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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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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An injective transformation is said to be an injection Note: Injective transformations need not be linear An injection guarantees that distinct codomain vectors
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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
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Then T is invertle if and only if for all y ∈ Rn, the system of linear equations A x = y has exactly one solution 2 Let T : R3 → R3 be the linear transformation given
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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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5 fév 2007 · of linear algebra is the characterization of the solutions to the set of m linear The linear map T : V → W is called injective if for all u, v ∈ V , the
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Hence its kernel consists of the 0 vector only, making ψ an injection (b) We apply the Rank-Nullity Theorem again If ψ is injective, the rank of its kernel is 0, and
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