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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?

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    $\begingroup$ Well since linear transformations are just a special type of affine transformation, they should preserve parallel lines. And in fact even the ratios of parallel lengths should be preserved. Also, particular for linear maps, they should fix one particular point called the origin. $\endgroup$ – user137731 Aug 2 '16 at 18:00
  • $\begingroup$ To answer the edit: that's a different definition of the word "linear". In that sense Callahan means like a line or more generally like a Euclidean subspace. $\endgroup$ – user137731 Aug 2 '16 at 18:10
  • $\begingroup$ @Bye_World Well, he zooms into the graph and then we just see parallel lines. So, he says that it's linear. And while I understand what he means and what you mean, I think I don't quite get it in its most general meaning. Could you please elaborate a bit more? $\endgroup$ – TheQuantumMan Aug 2 '16 at 18:14
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    $\begingroup$ "Zoom in" isn't a very precise notion. But the "most general" view (of a scalar function) I guess is that a function $f$ is differentiable at the point $a$ iff in some interval containing $a$ one can write $$f(a+h) = f(a) + f'(a)h + o(h)$$ which essentially says that near enough to the point $(a,f(a))$ on the graph, the function behaves very much like the straight line $h\mapsto f(a)+f'(a)h$. You can see more on this in my answer to this question. $\endgroup$ – user137731 Aug 2 '16 at 18:19
  • $\begingroup$ @Bye_World thanks a lot! $\endgroup$ – TheQuantumMan Aug 2 '16 at 19:27
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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.

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  • $\begingroup$ Thanks for the answer, although I have a question regarding the " equal distances on a single line to equal distances" part. You stated this from the very beginning and used this to show some things. But, how can one realize by himself that linear transformations do that? How is that obvious? $\endgroup$ – TheQuantumMan Aug 3 '16 at 9:53
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    $\begingroup$ @TheQuantumMan: Mark two equal line segments on the same line, and see that they are mapped to two equal line segments by the map? However thinking about it, maybe it can be removed if the demand that straight lines are mapped to straight lines is refined to the demand that they are mapped onto straight lines (that is, the map may not miss a single point of the line). Actually I didn't really like that specific condition too much anyway; if you look at the edit history, it was an afterthought after noticing that my original condition was too weak. $\endgroup$ – celtschk Aug 3 '16 at 10:07
  • $\begingroup$ @TheQuantumMan: Unfortunately it's not possible to remove that condition, because the alternative would not work in one dimension. $\endgroup$ – celtschk Aug 3 '16 at 15:16
  • $\begingroup$ But is that a condition or a consequence of how we define linearity? I could just draw it down by just basing by calculations on f(x+y)=f(x)+f(y) and f(ax)=af(x). Am I getting something wrong here? $\endgroup$ – TheQuantumMan Aug 3 '16 at 17:00
  • $\begingroup$ No, the common definition is exactly that. But I assumed that is not what you're after, as you can read that in any linear algebra book. Of course any two definitions which have the same consequences are equivalent. $\endgroup$ – celtschk Aug 3 '16 at 17:08
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It transforms parallelograms to (possibly degenerate) parallelograms, and preserves scaling.

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  • $\begingroup$ Sorry, I just edited the question to also include functions. Also, is this the most general definition of it? $\endgroup$ – TheQuantumMan Aug 2 '16 at 18:01
  • $\begingroup$ @TheQuantumMan There's no difference between a "transformation" and a "function". They are synonyms. $\endgroup$ – user137731 Aug 2 '16 at 18:02
  • $\begingroup$ Because a function is just something that transforms anything that you put into it, right? $\endgroup$ – TheQuantumMan Aug 2 '16 at 18:03
  • $\begingroup$ @TheQuantumMan It uniquely tranforms, yes. That bit is important to rule out the so-called "multivalued functions" which are not functions at all. $\endgroup$ – user137731 Aug 2 '16 at 18:04
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    $\begingroup$ Scaling is $f(tx) = t f(x)$ for scalars $t$. Parallelograms (given $f(0)=0$ from the scaling) is $f(x+y) = f(x) + f(y)$, i.e. the first parallelogram has vertices $[0,x,x+y,y]$ and is transformed to the parallelogram with vertices $[0, f(x), f(x+y), f(y)]$. $\endgroup$ – Robert Israel Aug 2 '16 at 18:21
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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 :)

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  • $\begingroup$ Yeah I thought about that. But I didn't came up with a better word for what I mean. $\endgroup$ – Yaddle Aug 2 '16 at 18:04
  • $\begingroup$ That's nice. I changed it :) $\endgroup$ – Yaddle Aug 2 '16 at 18:06
  • $\begingroup$ I added some examples. $\endgroup$ – Yaddle Aug 2 '16 at 18:12
  • $\begingroup$ I understand your examples and I can see how you mean "flat containing the origin". But, what is the thing that characterizes these examples? Maybe you should read the edit I made to the question to understand. Sorry for being a pain in the ass! $\endgroup$ – TheQuantumMan Aug 2 '16 at 18:17
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    $\begingroup$ @TheQuantumMan You can approximate differentiable functions through linear functions in small neighbourhoods, since if you zoom in sufficiently you can sometimes get a neighbourhood where the function has a small curvature. Hence a linear function (which has no curvature) is pretty "close" to the other function. $\endgroup$ – Yaddle Aug 2 '16 at 18:22
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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.

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