statement theorem
An injective endomorphism of a finite-dimensional space is surjective
Statement
Let be a field, let be a finite-dimensional vector space over , and let be a linear map. Then these conditions are equivalent:
- is injective.
- is surjective.
- is bijective.
Remarks
The proof uses the rank–nullity theorem: if and only if . It fails in infinite dimensions: on , multiplication by is injective and not surjective. In characteristic , the derivative on is surjective and not injective. For square matrices, this is the statement that has only the trivial solution if and only if has a solution for each .
Explanations
L1 human
For a linear map from a finite-dimensional space to itself, one to one and onto are the same thing. If no information is lost, then nothing in the target can be missed, because the dimensions match. This is like a map from a finite set to itself.
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What this needs
To read this node, read these 16 nodes in this order. The foundations come first, and the list ends with this node.
- no evidence level
- Vector space definition
- Linear combination definition
- Linear map definition
- Linear subspace definition
- Linear independence definition
- Image of a linear map definition
- Kernel of a linear map definition
- Linear span definition
- Spanning set definition
- Basis definition
- Finite-dimensional vector space definition
- Dimension definition
- Rank of a linear map definition
- Nullity definition
- Rank–nullity theorem statement
- An injective endomorphism of a finite-dimensional space is surjective statement
Sources
- Web https://en.wikipedia.org/wiki/Rank–nullity_theorem, Reformulations and generalizations
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