$f(X^1,...,X^n)=\det[X^1...X^n]$ is differentiable

Question

Let $f:\mathbb{R}^n\times\dots \times\mathbb{R}^n\rightarrow\mathbb{R}$ defined by $$ \ f(X^1,...,X^n)=\det[X^1...X^n] \\ $$ is a differentiable function and $Df_H(A)=\sum_{i=1}^n\det[A^1\dots A^{i-1}H^iA^{i+1}\dots A^n]$.

I wanted to start with showing $DF(A)$ is linear, but I couldn't show it because if I define $T(A)=\sum\det[A^1\dots H^i\dots A^n]$, I get $$ \ T(\alpha A)=\sum \det[\alpha A^1\dots\alpha H^i\dots\alpha A^n]=\alpha^n\sum[A^1\dots H^i\dots A^n]\ne \alpha T(A) \\ $$ After I show this, I couldn't either to prove that the limit $$ \ \lim_{H\rightarrow 0}\frac{f(A+H)-f(A)-T_H(A)}{\left \| H \right \|} \\ \ =\lim_{H\rightarrow 0}\frac{\det[A^1+H^1...A^n+H^n]-\det[A^1...A^n]-\sum\left \| H \right \|\cdot\det [A^1...H^i...A^n]}{\left \| H \right \|} \\ $$ equals to $0$.

Answer 1

Given a matrix $M$, let

• $M^i$ be its $i^\text{th}$ column
• $M^{(i)}$ be $M$ with the $i^\text{th}$ column removed,
• $M_{(j)}$ be $M$ with the $j^\text{th}$ row removed,

and the symbols $M^i_{(j)}$ and $M^{(i)}_{(j)}$ defined by combining the corresponding definitions.

First, perform one step of the cofactor expansion on the $i^\text{th}$ column of $Df_H(A)$:

\begin{align*} \sum_{i=1}^n\det[A^1,\dots A^{i-1},B^i,A^{i+1},\dots, A^n] &= \sum_{i=1}^n\sum_{j=1}^n (-1)^{i+j}b_{ij}\det[A^1_{(j)},\dots A^{i-1}_{(j)},A^{i+1}_{(j)},\dots, A^n_{(j)}] \\ &=\sum_{i=1}^n\sum_{j=1}^n (-1)^{i+j}b_{ij}\det(A^{(i)}_{(j)}) \\ &=\sum_{i,j} (\operatorname{adj}A)_{ji}b_{ij} \end{align*}

(the last equality is true because it is the definition of the adjugate Adj.)

Then, read the second half of Christoph's answer to a related question.