Linear function on $\Bbb R^n$ If $L$ is a linear function $\mathbb{R}^n$, how do I show that $\mathbf D L(p)=L$ for all points $P$ in $\mathbb{R}^n$. Why does this not contradict calculus on one variable, where the derivative of $x \mapsto mx + b$ is the constant $m$?
 A: The function $f(x) = mx + b$ is not a linear function, it is an affine function (linear functions preserve scalars and sums, i.e. $f(x+y)=f(x) + f(y)$ and $f(\lambda x ) = \lambda  f(x)$).
Given a linear function $f : \mathbb{R}^n \longrightarrow \mathbb{R}^n$, we know such a function is given by
\begin{equation*}
\begin{pmatrix}
x_1 \\
x_2 \\
\vdots \\
x_n
\end{pmatrix}
\mapsto f(x) = \begin{pmatrix}
f_1(x) \\
f_2(x) \\
\vdots \\
f_n(x)
\end{pmatrix} = Ax =
\begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
 \vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2 \\
\vdots \\
x_n
\end{pmatrix} = 
\begin{pmatrix}
a_{11} x_1 + a_{12} x_2 + \cdots+ a_{1n} x_n \\
a_{21} x_1 + a_{22} x_2 + \cdots + a_{2n} x_n \\
\vdots \\
a_{n1} x_1 + a_{n2} x_2 + \cdots + a_{nn} x_n
\end{pmatrix}
\end{equation*}
With that, we clearly have
\begin{equation*}
\frac{\partial f_i}{\partial x_j} (P) = a_{ij}
\end{equation*}
Hence $Df(P) = A$, the same linear function, as required.
A: In general, the derivative of $f$ mapping into the space $\Bbb R$ is defined as the (unique) linear function $A$ such that
$$
\lim_{h \to 0} \frac{f(x + h) - f(x) - A(h)}{\|h\|} = 0
$$
So, when $f: \Bbb R^n \to \Bbb R$, $A(h) = [f'(x)](h)$ is a linear function that takes vectors in the domain of $f$, namely $\Bbb R^n$.  

So, suppose we have a linear function $f:\Bbb R \to \Bbb R$, i.e $f(x) = ax$ for $a \in \Bbb R$.  We find that
$$
\lim_{h \to 0} \frac{f(x+h) - f(x) - ah}{h} = 0
$$
it follows that the derivative of $f$ is given by the map $f'(x): \Bbb R \to \Bbb R$ defined as $[f'(x)](h) = ah$.  Notice that this derivative does not change for different inputs $x$. That is, the function given by $f'(x)$ is constant.
It is common to identify this map $[f'(x)](h) = ah$ with the constant $a$, hence the usual statement that $f'(x) = a$. 
Similarly, defining $f:\Bbb R^n \to \Bbb R$ by $f(x) = Lx,$ we find 
$$
\lim_{h \to 0} \frac{f(x+h) - f(x) - Lh}{\|h\|} = 0
$$
So that $[f'(x)](h) = Lh$.  Note that this function does not change for different inputs $x$.  That is, the derivative of a linear function is indeed "constant".
