A First Course in Linear Algebra · Problem ILT-M60

Linear Transformations — Robert Beezer

Mathematics — Linear Algebra · injective-linear-transformations

Problem

Suppose UU and VV are vector spaces. Define the function Z:U→VZ: U \to V by Z(u)=0VZ(\mathbf{u}) = \mathbf{0}_V for every u∈U\mathbf{u} \in U. Then by Exercise LT.M60, ZZ is a linear transformation. Formulate a condition on UU that is equivalent to ZZ being an injective linear transformation. In other words, fill in the blank to complete the following statement (and then give a proof): ZZ is injective if and only if UU is \_\_\_\_\_\_\_\_\_\_. (See Exercise SLT.M60, Exercise IVLT.M60.)
Intuitive explanation

The zero map sends every vector to the zero vector, so it's injective only when there's nothing "extra" to map—when the domain itself is trivial. Think about injectivity: a function is injective when different inputs give different outputs. Here, all inputs give the same output (0V\mathbf{0}_V), so the only way to avoid collisions is if there's only one input to begin with. This means UU must contain only the zero vector, making it the trivial vector space.

Intuitive explanation

The zero map sends every vector to the zero vector, so it's injective only when there's nothing "extra" to map—when the domain itself is trivial. Think about injectivity: a function is injective when different inputs give different outputs. Here, all inputs give the same output (0V\mathbf{0}_V), so the only way to avoid collisions is if there's only one input to begin with. This means UU must contain only the zero vector, making it the trivial vector space.

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