Suppose $f:\Bbb R^n\to \Bbb R^m$ is a linear map. Show that $Df(a)=f(a)$.
Tried using limit definition:
$$\lim\limits_{h \to 0}\frac{\Vert f(a+h)-f(a)-f(a)h\Vert}{\Vert h\Vert}$$$$=\lim\limits_{h \to 0}\frac{\Vert f(a)+f(h)-f(a)-f(a)h\Vert}{\Vert h\Vert}$$ $$=\lim\limits_{h \to 0}\frac{\Vert f(h)-f(a)h\Vert}{\Vert h\Vert}$$
Want to show this is $0$ but can't see where to go from here. Unless I misunderstand what $Df(a)=f(a)$ means.
On
As @WillM. says in the comments, what you are trying to show is not true and in fact doesn't even make sense. $f(a)$ is an element of $\mathbb{R}^n$ whereas $Df(a)$ is a linear map from $\mathbb{R}^n$ to $\mathbb{R}^n$.
What you have is the following:
If $f: \mathbb{R}^n \to \mathbb{R}^m$, then it is said to be differentiable at the point $a \in \mathbb{R}^n$ if there is a linear map $T : \mathbb{R}^n \to \mathbb{R}^m$ such that $$ \lim_{h \to 0} \frac{|f(a+h) - f(a) - T(h)|}{|h|} = 0\qquad (h \in \mathbb{R}^n). $$ In this case, $T$ is called the derivative of $f$ at the point $a$. It is usually denoted by $Df(a)$, so $Df(a) : \mathbb{R}^n \to \mathbb{R}^m$ is a linear map satisfying $$ \lim_{h \to 0} \frac{|f(a+h) - f(a) - Df(a)(h)|}{|h|} = 0\qquad (h \in \mathbb{R}^n). $$
What you actually want to show is the following:
Suppose $f : \mathbb{R}^n \to \mathbb{R}^n$ is linear and $a \in \mathbb{R}^n$. Then $f$ is differentiable at $a$ and $Df(a) = f$, so for all $x \in \mathbb{R}^n$, we have $Df(a)(x) = f(x)$.
You have the right idea, as you have used the linearity of $f$ to write $f(a+h)$ as $f(a) + f(h)$. Apply the same idea to the corrected problem and you should be okay. Good luck.
Hint: Let $f(x) = Ax$. Let \begin{align} \frac{\|f(a+h)-f(a)-Ah\|}{\|h\|} = \frac{\|A(a+h)-Aa-Ah\|}{\|h\|} = 0. \end{align}