How to determine the partial derivative of a concatenation of a vector and a scalar?

Question

Suppose $\mathbf{x}$ is an $n$ dimensional vector and $y$ is 1 dimensional. We define $\mathbf{z}=(\mathbf{x},y)$ which is the concatenation of $\mathbf{x}$ and $y$. Given that $\mathbf{V}$ is a PSD matrix, how is it possible to find the derivative of $\| \mathbf{z}\|_{\mathbf{V}}$ (where $\| \mathbf{z}\|_{\mathbf{V}}\ =\sqrt{ \mathbf{z}^\top \mathbf{V}\mathbf{z}}$) with respect to $\mathbf{x}$ and $y$ separately as a closed form solution?

Answer 1

$ \def\R#1{{\mathbb R}^{#1}} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\frob#1{\left\| #1 \right\|_F} \def\qiq{\quad\implies\quad} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} \def\gradLR#1#2{\LR{\grad{#1}{#2}}} \def\m#1{\left[\begin{array}{c}#1\end{array}\right]} \def\mc#1{\left[\begin{array}{c|c}#1\end{array}\right]} $Consider a slightly more general case in which $y$ is also a vector $$\eqalign{ x\in\R{n}\quad y\in\R{m}\quad z\in\R{k}\qquad k=m+n }$$ Partition the identity matrix $$\eqalign{ I_k &= \mc{I_n&0\\0&I_m} = \mc{E&F} \\ E^TE &= I_n,\quad F^TF=I_m \\ E^TF &= 0,\quad\;\, F^TE=0,\qquad EE^T + FF^T=I_k \\ }$$ and use the partitions to relate the vectors $$\eqalign{ z &= Ex+Fy \qiq x=E^Tz,\;\;y=F^Tz \\ }$$ The same formula can also be applied to the respective gradients $$\eqalign{ g &= Ep+Fq \qiq p=E^Tg,\;\;q=F^Tg \\ \\ g &= \grad{\phi}{z}, \qquad p=\grad{\phi}{x}, \quad q=\grad{\phi}{y} \\ }$$ The remaining task is to calculate $g\,$ (which many previous posts on this site have done).