Differentiability of matrix-vector product map

Question

Let $E$ denote the set of endomorphisms (linear operators) on $\mathbb{R}^n$, and define $G : \mathbb{R^n} \times E \rightarrow \mathbb{R}^n$ by $$ G(x,X) = Xx. $$

Question: Is $G$ differentiable? If yes, what is its derivative?

I could not go very far in my attempt. For any $(y,Y), (h,H) \in \mathbb{R^n} \times E$ we have $$ G((y,Y)+(h,H)) - G(y,Y) = Yh + Hy + Hh. $$ The next step would be to guess a linear map $L_{(y,Y)}$ such that $$ \frac{\lVert G((y,Y)+(h,H)) - G(y,Y) - L_{(y,Y)}(h,H) \rVert}{\lVert (h,H) \rVert} \rightarrow 0$$ as $(h,H) \rightarrow 0$. However I am unable to come up with such a map.

Answer 1

Product rule. If $B$ is a continuous bilinear function then $B$ is differentiable and $$B'(x,y)\cdot (h,k) = B(x, k) + B(h, y).$$ Proof. Simply expand $B(x+h,y+k)=B(x,y)+B(x,k)+B(h,y)+B(h,k).$ Then, $\|B(h,k)\|\leq\|B\|\|h\|\|k\|=o(\|(h,k)\|).$ Q.E.D.

Apply simply this rule to $B(x,X)=Xx.$

Answer 2

HINT: suppose that $\langle\cdot |\cdot \rangle:\Bbb R^n\times\Bbb R^n\to\Bbb R$ is a bilinear product and $f,g:\Bbb R^m\to\Bbb R^n$ are Fréchet differentiable functions, then its easy to check from the definition of Fréchet derivative that

$$\partial[\langle f(x)|g(y)\rangle](h,v)=\langle\partial f(x)h|g(y)\rangle+\langle f(x)|\partial g(x)v\rangle\tag1$$

for $x,y,h,v\in\Bbb R^m$. Then note that the action of a matrix on a vector can be written as a sum of bilinear products, that is

$$Xv=([X]_1\cdot v,[X_2]\cdot v,\ldots,[X]_n \cdot v)\tag2$$

where the dot $\cdot$ stands for the dot product in $\Bbb R^n$ and $[X]_k$ is $k$-th file of the matrix $X$. Then using $(1)$ is easy to find $\partial[Xv](Y,h)$ for any $h,v\in\Bbb R^n$ and any $X,Y\in E$.