Relationship between properties of linear transformations algebraically and visually I learned from 3Blue1Brown's Linear Algebra videos that a 2-D transformation is linear if it follows these rules:


*

*lines remain lines without getting curved

*the origin remains fixed in place

*grid lines remain parallel and evenly spaced


I'm now going through linear algebra from a textbook, which lays out this definition of a linear transformation:


*

*T(u+v) = T(u) + T(v)

*T(cu) = cT(u)  


I'm wondering, is there a connection between these two ways of thinking of linear transformations? Do the visual ways of seeing 2-D linear transformations correspond to the formal definition when in 2-D?
 A: If we make reasonable algebraic definitions for the three points above, then the answer is yes: there is a correspondence between the geometric perspective and the algebraic perspective.
Let $f$ be a map that fixes the origin ($f(\vec{0}) = \vec{0}$) and sends lines of evenly spaced points to lines of evenly spaced points ($f(\vec{u} + c\vec{v}) = \vec{u}' + c\vec{v}'$). Then we have
$$
f(c\vec{v}) = c\vec{v}'
$$
If we take $c=1$ then we get $f(\vec{v}) = \vec{v}'$. So $f(c\vec{v}) = cf(\vec{v})$.
Now consider
$$
f(c_1\vec{u} + c_2\vec{v}) = \vec{u}' + c_2\vec{v}'
$$
If we set $c_2 = 0$, we get
$$
f(c_1\vec{u}) = c_1 f(\vec{u}) = \vec{u}'
$$
So we have
$$
f(c_1 \vec{u} + c_2 \vec{v}) = c_1f(\vec{u}) +c_2\vec{v}'
$$
But if $c_1 = 0$, we then have
$$
f(c_2 \vec{v}) = c_2 f(\vec{v}) = c_2 \vec{v}'
$$
So $f(c_1 \vec{u} + c_2 \vec{v}) = c_1 f(\vec{u}) + c_2f(\vec{v})$. So $f$ is linear. Therefore, if $f$ fixes the origin and sends lines of evenly spaced points to evenly spaced points, then $f$ is linear.
Can you show the converse ?
