Linear mappings/Determination theorem/Projection image/Section

Behind the following statement (the determination theorem) is the important principle, that in linear algebra (of finite dimensional vector spaces), the objects are determined by finitely many data.

Theorem

Let ${}K$ be a field, and let ${}V$ and ${}W$ be ${}K$-vector spaces. Let ${}v_{i}$, ${}i\in I$, denote a basis of ${}V$, and let ${}w_{i}$, ${}i\in I$, denote elements in ${}W$. Then there exists a unique linear mapping

$f\colon V\longrightarrow W$

with

$f(v_{i})=w_{i}{\text{ for all }}i\in I.$

Proof

Since we want ${}f(v_{i})=w_{i}$ and since a linear mapping respects all linear combinations, that is

${}f{\left(\sum _{i\in I}s_{i}v_{i}\right)}=\sum _{i\in I}s_{i}f{\left(v_{i}\right)}\,$

holds, and since every vector ${}v\in V$ is such a linear combination, there can at most exist one such a linear mapping.
We define now a mapping

$f\colon V\longrightarrow W$

in the following way: we write every vector ${}v\in V$ with the given basis as

${}v=\sum _{i\in I}s_{i}v_{i}\,$

(where ${}s_{i}=0$ for almost all ${}i\in I$) and define

${}f(v):=\sum _{i\in I}s_{i}w_{i}\,.$

Since the representation of ${}v$ as such a linear combination is unique, this mapping is well-defined.
Linearity. For two vectors ${}u=\sum _{i\in I}s_{i}v_{i}$ and ${}v=\sum _{i\in I}t_{i}v_{i}$, we have

${}{\begin{aligned}f{\left(u+v\right)}&=f{\left({\left(\sum _{i\in I}s_{i}v_{i}\right)}+{\left(\sum _{i\in I}t_{i}v_{i}\right)}\right)}\\&=f{\left(\sum _{i\in I}{\left(s_{i}+t_{i}\right)}v_{i}\right)}\\&=\sum _{i\in I}(s_{i}+t_{i})f{\left(v_{i}\right)}\\&=\sum _{i\in I}s_{i}f{\left(v_{i}\right)}+\sum _{i\in I}t_{i}f(v_{i})\\&=f{\left(\sum _{i\in I}s_{i}v_{i}\right)}+f{\left(\sum _{i\in I}t_{i}v_{i}\right)}\\&=f(u)+f(v).\end{aligned}}$

The compatibility with scalar multiplication is shown in a similar way, see exercise.

$\Box$

In particular, a linear mapping ${}\varphi \colon K^{n}\rightarrow K^{m}$ is uniquely determined by ${}\varphi (e_{1}),\ldots ,\varphi (e_{n})$.

Example

In many situations, a certain object (like a cube) in space ${}\mathbb {R} ^{3}$ shall be drawn in the plane ${}\mathbb {R} ^{2}$. One possibility is to work a projection. This is a linear mapping

$\mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{2}$

which is given (with respect to the standard bases ${}e_{1},e_{2},e_{3}$ and ${}f_{1},f_{2}$) by

$e_{1}\longmapsto f_{1},\,e_{2}\longmapsto af_{1}+bf_{2},\,e_{3}\longmapsto f_{2},$

where the coefficients ${}a,b$ are usually chosen in the range ${}[{\frac {1}{3}},{\frac {1}{2}}]$. Linearity has the effect that parallel lines are mapped to parallel lines (unless they are mapped to a point). The point ${}(x,y,z)$ is mapped to ${}(x+ay,by+z)$. The image of the object under such a linear mapping is called a projection image.