Bidual space and maps
Question
Solution
Short
Video
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Let V be a finite dimensional if infinite T is not surjective but always injective vector space over K. Then exists a canonical isomorphism T: Vlongrightarrow V^*^* given gy: forall vin V Tv: V^*longrightarrow K is the map Tvl:lv forall lin V^*. In other words: vin V longmapsto lin V^* longmapsto lvin K
Solution:
Proof. We claim T is a linear map. Indeed if vin P alphain K we have Talpha vlin V^*longmapsto lalpha vin K lin V^*longmapsto alpha lvin K alpha lin V^*longmapsto lvin K alpha Tv Similarly one shows that Tv'+v''Tv'+Tv''. We now claim that T is injective. To see this we'll show that textKerT i.e. Tv. Longrightarrow lv forall lin V^*. Choose a basis mathcalBv_...v_n for V. Write vc_v_+...+c_nv_n. Apply the above to lv_k^* and we obtain v_k^*c_v_+...+c_nv_n. But v_k^*c_v_+...+c_nv_nc_k. Longrightarrow c_k. And this holds forall leq kleq n Longrightarrow v. This proves T is injective. To finish the proof note that textdimV^*^*textdimV^*textdimV. So since T:Vlongrightarrow V^*^* is injective it must be an isomorphism. Doesn't hold for infinite dimensional vector spaces. More specifically T:Vlongrightarrow V^*^* is injective but not surjective. In fact when V is infinite dimensional V^*^* is not isomorphic to V
Let V be a finite dimensional if infinite T is not surjective but always injective vector space over K. Then exists a canonical isomorphism T: Vlongrightarrow V^*^* given gy: forall vin V Tv: V^*longrightarrow K is the map Tvl:lv forall lin V^*. In other words: vin V longmapsto lin V^* longmapsto lvin K
Solution:
Proof. We claim T is a linear map. Indeed if vin P alphain K we have Talpha vlin V^*longmapsto lalpha vin K lin V^*longmapsto alpha lvin K alpha lin V^*longmapsto lvin K alpha Tv Similarly one shows that Tv'+v''Tv'+Tv''. We now claim that T is injective. To see this we'll show that textKerT i.e. Tv. Longrightarrow lv forall lin V^*. Choose a basis mathcalBv_...v_n for V. Write vc_v_+...+c_nv_n. Apply the above to lv_k^* and we obtain v_k^*c_v_+...+c_nv_n. But v_k^*c_v_+...+c_nv_nc_k. Longrightarrow c_k. And this holds forall leq kleq n Longrightarrow v. This proves T is injective. To finish the proof note that textdimV^*^*textdimV^*textdimV. So since T:Vlongrightarrow V^*^* is injective it must be an isomorphism. Doesn't hold for infinite dimensional vector spaces. More specifically T:Vlongrightarrow V^*^* is injective but not surjective. In fact when V is infinite dimensional V^*^* is not isomorphic to V