Derivative of component wise power of a vector

Question

let $v = [v_1,v_2,...,v_n]^T$. Define the component wise a-th power of $v^a$ be $[v_1^a,v_2^a,...,v_n^a]^T$. What is $\frac{dv^a}{dv}$ and how to arrive at it?

Answer 1

Matrix multiplication of diagonal matrices can replace component-wise multiplication of vectors.
In particular, if $\def\o{{\tt1}}\o$ is the all-ones vector and $\def\Diag{\operatorname{Diag}} V=\Diag(v)\:$ then $$\eqalign{ v^{(n)} &= V^n\,\o \\ dv^{(n)} &= dV^{n}\:\o = nV^{n-1}\:dV\,\o = nV^{n-1}\:dv \\ \frac{\partial v^{(n)}}{\partial v} &= nV^{n-1} \\ }$$ As a check, note that setting $n=\tt1$ recovers the well-known result $$\frac{\partial v}{\partial v} = I$$

Update

The Diag() function cannot be represented using only matrix products, but if you allow elementwise/Hadamard products then $$\Diag(v) \;=\; I\odot\left(v\o^T\right) \;=\; I\odot\left(\o v^T\right)$$ Another possibility is to extend the Kronecker delta symbol to three indices $$\def\d{\vec\delta} \d_{ijk} = \begin{cases} \o\qquad {\rm if}\;\;i=j=k \\ 0\qquad {\rm otherwise} \\ \end{cases}$$ Then you can use a matrix product (aka dot product) to write $$\Diag(v) \;=\; \d\cdot v \;=\; v\cdot\d$$