Always exist a linear transformation from an infinite dimensional vector space X to X that is a surjective but injective?

Question

In the ring of all linear transformation of an infinite dimensional vector space over a division ring, always exist a linear transformation which is surjective but no injective?

Answer 1

There exist a set $B$ and a sequence of distinct vectors $x_1,x_2,..$ not in $B$ such that $\{x_1,x_2,...\} \cup B$ is a basis for the vector space. Define $Tx_1=0$ and $Tx_{n+1}=x_n$ for $n \geq 1$. Also let $Tx=x$ for $x \in B$. We can now extend $T$ to a linear map from $X$ in to itself. This map is surjective but nor injective (since $Tx_0=0=T0$ and $x_0 \neq 0$).