The discussion revolves around the Fredholm Alternative Theorem in the context of linear transformations, specifically examining the relationship between the surjectivity of a linear operator \(T\) and its nullity in an n-dimensional vector space. Participants explore the implications of the theorem and how to prove the statements regarding solutions to the equation \(T(x) = b\).
Participants generally agree on the implications of the Rank-Nullity Theorem regarding the relationship between surjectivity and nullity. However, the discussion includes multiple perspectives on how to articulate and prove these implications, indicating some disagreement on the clarity and correctness of earlier posts.
Some participants express uncertainty about the proofs and the logical connections between the statements, particularly regarding the application of the Rank-Nullity Theorem. There are also indications that earlier posts may not have addressed the questions posed adequately.
Swati said:Let T:V->V be a linear operator on an n-dimensional vector space. Prove that exactly one of the following statements holds:
(i) the equation T(x)=b has a solution for all vectors b in V.
(ii) Nullity of T>0
Use the rank-nullity theorem again, to show that T is surjective.Swati said:how to proof if second statement is true then first statement is false.
Yes of course. I should have said: Use the rank-nullity theorem again, to show that T is NOT surjective.Deveno said:if the SECOND statement is true, T CANNOT be surjective:
by the rank-nullity theorem:
dim(V) = rank(T) + nullity(T).
if nullity(T) > 0, then rank(T) < dim(V), so that:
dim(im(T)) < dim(V).
thus there is some b in V not in im(T).
(i only posted this because Opalg's post answers the wrong question).