Krdytkyu

Suppose $f:Bbb R^nto Bbb R^m$ is a linear map. Show that $Df(a)=f(a)$.

Tried using limit definition:

$$limlimits_{h to 0}frac{Vert f(a+h)-f(a)-f(a)hVert}{Vert hVert}$$$$=limlimits_{h to 0}frac{Vert f(a)+f(h)-f(a)-f(a)hVert}{Vert hVert}$$ $$=limlimits_{h to 0}frac{Vert f(h)-f(a)hVert}{Vert hVert}$$

Want to show this is $0$ but can't see where to go from here. Unless I misunderstand what $Df(a)=f(a)$ means.

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}

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
$$
lim_{h to 0} frac{|f(a+h) - f(a) - T(h)|}{|h|} = 0qquad (h in mathbb{R}^n).
$$
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
$$
lim_{h to 0} frac{|f(a+h) - f(a) - Df(a)(h)|}{|h|} = 0qquad (h in mathbb{R}^n).
$$

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.