Question on Matrix Derivatives When differentiating with respect to a matrix, is it possible to rewrite the derivative operator via some transformation so that you're differentiating with respect to the diagonal eigenvalue matrix instead?
I.e., is something like this possible? 
\begin{equation}\frac{\partial }{\partial A}=Q'\frac{\partial}{\partial \Lambda}Q\end{equation} Where $Q$ is some transformation and $\Lambda$ is the diagonal matrix of eigenvalues. 
The reason I ask is because I'm working on something like this 
\begin{equation}f(\Lambda)=e^{\sum_{i}\lambda_{i}^{2}v_{i}}=e^{\mathrm{tr} \Lambda^{2}V}\end{equation}
I can rewrite this as 
\begin{equation}f(A)=e^{\mathrm{tr}UA^{2}U^{\dagger}V}=e^{\mathrm{tr}A^{2}U^{\dagger}VU}\end{equation}
I'm not sure if differentiating with respect to $A$ is ok because of the unitary transformations $U$ which technically depend on $A$.
 A: Let $f:M_n(\mathbb{R})\rightarrow \mathbb{R}$ be a $C^1$ function. Let $A$ be a diagonalizable matrix $A=PDP^{-1}$ with $A=[a_{i,j}],P=[p_{i,j}],P^{-1}=[p'_{i,j}]$ and $P$ is a fixed invertible matrix. Let $(E_{i,j})$ be the canonical basis of $M_n(\mathbb{R})$. Then $\dfrac{∂D}{∂a_{i,j}}=P^{-1}E_{i,j}P\;$  ($D$ is diagonal but a perturbation of $D$ is not). We consider the scalar product on $M_n(\mathbb{R})$ defined by 
$<U,V>=trace(UV^T)=\sum_{k,l}u_{k,l}v_{k,l}$ where $U=[u_{k,l}],V=[v_{k,l}]$.
Proposition:  $\dfrac{∂f}{∂a_{i,j}}=< \dfrac{∂f}{∂D},\dfrac{∂D}{∂a_{i,j}}>$.
Proof: $\dfrac{∂f}{∂a_{i,j}}=\sum_{k,l}\dfrac{∂f}{∂d_{k,l}}\dfrac{∂d_{k,l}}{∂a_{i,j}}=\sum_{k,l}(\dfrac{∂f}{∂D})_{k,l}(\dfrac{∂D}{∂a_{i,j}})_{k,l}$
For instance(the one of user?? few days ago), let $A=\begin{pmatrix}1&1\\0&-1\end{pmatrix},D=\begin{pmatrix}1&0\\0&-1\end{pmatrix},P=\begin{pmatrix}1&-1/2\\0&1\end{pmatrix}$ and let $f(A)=\sup(spectrum(A))$. Then $\dfrac{∂f}{∂A}=\begin{pmatrix}1&0\\1/2&0\end{pmatrix},\dfrac{∂f}{∂D}=\begin{pmatrix}1&0\\0&0\end{pmatrix}$. The formula above gives the link between these $2$ derivatives. Recall that $f$ is $C^1$ because the $\sup $ is reached by a sole eigenvalue.
