How Do You Determine if a Linear Transformation is Injective?

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SUMMARY

To determine if a linear transformation is injective, one must analyze its kernel. For a linear map from vector space V to itself, the determinant can be computed to ascertain whether the kernel is zero. A function f is injective if f(x) = f(y) implies x = y. Specifically, a linear function is injective if and only if its kernel is trivial, containing only the zero vector, as stated by Matt Grime.

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I was just wondering how you know if linear transformations injective?
 
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You work out its kernel. If it's a map from V to V then you can work out its determinant which tells you if the kernel is zero or not (but not what it is).
 
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thanks
 
A function, f, in general, is injective if f(x)= f(y) implies x= y. If f is linear, then f(x)= f(y) gives f(x)- f(y)= f(x-y)= 0 while x= y is the same as x- y= 0. That is why a linear function is injective if and only if its kernel is trivial: {0}, as matt grime said.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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