Differentiability of Linear Maps I am wondering whether all linear mappings have first-order partial derivatives (or stronger properties such as being continuously differentiable at all orders). Formally, suppose $A$ is an $m \times n$ matrix and define the mapping $F: \mathbb{R}^{n} \to \mathbb{R}^{m}$ by $$F(x) = Ax \text{ for every } x \in \mathbb{R}^{n}$$
It is a well known result that the above identity implies that $F$ is linear. Given this, do we immediately know that $F$ has first order partial derivatives? My thought process so far is that because $F$ must be linear in each of its components, and because a linear function is differentiable, then $F$ has first order partial derivatives. Am I on the right track?
 A: The derivative at each $x\in \mathbb{R}^{n}$ is the same namely, $A$. Hence the derivative map $D:\mathbb{R}^{n}\rightarrow L(\mathbb{R}^{n},\mathbb{R}^m)$is continuous since it is constant. This says that $F$ is continuously differentiable.
Apply the theorem that says $F$ is continuously differentiable iff the first partial derivatives exist and are continuous.
A: What you're after giving is pretty much a definition of the derivative in higher dimensions. Recall that in $\mathbb{R}$ we define the derivative as $f(x+h)=f(x)+f'(x)h+o(h)$. This is just a linear approximation. 
The same thing happens in higher dimenions: $$F(x+h)=F(x)+F'(x)h+o(h)$$ where $F'(x)$ is now an appropriate matrix. So in fact in you're example, not only is $F$ differentiable, but $$F'(0)=A$$
A: Your intuition is correct. I'll use $(v)_i$ to mean the $i$-th component of the vector $v$. We have $(F(x))_i=(a_1x_1+\cdots +a_mx_m)$ for some constants $a_i$, $1\leq i\leq m$ which is linear in each variable and so has all partial derivatives.
A: You are correct.  If you write your function F in terms of its components it sends $(x_1,...,x_n)$ to the vector with first component $A_{11}x_1+A_{12}x_2 +...+A_{1n}x_n$, second component $A_{21}x_1+A_{22}x_2 +...+A_{2n}x_n$, etc., where $A_{ij}$ is the entry of your matrix $A$ in the i-th row and j-th column.  The derivative of a vector valued function is defined componentwise, and each of the components is differentiable.
A: The derivative of $F$ at every point is just $A$ itself.
Let $e_1, e_2, \ldots$ be standard basis vectors.  $F$ can be written as
$$F(x) = (x \cdot e_1) A(e_1) + (x \cdot e_2) A(e_2) + \ldots$$
Let $a$ be a vector, and let $\nabla$ be the standard differentiation operator with respect to $x$.  The derivative of $F$ in the direction of $a$ is then
$$\begin{align*} (a \cdot \nabla) F(x) &= (a \cdot \nabla) (x \cdot e_1) A(e_1) + (a \cdot \nabla) (x \cdot e_2) A(e_2) + \ldots \\ & = (a \cdot e_1) A(e_1) + (a \cdot e_2) A(e_2) + \ldots \\ &= F(a) = A(a) \end{align*}$$
For linear functions, it can be thought that any scalar differential operator (like $a \cdot  \nabla$) can freely come inside a linear function to differentiate only its argument.
