TITLE: Derivative of Linear Map
QUESTION [10 upvotes]: 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!

REPLY [8 votes]: 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:
$$\begin{align}
df_x(h) &= \lim_{t \to 0} \frac{f(x+th) - f(x)}{t} \\
        &= \lim_{t \to 0} \frac{f(x) + f(th) - f(x)}{t} \\
        &= \lim_{t \to 0} \frac{f(x) + tf(h) - f(x)}{t} \\
        &= \lim_{t \to 0} \frac{tf(h)}{t} \\
        &= \lim_{t \to 0} f(h) \\
        &= f(h)
\end{align}$$