Matrix derivative of $|I-2iT\Sigma|^{-\frac{n}{2}}$ 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?
 A: $
\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.
$$\eqalign{
A:B &= B:A \\
A:B &= A^T:B^T \\
C:\LR{AB} &= \LR{CB^T}:A \\&= \LR{A^TC}:B \\
}$$
