Restriction of a differentiable map $R^3\rightarrow R^3$ to a regular surface is also differentiable. This is again an excercise from Do Carmo's book. 
Prove: if $f:R^3 \rightarrow R^3$ is a linear map and $S \subset R^3$ is a regular surface invariant under $L,$ i.e, $L(S)\subset S$, then the restriction $L|S$ is a differentiable map and $$dL_p(w)=L(w), p\in S,w\in T_p(S).$$
My attempt: Since any linear map on $R^3$ can be represented by a linear transformation matrix , it must be differentiable.  By definition I have to show that for any local parametrization of S say $(U,x)$, map defined by $x^{-1}\circ L \circ x:U\rightarrow U  $ is differentiable locally. Now, both $x$ and $L$ are differentiable , however , $x^{-1}$ is not necessarily differentiable.
Moreover, example 3, page 74 of Do Carmo's says : Let $S_1$ and $S_2$ be regular surfaces. Assume that $S_1\subset V \subset R^3$ where $V$ is an open subset of $R^3$, and that $\phi:V \rightarrow R^3$ is a differentiable map such that $\phi(S_1)\subset S_2$. Then the restriction $\phi|S_1: S_1\rightarrow S_2$ is a differentiable map.
This fact is left without proof, but I think it might be useful for the question.
Can anyone give me some help ? Thanks in advance
 A: First of all, if $x:U\subset \mathbb R^2\rightarrow S$ is a parametrization, then $x^{-1}: x(U) \rightarrow \mathbb R^2$ is differentiable: indeed, following the very definition of a differentiable map from a surface, $x$ is a parametrization of the open set $x(U)$ and since $x^{-1}\circ x$ is the identity map, it is differentiable.
Now, let $p$ be a point on the surface $S$, $x:U\subset \mathbb R^2\rightarrow S$ be a parametrization s.t. $x(0)=p$ and $y:V\subset \mathbb R^2\rightarrow S$ be another parametrization s.t. $L(p)=y(0)$. 
To make it clear, let's say that $x(u,v)=(x_1(u,v),x_2(u,v),x_3(u,v))$ and $y^{-1}(x,y,z)=(\varphi_1(x,y,z),\varphi_2(x,y,z))$ then the map $L\circ x:U\rightarrow S$ is given by : $$L\circ x (u,v)=\begin{pmatrix} a&b&c\\d&e&f \\g&h&i\end{pmatrix}\begin{pmatrix} x_1(u,v) \\ x_2(u,v) \\ x_3(u,v) \end{pmatrix}$$
So $f(u,v)=y^{-1}\circ L \circ x(u,v)$ looks like $$f(u,v)=y^{-1}\circ L \circ x(u,v)=\\\ \begin{pmatrix}\varphi_1(ax_1(u,v)+bx_2(u,v)+cx_3(u,v),\cdots,gx_1(u,v)+hx_2(u,v)+ix_3(u,v)) \\ \varphi_2(gx_1(u,v)+hx_2(u,v)+ix_3(u,v),\cdots,gx_1(u,v)+hx_2(u,v)+ix_3(u,v))\end{pmatrix}$$
which is clearly differentiable.
Moreover, you can easily check using the chain rule that $$df_0=d(y^{-1})_{L(p)}\circ L \circ dx_0.$$
Roughly speaking, this map does : $$\mathbb R^2 \underset{dx}{\longrightarrow} T_pS \underset{L}{\longrightarrow} T_{L(p)}S\underset{dy^{-1}}{\longrightarrow} \mathbb R^2$$
which means that you send a vector of $\mathbb R^2$ onto $T_pS$ using the parametrization $x$ (it always gives you a good basis of the tangent space), then L acts and you read the information again using the second parametrization $y$ that takes the new vector onto $\mathbb R^2$.
So $L$ is nothing else but the derivative of $L:S\rightarrow S$ as a map between two surfaces.
In fact, this has to be expected because you might know that the derivative of a linear map between two vector spaces does not depend on the point and is equal to itself, so it has to be the same for surface or submanifold in general.
