Direct sums and isomorphisms
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Exercise:
Let V be a vector space over K Usubseteq V a subspace and U'subseteq V another subspace shich is a complement of U in V recall: this means U+U'V and Ucap U'. Then the map phi:Uoplus U'longrightarrow V phiuu':u+u' is a linear isomorphism.
Solution:
Proof of linearity. Linearity of phi follows directly from the dfinitions. Proof of injectivity of phi. Suppose phiuu' Longrightarrow u+u' Longrightarrow u-u'. But U' is a complement of U hence Ucap U' by definition. So U-U implies that UU' Longrightarrow uu'. This proves that textKerphi_Uoplus U' hence phi is injective. Proof of surjectivity of phi. Since U' is a complement of U in V we have by definition that U+U'V. Let vin V exists uin U u'in U' s.t. u+u'v Longrightarrow phiuu'v. This shows textImphiV i.e. phi is surjective. When U' is a complement of U in V then U+U'V and we also have a canonical isomorphism Uoplus U' longrightarrow V. But formally speaking Uoplus U^ is a different vector space than V.
Let V be a vector space over K Usubseteq V a subspace and U'subseteq V another subspace shich is a complement of U in V recall: this means U+U'V and Ucap U'. Then the map phi:Uoplus U'longrightarrow V phiuu':u+u' is a linear isomorphism.
Solution:
Proof of linearity. Linearity of phi follows directly from the dfinitions. Proof of injectivity of phi. Suppose phiuu' Longrightarrow u+u' Longrightarrow u-u'. But U' is a complement of U hence Ucap U' by definition. So U-U implies that UU' Longrightarrow uu'. This proves that textKerphi_Uoplus U' hence phi is injective. Proof of surjectivity of phi. Since U' is a complement of U in V we have by definition that U+U'V. Let vin V exists uin U u'in U' s.t. u+u'v Longrightarrow phiuu'v. This shows textImphiV i.e. phi is surjective. When U' is a complement of U in V then U+U'V and we also have a canonical isomorphism Uoplus U' longrightarrow V. But formally speaking Uoplus U^ is a different vector space than V.