Let $\mathsf{SPD}_n$ denote the set of all real symmetric and positive definite $n\times n$ matrices. This set is convex so for every $A,B\in\mathsf{SPD}_n$ there exists a smooth path $\varphi:[0,1]\rightarrow\mathsf{SPD}_n$, i.e. $\varphi$ is a smooth function with $\varphi(0) = A$ and $\varphi(1) = B$.
Let $f:\mathsf{SPD}_n\rightarrow \mathbb R$. I am interested in the derivative: $\frac{\partial}{\partial t}f(\varphi(t))$. By the usual chain rule, this is $$\mathfrak Df(\varphi(t))\cdot\dot\varphi(t),$$ where $\mathfrak Df$ is the derivative of $f$ with respect to its argument (i.e. a matrix) and $\dot\varphi$ the derivative along the path. Note that $\cdot$ denotes the Frobenius inner product. This is my setting.
Since $\varphi(t)$ is a symmetric matrix for every $t$, there should exist orthonormal matrices $V(t)$ and diagonal matrices $\Lambda(t)$ such that $\varphi(t) = V(t)\Lambda(t)V(t)'$. Is there any chance to restate my setting (perhaps under additional assumptions) such that $$\frac{\partial}{\partial t}f(\varphi(t)) = \mathfrak Df(\varphi(t))\cdot\dot\psi(t),$$ where $\dot\psi$ is a vector valued function and $$ \mathfrak Df(M) = \left[\begin{array}{c} \frac{\partial}{\partial\lambda_1}f(M) \\ \frac{\partial}{\partial\lambda_2}f(M) \\\vdots \\ \frac{\partial}{\partial\lambda_n}f(M)\end{array}\right] $$ now is the vector of derivatives wrt to the eigenvalues of the argument of $f$? Consequently, $\cdot$ denotes now the usual Euclidean inner product.
In words: can $\frac{\partial}{\partial t}f(\varphi(t))$ can be represented by $$\text{derivative of $f$ wrt eigenvalues of argument} \times \text{something} $$ ?
Background: The above is trivially true if I consider the set of diagonal matrices (because eigenvalues and diagonal elements elements coincide). Now I want to adapt the setting for spd matrices. In my application, $A$ and $B$ are covariance matrices. $A$ is the sample covariance and $B$ is a "target" covariance. On the way (i.e. along $\varphi$) from $A$ to $B$ dimension reduction should be performed. The dimension reduction works similarily to PCA which is why I need to take the derivative wrt eigenvalues: if $\left\vert\frac{\partial}{\partial\lambda_j(t)}f(\varphi(t)\right\vert<\xi$ for some threshold $\xi>0$, "dimension" $j$ is ignored.
