Check that if F(X)=aX then DF(X)=aI I have the following problem:
Check that if $\ F: \mathbb{R}^n → \mathbb{R}^n$ is given by $\ F(X)=aX$ then $\ DF(X)=aI$, where $\ I$ is the identity matrix and $\ a$ in $\ R$.
I'm struggling with this, don't know where to start from, need some help, thanks!
 A: Note that by definition you have:
$$
(f_1(\textbf{x}),...,f_n(\textbf{x}))=F(\textbf{x})=F(x_1,...,x_n)=a(x_1,...,x_n)
$$
So that $f_i(x_1,...,x_n)=ax_i$, and therefore $D_if_i\equiv a$, and $D_jf_i\equiv0$ for $j\neq i$.
As a result we get for all $\textbf{y}\in\mathbb{R^n}$:
$$
F'(\textbf{x})\textbf{y}=\left(\sum_1^nD_jf_1(\textbf{x})y_j,...,\sum_1^nD_jf_n(\textbf{x})y_j\right)=\left(D_1f_1(\textbf{x})y_1,...,D_nf_n(\textbf{x})y_n\right)=a\textbf{y}
$$

More  generally, whenever $F$ is a linear transformation, we have $F'(\textbf{x})=F$ for all $\textbf{x}$. Indeed, in that case:
$$
\frac{\|F(\textbf{x}+\textbf{h})-F(\textbf{x})-F(\textbf{h})\|}{\|\textbf{h}\|}=0
$$
for all $\textbf{x}$ and $\textbf{h}\neq\textbf{0}$.
A: A slightly different take on this one:
You have to start with a definition of $DF$.  A pretty common one is:
Definition:  Let $F:U \to V$ be a function, where $U, V$ open are subsets of $\Bbb R^n$; $U, V \subset \Bbb R^n$.  Then we say $F$ is differentiable at a point $x \in U$ provided there exists a real $\epsilon > 0$ and a  linear map $DF(x):  \Bbb R^n \to \Bbb R^n$ such that, for all $h \in \Bbb R^n$ with $\Vert h \Vert < \epsilon$ we have
$F(x + h) = F(x) + DF(x)h + h \theta(h), \tag{1}$
where $\theta(h) \to 0$ as $h \to 0$.
Though this definition may seem a bit odd and unwieldy to those only familiar with the basics of calculus, it is easy to see that it yields the same result as the more elementary definition in the case $n = 1$, which is:
$DF(x) = \lim_{h \to 0} \dfrac{F(x + h) - F(x)}{h}. \tag{2}$
We illustrate this equivalence by means of some simple examples:  maintaining the choice $n = 1$, we take
$F(x) = x^2; \tag{3}$
then
$F(x + h) = (x + h)^2 = x^2 + 2xh + h^2 = F(x) + 2xh + h\theta(h)  \tag{4}$
with $\theta(h) = h$.  According to (4), we have
$DF(x) = 2x, \tag{5}$
in keeping with the usual result obtained from (2):
$DF(x) = \lim_{h \to 0} \dfrac{ x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \to 0} \dfrac{2xh + h^2}{h} =  \lim_{h \to 0}(2x +h) = 2x; \tag{6}$
likewise, taking
$F(x) = x^3, \tag{7}$
we see that
$F(x + h) = (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 = F(x) + 3x^2h + h(3xh + h^2), \tag{8}$
so with $\theta(h) = 3xh + h^2$ we have
$DF(x) = 3x^2, \tag{9}$
again in agreement with ordinary calculus.  As a final example, it is instructive for the present question to look at
$F(x) = ax + b, \tag{10}$
that is, a simple linear function of $x$.  Then
$F(x + h) = a(x + h) + b = ax + ah + b = ax + b + ah = F(x) + ah; \tag{11}$
here we have $\theta(h) = 0$ and so
$DF(x) = a \tag{12}$
for all $x$.
We return to the case of general $n$ and examine
$F(X) = AX + B, \tag{13}$
where $A$ is an $n \times n$ matrix and $B \in \Bbb R^n$.  We see
$F(X + h) = A(X + h) + B = AX + B + Ah = F(X) + Ah; \tag{14}$
thus
$DF(X) = A \tag{15}$
for all $X \in \Bbb R^n$; as in (10)-(12) we have that the derivative of a linear function is simply the coefficient of the independent variable; in this case it happens to be the matrix $A$.  When $B = 0$, we specialize (13) to
$F(X) = AX, \tag{16}$
finding by (15) that the derivative of a linear map is in fact that map itself.  
Applying these remarks to the question at hand, we see that
$F(X) = aX = aIX; \tag{17}$
$aI$ is a linear map, so we have
$DF(X) = aI \tag{18}$
for all $X \in \Bbb R^n$ as well.
