I've recently been studying linear algebra and I'm beginning to wonder if one way to tell if a function is a linear transformation is whether it's injective, or, in other words, to test the truth value of the claim "if a function is a linear transformation, then it is injective."
To be clear, I will state a rough definition of linear transformations:
A linear transformation is a function $f:X \to Y$ where $X$ and $Y$ are vector spaces, satisfying the following:
For all vectors $\hat{x},\hat{y} \in X, $ $f(\hat{x}+\hat{y})=f(\hat{x})+f(\hat{y})$
For all scalars $c$, $f(c \cdot \hat{x})=c \cdot f(\hat{x})$
One thing I'm noticing is that it would appear that all of the functions I can imagine that fail to satisfy these conditions are not injective.
Non-Injective:
$ \bullet \, \sin(x+y)\neq \sin(x)+ \sin(y) \hspace{1cm} \sin(cx) \neq c \cdot \sin(x)$
$\bullet \, |x+y| \neq |x|+|y| \hspace{1cm} |cx| \neq c \cdot|x|$
Now consider the functions that do satisfy the definition:
Injective:
$\bullet$ Any line passing through the origin: $\, f(x)=a \cdot x; \, a(x+y)=a \cdot x + a \cdot y$ and $a \cdot (cx) = c (a \cdot x)$
I suppose that injectivity often helps for linear transformations, but I would imagine that it is not a necessity. For example, the function $f(x)=x^3$ is injective but surely $(x+y)^3 \neq x^3 + y^3$
It would appear I've answered my own question in that "linear transformations need not be injective," but is there a proof of the notion that "all non-injective functions fail to be linear transformations?"
Best Answer
The most basic example of a non-injective linear map is $z : \mathbb{R} \to \mathbb{R}$ defined by $z(x) = 0$.
The theory of when a linear map is injective is studied extensively in linear algebra. There are many theorems about when such a map is injective when the vector spaces are finite-dimensional.