TITLE: Determinant for the element-wise derivative of a matrix
QUESTION [1 upvotes]: Let $M$ a $d\times d$ matrix with entries $$M_{ij}=x^{\alpha_j+i-j},$$ with $\alpha_j$ a non-negative integer. It is possible to write an expression for $det\left(\frac{d^{k}}{dx^k}M\right)$ in terms of $det(M)$? or for the particular cases $\alpha_j=0,\forall j\in \lbrace 0,...,d\rbrace$ and $k=1$?

REPLY [0 votes]: We can write
$$
M = \pmatrix{x^1\\ \vdots \\ x^d} \pmatrix{x^{\alpha_1 - 1} & \cdots & x^{\alpha_d - d}} = v(x)[w(x)]^T.
$$
We can see that $M$ has rank (at most) $1$, which means that for all $d > 1$, we will have $\det(M) = 0$. On the other hand, we find that
$$
M'(x) = [v(x)w^T(x)]' = v'(x) w(x)^T + v(x)w'(x)^T\\
= \pmatrix{1\\2x\\ \vdots \\ dx^{d-1}}w(x)^T + v(x) \pmatrix{(\alpha_1 - 1)x^{\alpha_1 - 2} & \cdots & (\alpha_d - d)x^{\alpha_d - d - 1}}.
$$
We can see that $M'(x)$ has rank (at most) $2$, which means that for all $d > 2$ we will have $\det M'(x) = 0$.
Similarly, we can generally conclude that $M^{(k)}(x)$ will have a zero determinant whenever $k < d-1$.
It is clear that if $k$ $M^{(k)}(x)$ is not constant, then it cannot be written as a function of the derivatives of $M$ of order less than $d - 1$, since these are all constant. However, we can see that there are some values for which $M^{(k)}$ no longer has constant zero determinant. For instance,
$$
M(x) = \pmatrix{1 & x\\ x^{-1} & 1} \implies M'(x) = \pmatrix{0&1\\-x^{-2} & 0}.
$$
We see that $\det(M'(x)) = x^{-2}$ cannot be written as a function of $\det(M(x)) = 0$.
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