Image of Projection in Plane
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Theorem
Let $M$ and $N$ be distinct lines in the plane.
Let $\pr_{M, N}$ be the projection on $M$ along $N$:
- $\forall x \in \R^2: \map {\pr_{M, N} } x =$ the intersection of $M$ with the line through $x$ parallel to $N$.
Then $M$ is the image of $\pr_{M, N}$.
Proof
Let $x \in \R^2$ be arbitrary.
By definition, the image of $x$ is the intersection of $M$ with the line through $x$ parallel to $N$.
Therefore $\map {\pr_{M, N} } x \in M$.
Hence:
- $\Img {\pr_{M, N} } \subseteq M$.
Now consider $y \in M$.
By Playfair's axiom there exists exactly one straight line $L$ parallel to $N$ passing through $y$.
Hence $y$ is the image of every point on $L$.
As $y$ is arbitrary, it follows that each point on $M$ is the image of some straight line in $\R^2$ under $\pr_{M, N}$.
That is:
- $M \subseteq \Img {\pr_{M, N} }$
The result follows by definition of set equality.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.5$