Derivative of trace

Question

I had never even heard of matrix calculus before yesterday but need it for a simple calculation and am stumped. If I can solve the following problem, I think I'll be fine for my larger calculuation. Let $F: \text{GL}(2,\mathbb{C}) \to \mathbb{R}$ via $$ g \mapsto \text{tr}((gAg^{-1})^*gAg^{-1}) $$ For some fixed matrix $A$. So this is like an inner product, but conjugated. My question is, what is the derivative of this? I want to take the derivative at the identity, so I should get a map: $$ dF\vert_{id}: T_0\text{GL}(2,\mathbb{C}) \cong \text{End}(2,\mathbb{C}) \to T_o\mathbb{R} \cong \mathbb{R}. $$ In particular, given some generic matrix $M$ (some endomorphism), What is $dF\vert_0 (M)$?

Answer 1

$F:GL_n(\mathbb{C})\rightarrow \mathbb{R}$ has no derivative (as a complex function) except if $A$ is a scalar matrix.

The reason is that $u\in GL_n(\mathbb{C})\rightarrow tr(u^*u)\in \mathbb{R}$ has no derivative (as a complex function). For example, the case $n=1$ reduces to $u\in \mathbb{C}\rightarrow \overline{u}u\in \mathbb{R}$, that is not an holomorphic function.

Yet, the function $\phi:u\in M_n(\mathbb{R})\rightarrow tr(u^Tu)\in\mathbb{R}$ has a derivative: $D\phi_u:h\in M_n(\mathbb{R})\rightarrow 2tr(u^Th)$.

In the same way, the function

$F:g\in GL_n(\mathbb{R})\rightarrow u=gAg^{-1}\rightarrow \phi(u)=tr(u^Tu)\in\mathbb{R}$ has also a derivative:

$DF_g:k\in M_n(\mathbb{R})\rightarrow 2tr(u^TDu_g(k))$ where $Du_g(k)=kAg^{-1}-gAg^{-1}kg^{-1}$.

In particular, if $g=I_n$, then $u=A$ and $DF_{I_n}:k\rightarrow 2tr(A^T(kA-Ak))=2tr((AA^T-A^TA)k)$.

EDIT. Answer to @user46348. Proceed as for a product

$D(gAg^{-1})=(Dg)Ag^{-1}+(gA)(Dg^{-1})$ where $Dg(k)=k,Dg^{-1}(k)=-g^{-1}kg^{-1}$.