I was wondering what is the geometric meaning or intuition behind a transformation and function(separately)being linear.
An example(or graph) illustrating the characteristics of a linear function/map would be much appreciated.
Thanks in advance.
EDIT: Also, I have read that if we "zoom in" the graphs of some functions, we see that they "become linear around the point of magnification"(you can find it in Callahan's Advanced Calculus". What does that mean?
A linear transformation maps straight lines continuously to straight lines, equal distances on a single line to equal distances, and additionally the origin to the origin. Without the third condition, it's called an affine transformation.
Proof that those conditions are sufficient to recover the usual definition:
Be $V$ and $W$ vector spaces and $f:V\to W$ a function that fulfils those definitions. Be $v_1$ and $v_2$ vectors in $V$, and $w_i = f(v_i)$.
Since the origin is preserved, $f(0_V) = 0_W$.
Now consider the straight line $\{\lambda v_1: \lambda\in\mathbb R\}$. Since straight lines are mapped to straight lines, and a line is fixed by two points, we know that the image of the line is $\{\mu w_1: \mu\in\mathbb R\}$, where $\mu(\lambda)$ is by assumption a continuous function. We also know $\mu(1)=1$ because $w_1 = f(v_1)$.
Now since on a single line, equal lengths are mapped to equal lengths, we know that $f(\lambda v_1+v_1)=\mu(\lambda)w_1+w_1$. From that we can derive that for integer $n$, $f(n v_1)=n w_1$, and with an analogous argument that for any rational number $q$, $f(q v_1) = q w_1$. Continuity then gives us $\mu(\lambda)=\lambda$, that is, $f(\lambda v_1) = \lambda w_1$. Of course since $v_1$ is arbitrary, this is true for every vector.
Now consider the straight line going through $2\alpha v_1$ and $2\beta v_2$. This line is given by $\lambda 2\alpha v_1 + (1-\lambda) 2 \beta v_2$. This line is mapped to the straight line through $2\alpha w_1$ and $2\beta w_2$, given by $\mu(\lambda) 2\alpha w_1+(1-\mu(\lambda)) 2\beta w_2$. Clearly $\mu(0)=0$ and $\mu(1)=1$.
Now consider specifically $\lambda=\frac12$, that is, the point $\alpha v_1 + \beta v_2$. That point has equal distance from $2\alpha v_1$ and $2\beta v_2$. Therefore since equal distances on a line are mapped to equal distances, the image point also has to have equal distance from $2\alpha w_1$ and $2\alpha w_2$, that is, it must be the point $\alpha w_1 + \beta w_2$.
Therefore we have $$f(\alpha v_1 + \beta v_2) = \alpha w_1 + \beta w_2 = \alpha f(v_1) + \beta f(w_2)$$ But this is the conventional definition of a linear function.
In calculus a function $f:\>{\mathbb R}\to{\mathbb R}$ is linear if it is given by an expression of the form $f(x)=ax+b$ with constants $a$ and $b$. The essential feature of such a function is that when it is evaluated at two points $x_0$, $x_1$ the increment $\Delta f=f(x_1)-f(x_0)$ of the output is the constant multiple $a\>\Delta x$ of the increment $\Delta x=x_1-x_0$ of the input. In many cases one has in fact $b=0$. In linear algebra, as well as in functional analysis, this is indeed a compulsory condition for linearity.
If a transformation $f:\>X\to Y$ between spaces $X$ and $Y$ is at stake then the linearity of $f$ can only been felt when looking at arbitrary elements $x$, $y\in X$ and their images $f(x)$, $f(y)$. If $f$ is linear then $y=\lambda x$ should imply $f(y)=\lambda f(x)$, and $z=x+y$ should imply $f(z)=f(x)+f(y)$. In other words: Linear relations between points (vectors) should be preserved. It is common to write the stated conditions in the form $$f(\lambda x)=\lambda\>f(x),\qquad f(x+y)=f(x)+f(y)\ ,$$ in order to keep the number of introduced variables to a minimum.
It transforms parallelograms to (possibly degenerate) parallelograms, and preserves scaling.
If you have a linear transformation on a space $X$ then the image is a subspace of the space $X$. Geometrically that means that the image of the transformation is a flat that contains the origin in the space.
Examples: The easiest example is the 0-transformation.
Let $X = \mathbb R ^n$ and $f: X \to X, v \mapsto 0$ then $Im(f) = \{0\}$ and thus just a point.
The next easiest example is the identity:
Let $X = \mathbb R ^n$ and $f: X \to X, v \mapsto v$ then $Im(f) = X$ and thus the whole space.
A more complex example is:
Let $X = \mathbb R ^n$ and $f: X \to X, v \mapsto (v_1, 0 , \dots, 0)$ then $Im(f) = \{(v_1, 0 , \dots, 0) \mid v_1 \in \mathbb R\}$ and thus a line.
Hope that helps you :)
