If $L(X)$ is linear then $L'(X)=L(X)$ Let $L:\mathbb R^n\to \mathbb R^m$ be a linear map, Prove that $L'(X)=L(X)$ for all $X\in \mathbb R^n$.
This is what I've done so far,
let $X\in \mathbb R^n$ then we know that $$DL(X)=L'(X)=J_L(X)X$$
Where $J_F(X)$ is just the Jacobian matrix of $L$, now to finish the proof we have to show that the jacobian matrix of a linear map is just the matrix representation of it, but I have no idea how to do it
note that if we can show this then we just have $$L(X)=AX=J_L(X)X=L'(X)$$
and this finishes the proof.
 A: Let $X, V \in \mathbb{R}^{n}$ and $t \in \mathbb{R}$, then the derivative of $L$ at $X$ is
$$L^{\prime}(V) := \lim_{t \rightarrow 0} \frac{L(X + tV) - L(X)}{t} = \lim_{t \rightarrow 0} \frac{L(X) + tL(V) - L(X)}{t}  = L(V),$$
since $L$ is a linear operator.
A: As DonAntonio says, this result is trivial. You don't need to consider the Jacobian at all, just the definition of differentiability. Say $x_0 \in \mathbb{R}^n$. We say the derivative at this point is the unique $L_{x_0}'$ such that
$$L(x) = L(x_0) + L_{x_0}'(x-x_0) + \alpha(x-x_0)$$
where we have the usual limit condition involving $\alpha$.
To prove your result, simply define $\alpha$ to be zero. Define $L_{x_0}' = L$. Then since $L$ is linear, $L(x) = L(x_0) + L(x-x_0) = L(x_0)+L(x)-L(x_0) = L(x)$. This proves $L'$ is $L$ for every $x_0 \in \mathbb{R}^n$
A: The real result is that the differential $\mathrm dL$  of a linear map $L$ is $L$ itself.
This is true by definition, since the differential of a map $f:\mathbf R^n\longrightarrow\mathbf R^m$ at a point $x_0\in\mathbf R^n$ , if it exists, is  the linear map $\ell:\mathbf R^n\longrightarrow\mathbf R^m$ such that
$$f(x)=f(x_0)+\ell(x-x_0)+o\bigl(\|x-x_0\|\bigr).$$
