Derivative of a minor

Question

From matrix $Y_{T \times I}$ I can can calculate the covariance as follows: $$ S=\frac{1}{I}{Y^tY} $$ and a singular entry of the inverse like so (as explained in this post): $$ s_{i,j}^{-1} = \frac{(-1)^{i+j}M_{j,i}}{\det(S)} $$ where $M_{j,i}$ is a minor taken from $S^T$. What would $\frac{\delta(s_{i,j}^{-1})}{\delta{y_{t,i}}}$ be?

Answer 1

Let $S = \alpha Y^TY$
Then $$\eqalign{ \alpha S^{-1}Y^T &= (Y^TY)^{-1}Y^T = Y^+ \cr \alpha YS^{-1} &= Y(Y^TY)^{-1} = (Y^+)^T \cr }$$ where $Y^+$ is the pseudoinverse of $Y$.

Taking the requested differential $$\eqalign{ dS^{-1} &= -S^{-1}\,dS\,S^{-1} \cr &= -S^{-1}\,\alpha(dY^T\,Y+Y^T\,dY)\,S^{-1} \cr &= -S^{-1}\,dY^T\,\alpha YS^{-1} - \alpha S^{-1}Y^T\,dY\,S^{-1} \cr &= -S^{-1}\,dY^T\,(Y^+)^T - Y^+\,dY\,S^{-1} \cr\cr dS^{-1}_{ij} &= -S^{-1}_{im}\,dY^T_{mn}\,(Y^+)^T_{nj} - Y^+_{im}\,dY_{mn}\,S^{-1}_{nj} \cr &= -S^{-1}_{im}\,dY_{nm}\,Y^+_{jn} - Y^+_{im}\,dY_{mn}\,S^{-1}_{nj} \cr }$$ And applying the basic gradient for these situations:
$$\frac{\partial Y_{ij}}{\partial Y_{kl}} = \delta_{ik}\delta_{jl}$$ Yields $$\eqalign{ \frac{\partial S^{-1}_{ij}}{\partial Y_{kl}} &= -S^{-1}_{im}\,\delta_{nk}\delta_{ml}\,Y^+_{jn} - Y^+_{im}\,\delta_{mk}\delta_{nl}\,S^{-1}_{nj} \cr &= -S^{-1}_{il}Y^+_{jk} - Y^+_{ik}S^{-1}_{lj} \cr }$$ I know that you were hoping for something in terms of the minors, but nothing like that occurs to me.