hi there. after some (mild and sporadic) thinking about the question of finding a vector space V algebraically isomorphic to its linear space of endomorphisms, i’m getting to the point of saying that there’s no such V (this is a follow up of this post).

The reason is the following: if B is a basis of V, choose a vector w in it. Then consider the set P of nonempty parts of B, which do not contain w. Then P has cardinality strictly greater than B. Now consider the map F: P –> End(V) which associates to each N in P the (unique) endomorphism extending the function DN:B->V given by

D(v)=dN(v) .v + dW(v).w

where dN is the characteristic function of N in B, and dW the characteristic function of {w} in B. The mapping F is clearly injective. Now we notice that  every finite subset of Im(F) is linearly independent over k. Then Im(F) is linearly independent over k, it can be extended  to a basis of End(V).

This means that there’s a basis of End(V) with cardinality strictly greater than B, where B is a given basis of V. We conclude that ANY basis of End(V) has cardinality strictly greater than ANY basis of V.

Then, we have that V is never isomorphic to End(V), for ANY vector space V (over ANY field k).