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If a linear transformation doesn't preserve the origin of vector space, parallelism, collinearity. Does this mean the transformation automatically doesn't hold the properties of linear transformation?

  1. T(x + y) = T(x) + T(y)

  2. T(cx) = c T(x)

Because it seems that the idea of linear transformation preserve origin, collinearity, parallelism comes as a consequence of these properties which seems to be the case based on the link below.

https://www.quora.com/Why-do-a-transformation-from-V-to-W-that-maps-all-lines-to-lines-but-moves-the-origin-is-not-considered-a-linear-one

Or is there transformation that doesn't preserve the origin but is considered linear transformation.

And here is my second question :

I found a question Is a map that preserves lines and fixes the origin necessarily linear? but I don't understand the accepted answer as I don't come from math background. Does this mean that a transformation that preserves origin, lines, etc isn't necessarily a linear transformation? if so could anyone provide an example of such function in R2 or R3 ?

MellowPotato
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  • As you can see from the definition every linear transformation has to preserve the origin and lines, for a counterexample for the other direction @Joonas Ilmavirta 's answer to the quoted question gives an example in $\mathbb{R}^2\rightarrow\mathbb{R}^2$ – Peter Melech Aug 02 '20 at 11:13
  • @PeterMelech thank you for answering, I'm currently reading a book called "Mathematics for 3D Game Programming and Computer Graphics, Third Edition", in there the book explains the concept of linear transformation but only emphasize the additivity, multiplication properties and doesn't mention additional properties like preserving origin. Hence why I'm confused and asked this question because it comes off more as an intuition from the properties of additivity and multiplication rather than a definition. Thank you for clarifying. – MellowPotato Aug 02 '20 at 12:13

1 Answers1

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A linear transformation must always send $0$ to $0$: Observe that $T(0) = T(0 + 0) = T(0) + T(0)$, so adding $-T(0)$ to both sides gives $0 = T(0)$.

In other words, if it doesn't send $0$ to $0$, it isn't a linear transformation.

twosigma
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