A linear transformation is injective if and only if its kernel is the trivial subspace {0} Example If V and W are finite-dimensional vector spaces with the same dimension, then a linear map T : V → W is injective if and only if it is surjective In particular, ker(T) = {0} if and only if T is bijective
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[PDF] Chapter 16 Transformations: Injectivity and Surjectivity - Isoptera
Note: Injective transformations need not be linear An injection guarantees that distinct codomain vectors “came from” dis- tinct domain vectors For example, an
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assumption u1 − u2 = 0, hence u1 = u2 and hence s is injective • Let s : U −→ V be a linear transformation, which is bijective That is there is the inverse
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The linear mapping R3 → R3 which rotates every vector by θ around the x-axis Solution note: Invertible (hence surjective and injective) The inverse rotates by −θ
[PDF] A Characterization of Injective Linear Transformations - ResearchGate
We prove a characterization of the injective linear transformations on real vector spaces: Let X and Y be an m-dimensional and an n-dimensional real vector
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(7) A linear transformation T : Rm → Rn is bijective if the matrix of T has full row rank and full column rank Thus forces m = n, and forces the (now square) matrix to
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for injective linear transformation monomorphism embedding 4) for surjective linear transformation epimorphism 5) for bijective linear transformation
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(Fundamental Theorem of Linear Algebra) If V is finite dimensional, then both kerT and R(T) are If dimV = dimW, then T is injective if and only if T is surjective
[PDF] Linear Transformations
Let T : V → W be a linear transformation and let U be a subset of V The If f : A → B is a function that is both surjective and injective, then there exists a function
[PDF] Slide 1 Linear Transformations • Domain, range, and null spaces
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