Derivate of powers of stochastic matricies with respect to its elements

Question

Let $\mathbf{X} \in \mathbb{R}^{k \times k}$ be a square stochastic matrix (i.e $\mathbf{1}\mathbf{X} = \mathbf{1}$), are there any useful identities that can possibly be used to simplify the expression $\frac{\partial \mathbf{X}^n}{\partial X_{ij}}$, where $n \in \mathbb{Z}^+$ raises $\mathbf{X}$ to the $n$-th power and $X_{ij}$ is the element corresponding to the index $i,j$ in matrix $\mathbf{X}$.

During my search on StackExchange, I have come across greg's answer, stating that $$\frac{\partial F}{\partial\mathbf{X}_{ij}} = \sum_{k=1}^\infty \alpha_k(\sum_{l=1}^k\mathbf{X}^{k-l}\mathbf{E}_{ij}\mathbf{X}^{l-1})$$ where $$F(\mathbf{X}) = \sum_{k=0}^\infty \alpha_k \mathbf{X}^k$$ and $$\mathbf{E}_{ij} = e_ie_j^T$$

where $e_i$ is a cartesian basis vector and $\mathbf{E}_{ij}$ is the single-entry matrix.

Would anyone please tell me how this identity is derived or the name of this identity/theorem? Thank you so much!

Answer 1

In the same post that you linked to, I believe that Ruy's answer shows how that identity is derived (for fixed $k$). For $F(X) = X^n$, note that $\frac{\partial X^n}{\partial X_{ij}} = F'(X)(E_{ij}) = \left. \frac{\mathrm{d}}{\mathrm{d}t} \right|_{t=0} (X + t E_{ij})^n$: this is to be interpreted in the multivariable calculus way as the directional derivative of $F$ in the direction $E_{ij} = e_i e_j^T$ (i.e. the standard basis direction for the $(i, j)$th coordinate). Thus, as the answer shows, this evaluates to $$ E_{ij} X^{n-1} + X E_{ij} X^{n-2} + X^2 E_{ij} X^{n-3} + \dots + X^{n-2} E_{ij} X + X^{n-1} E_{ij} = \sum_{\ell=1}^n X^\ell E_{ij} X^{n-1-\ell} , $$ which is the same as the term in the infinite series version you are referencing up to a change of variables. (Why? Expand out $(X + t E_{ij})^n$; the only terms remaining after differentiating with respect to $t$ and sending $t$ to zero are the ones with a single $E_{ij}$ term in it.)