Dual of the annihilator isomorphic to quotient space

Asked Jun 19 '23 at 23:01

Active Jun 19 '23 at 23:01

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There are many resources online detailing the isomorphism between the annihilator $U^0:=\{f\in V':f(U)=0\}$ and the dual of the quotient space, $(V/U)':=Hom(V/U,\mathbb{R})$.

However, my lecture notes state without proof that we also have $V/U\cong (U^0)'$ - an isomorphism between the quotient space itself and the dual of the annihilator.

Could anyone please provide this isomorphism?

asked Jun 19 '23 at 23:01

hegash

1

For finite dimensional vector spaces, it follows by taking duals and remembering that there is a canonical isomorphism between a finite dimensional vector space and its double dual. For infinite dimensional vector spaces it is false, as witnessed by taking $U$ to be the trivial subspace, and remembering that thar the algebraic dual of an infinite dimensional vector space always has strictly larger dimension than the space. – Arturo Magidin Jun 19 '23 at 23:07
$(v+U)(f)=f(v)$. – Kavi Rama Murthy Jun 19 '23 at 23:09
Just use bidual iso ? – julio_es_sui_glace Jun 19 '23 at 23:13

Dual of the annihilator isomorphic to quotient space

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