = \frac{d(|I-2iT\Sigma|^{-\frac{n}{2}})}{d(|I-2iT\Sigma|)}\cdot\frac{d|I-2iT\Sigma|}{dT} = -\frac{n}{2}|I-2iT\Sigma|^{-\frac{n}{2}-1}\cdot|I-2iT\Sigma|\text{vec}^T((I-2iT\Sigma)^{T})^{-1} =\quad -\frac{n}{2}|I-2iT\Sigma|^{-\frac{n}{2}}\cdot\text{vec}^T((I-2iT\Sigma)^{T})^{-1}$$ but I don't know how to simplify this equation. Can anyone help me with this?

Answer 1

$ \def\p{\partial} \def\LR#1{\left(#1\right)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\fracLR#1#2{\LR{\frac{#1}{#2}}} $For typing convenience, define the matrix variables $$\eqalign{ S &= 2i\Sigma, \qquad A &= I-TS \\ }$$ Write the function, take the logarithm, calculate the differential, and rearrange to recover the gradient $$\eqalign{ f &= \det\!\LR{A}^{-n/2} \\ \log(f) &= -\fracLR{n}{2}\log\LR{\det(A)} \\ \fracLR{df}{f} &= -\fracLR{n}{2}\LR{A^{-T}:dA} \\ df &= -f\fracLR{n}{2}A^{-T}:\LR{-dT\,S} \\ &= +\fracLR{nf}{2}A^{-T}S^T:dT \\ \grad{f}{T} &= \fracLR{nf}{2}A^{-T}S^T \\ \\ }$$

---

In the above, a colon is used as a concise notation for the trace $$\eqalign{ A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; \trace{A^TB} \\ A:A &= \|A\|^2_F \\ }$$ The properties of the underlying trace function allow the terms in a such a product to be rearranged in many different but equivalent ways, e.g. $$\eqalign{ A:B &= B:A \\ A:B &= A^T:B^T \\ C:\LR{AB} &= \LR{CB^T}:A \\&= \LR{A^TC}:B \\ }$$