Derivative of Linear Map I'm reading Allan Pollack's Differential Topology and got stuck on this argument: In the second paragraph of page 9, section 1.2 he said 
"Note that if $f:U\to \mathbf{R^m}$ is itself a linear map $L$, then $df_x=L$ for all $x\in U$. In particular, the derivative of the inclusion map of $U$ into $\mathbf{R^n}$ at any point $x\in U$ is the identity transformation of $\mathbf{R^n}$."
I don't quite understand his first sentence. How could $f=L$ and $L$ linear imply $df_x=L$? A counter example is $f(x)=x$ linear, then $f'(x)=1 \neq L$.
Thanks a lot for everyone's help!
 A: Given a vector space $V$ over the numeric field $\mathbb{R}$, the differential of a map $f:V\rightarrow V:\vec{x}\mapsto f(\vec{x})$ is defined as the map:
$$
df:V\rightarrow L(V,V):\vec{x}\mapsto df(\vec{x})\in L(V,V)
$$
where $L(V,V)$ is the set of all linear maps from $V$ to $V$, for what we have:
$$
\vec{v}\in V,\ f(\vec{x}+t\vec{v})=f(\vec{x})+t\cdot df(\vec{x})(\vec{v})+o(t)
$$
where, as usual $\lim_{t\rightarrow 0}\frac{o(t)}{t}=0$.
If we consider a function $f:V\rightarrow\mathbb{R}$, this formula tells us that, near to a point $\vec{x}\in V$, we can approximate the value of $f$ with the value of a map linear on the vector $\vec{v}$ thought as a vector in the tangent space to the image $f(V)\subset V\times\mathbb{R}$ attached at the point $(\vec{x},f(\vec{x}))$.
Given a linear map $L:V\rightarrow V$ the definition of linearity is:
$$
\forall\alpha,\beta\in\mathbb{R},\ L(\alpha\vec{u}+\beta\vec{v})=\alpha L(\vec{u})+\beta L(\vec{v})
$$
By the definition of differential given above, we can write:
$$
L(\vec{x}+t\vec{v})=L(\vec{x})+t\cdot dL(\vec{x})(\vec{v})+o(t)
$$
By the definition of linearity we have that the left hand side can be written as $L(\vec{x})+t\cdot L(\vec{v})$.
So we have:
$$
L(\vec{x})+t\cdot L(\vec{v})=L(\vec{x})+t\cdot dL(\vec{x})(\vec{v})+o(t)
$$
comparing the sides of this relation we have that $o(t)=0$ and:
$$
dL(\vec{x})(\vec{v})=L(\vec{v})\quad\Rightarrow\quad \boxed{dL(\vec{x})=L}
$$
A: Formally, the situation is like this: there is a function $\phi:{\cal M}\to{\cal N}$ (between manifolds), and we want something to describe infinitesimal change. We associate to each $x\in{\cal M}$ a linear map, the differential, which is a transformation of tangent spaces $df_x:T_x{\cal M}\to T_{f(x)}{\cal N}$.
$\hskip 0.6in$ 
Image source: Wikipedia.
In Euclidean space, points themselves may be viewed as vectors and hence added or subtracted, and so the manifold and all of its tangent spaces may be identified together. That is, every tangent vector exists as a point in the original space (codomain). If $f:{\bf R}^n\to{\bf R}^m$ is differentiable, then the differential is the "directional derivative" as a linear function of the "direction." Explicitly, the matrix of this linear map $df_x$ is given by the Jacobian. The write-up says that the Jacobian matrix of the linear map $L:x\mapsto Ax$ is just $A$, hence $dL_x=L$ is an equality of maps for all $x$.
In the one-dimensional case, $f:{\bf R}\to{\bf R}$, the matrix of a linear map is $1\times1$, so essentially just a scalar value. The scalar is in fact $f'(x)$, so the differential is $df_x:v\mapsto f'(x)v$. In particular, $f(x)=x$ implies $f'(x)=1$ so $df_x:v\mapsto1v$ is the identity map, i.e. the same as $f$.
A: On top of the other answer mentioned here, I found it useful to look at the definition of $df_x$. According to Milnor:
$$ df_x(h) = \frac{f(x+th) - f(x)}{t}$$
As $f$ is linear:
$$\begin{align}
df_x(h) &= \lim_{t \to 0} \frac{f(x+th) - f(x)}{t} \\
        &= \lim_{t \to 0} \frac{f(x) + f(th) - f(x)}{t} \\
        &= \lim_{t \to 0} \frac{f(x) + tf(h) - f(x)}{t} \\
        &= \lim_{t \to 0} \frac{tf(h)}{t} \\
        &= \lim_{t \to 0} f(h) \\
        &= f(h)
\end{align}$$
