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On Wikipedia, it says

When $f$ is a function from an open subset of $mathbbR^n$ to $mathbbR^m$, then the directional derivative of $f$ in a chosen direction is the best linear approximation to f at that point and in that direction.

I just want to check that linear functions from $mathbbR^n$ to $mathbbR^m$, are defined as functions of the form $f(x) = ax+b$ where $a$ is a scalar and $b$ is a vector?

Also, it seems like functions of the form above just enlarge/shrink and shift. Is this correct? I thought that if anything was going to be a counterexample, it was going to be an off center circle; under the transformation x $mapsto$ 2x, I thought an off-center circle might map to an ellipse; but this doesn't seem to be the case. For example, if $(x, y)$ satisfies $(x-2)^2 + (y-2)^2 = 1$, then multiplying both sides by $2^2$ gives $(2x-4)^2 + (2y-4)^2 = 4$; so $(2x, 2y)$ satisfies $(X^2-4)^2 + (Y-4)^2 = 4$, which is still a circle with center at $(4, 4)$, as expected.

A linear function in this context is a map $f: mathbbR^n to mathbbR^m$ such that the following conditions hold:

1. 
   $f(x+y)=f(x)+f(y)$ for every $x,y in mathbbR^n$
   


2. 
   $f(lambda x)=lambda f(x)$ for every $x in mathbbR^n$ and $lambda in mathbbR$.

It can be shown that every such function has the form $f(x)=Ax$ where $A in mathbbR^m times n$ is an $m times n$ matrix. If $f$ has the form $f(x)=Ax + b$ for some $bin mathbbR^m$, then it is called an affine linear function.

This generalises the notion of a linear map $f: mathbbR to mathbbR$ of the form $f(x)=ax+b$, where $a,b$ are real numbers, which is probably what you had in mind. A linear affine map is a linear map, if and only if $b=0$. Note that your example is a special affine linear map from $mathbbR^n to mathbbR^n$ (the dimensions have to match).

An example of a linear function from $mathbbR^3$ to $mathbbR^3$ would be $$f(x,y,z) = beginpmatrix
1 & 2 & 7\
5& 3 & 7\
3& 8& 2
endpmatrix beginpmatrix
x\
y\
z
endpmatrix.$$

Your example in the case of $mathbbR^3$ is of the form

$$f(x,y,z) = beginpmatrix
a& 0 & 0\
0& a & 0\
0& 0& a
endpmatrix beginpmatrix
x\
y\
z
endpmatrix + beginpmatrix
b_x\
b_y\
b_z
endpmatrix,$$
for some $a in mathbbR$ and $(b_x, b_y, b_z) in mathbbR^3$.

In the case of a differentiable function at a point $x_0 in mathbbR^m$ $f: mathbbR^m to mathbbR^n$ we want to approximate the function by an affine linear map, that is locally around $x_0$ we have $f(x) approx A(x-x_0) + f(x_0)$, where $A in mathbbR^n times m$. The offset $f(x_0)$ ensures that the approximation takes the value $f(x_0)$ at the point $x_0$, and the matrix $A$ describes how the function changes linearly around $x_0$. The idea is that linear maps are really easy to handle using the tools of linear algebra.

This is in general not the form of a linear function. A function $f: mathbbR^n rightarrow mathbbR^m$ is linear if the following two equalities hold for all $alphainmathbbR$ and $x, yin mathbbR^n$:

$i)$ $f(x + y) = f(x) + f(y)$

$ii)$ $f(alpha x) = alpha f(x)$.

It turns out that all such functions are of the form $f(x) = Ax$ for some matrix $AinmathbbR^mtimes n$ (that is, a matrix with $m$ rows, $n$ columns).

One key difference with your proposed form is that linear functions always go through the origin, that is $f(0) = 0$, where $0$ is the zero vector (rather than the scalar). This is not the case if $bneq 0$ in your proposed form. For $f: mathbbR^2 rightarrow mathbbR$ you should think of a plane through the origin as the graph, rather than a line.

I just want to check that linear functions from Rn to Rm, are defined as functions of the form f(x)=ax+b where a is a scalar and b is a vector?

No. In fact, a linear function is one with the property that $f(ax) = af(x)$ for any $x$ is whatever vector space it's defined on and any $a$ in the scalar field of that vector space. In that case, that is precisely those of the form $f(x) = Ax$ for some matrix $A$.

Also, it seems like functions of the form above just enlarge/shrink and shift. Is this correct?

No, because of the above. For an example involving a circle, take $n = 2$, $m = 2$ and $A = left(array2&0\0&1right)$. This turns the unit circle into an ellipse. More generally, note that $n$ and $m$ do not have to be the same. For example, there's the linear map
beginalign*f&: mathbbR^3tomathbbR\&:left(arrayx\y\zright)mapsto x+y+z,endalign* which collapses everything down to a diagonal line (but not in the most "natural" way).

In single variable calculus the best linear approximation to a function $f$ at a point $p$ is
$$
g(x) = f(p) + f'(p)(x-p).
$$
You can see why that's close to $f(x)$ when $x$ is close to $p$ by looking at the definition of the derivative, and thinking about the tangent line.

In several variables $p$ and $x$ will be vectors. That formula will still be correct if you change "$f'(p)$" to "the directional derivative of $f$ at $p$ in the direction from $p$ to $x$".

As the other answers say, most of what you "want to check" isn't right.

In general, the derivative is the best local linear approximation to a function at a point. A differentiable function $f: mathbbR^n rightarrow mathbbR^m$ at $x=x_0$ is locally approximated by a vector space homomorphism $Df_x_0 in cal L(mathbbR^n, mathbbR^m)$, and it is in this sense that you must understand "linear".

In the direction $v in mathbbR^n$, the directional derivative is simply $Df_x_0(v)$ because the derivative contains all information about all local rates of change in all directions.

Basically what happens is that you attach a copy of $mathbbR^m+n$ to $x_0$, and you approximate the curvy graph of $f$ by the flat (linear) graph of $Df(x_0)$. This is called the tangent space to the graph of $f$ at $x=x_0$. If you balance a piece of cardboard on a beach ball, you have a good model for this. The origin is where the cardboard touches the ball, which is why you don't get an additive constant.

If you draw a line on your piece of cardboard through the point where it touches, you get a model for the directional derivative in the direction of your point. Rotate your cardboard tangent plane around that point, and you get different directional derivatives.