Find an example of a vector space V over K and a linear map L:V ?V such that L i

Question

Find an example of a vector space V over K and a linear map L:V ?V such that L is injective but not surjective. Hint: By a theorem proven in class, V has to be infinite dimensional.

Explanation / Answer

Now, I will sketch a proof for the validity of this assertion, however, I think it relies on finite-dimensional reasoning and that's why I'm not quite sure if there's some subtle assumption I am overlooking there. I will in fact, assume that the vector spaces are finite-dimensional, but even with this assumption, I fail to see if there's any limitation to do the same for infinite-dimensional spaces (interpreting that dim(E) R^n by setting these elements to the first elements of the basis of R^n, which come a in number of dim(Ker(A)) (assuming finite dimensions) (because they are LI, I can do this without any contradictions). the remaining elements of the basis of E are mapped to the next elements of the basis of R^n (they are a total of dim(E) - dim(Ker(A))). In this way i define an injective (not necessarily surjective, because there are dim(F) - dim(E) elements of the basis of R^n that were not reached) linear transformation T. Now, define the linear transformation S:R^n -> F by setting the first elements of R^n to zero so that A(x) = TS(x) for x being the elements of thebasis of E that originally went to zero. for the rest of elements of the basis of E that correspond to the next elements of the basis of R^n we map them to what they were originally. In this way A(x) = TS(x) for all elements of the basis of E, that is, A = TS. But I still have room to cover all of F. I define the next dim(F)-dim(E) elements of the basis of R^n to be sent to the last corresponding elements of the basis of F. now we only need to cover the dim(Ker(A)) elements of F that were not reached because their corresponding ones were sent to zero. this can be covered by the last elements of the basis of R^n we have.

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