Sets cardinality in linear algebra Let's suppose that $S$ and $T$ are sets, $S$ is finite, cardinality of $T$ is larger than that of $S$ (include the possibility where T could be an infinite set), Then, define $f: \mathbb{R}\langle T\rangle \rightarrow \mathbb{R}\langle S\rangle $ is a linear map. Show that $kerf \neq 0$.
My approach to the problem reduces to $T$ being finite. Then, when I tried the case of $T$ being infinite I tried to define a bijection to polynomials (where I proved something similar) , but I didn't succeed. Do you have any ideas in how to proceed. Thanks
 A: A very fundamental theorem in Linear Algebra (the "Rank-Nullity theorem):
  If f is a linear transformation from S to T, K is the kernel of f (the set of all s in S such that f(s)= 0) and I is the image of f (the set of all t in T such there exist s in S with f(s)= t) then dim(K)+ dim(I)= dim(T).
A: By definition, what you have is a linear transformation $f\colon V\to W$ between vector spaces over $\mathbb R$ (or any other field for that matter), and it is given that the dimension of $V$ is larger than the dimension of $W$. You may find it easier to prove the more general claim that any such linear transformation must have a non-trivial kernel. The rank-nullity theorem works fine with infinite cardinalities. 
A: If $T'\subseteq T$ is any subset, then $\mathbb{R}\langle T'\rangle$ is a subspace of $\mathbb{R}\langle T\rangle$, and you can restrict the map $f$ to a linear map $\mathbb{R}\langle T'\rangle\to \mathbb{R}\langle S\rangle$.  So if $T$ is infinite, you can reduce to the finite case by choosing $T'$ to be any finite subset of $T$ that has more elements than $S$.
