Why is a linear transformation of a cauchy sequence in a normed space also cauchy?

Suppose we have a cauchy sequence $\{a_n\}$ in a normed vector space $V$. Given a linear transformation $T:V \rightarrow V$, is the sequence $\{T(a_n)\}$ also cauchy? Or is it true only for finite dimensional normed spaces? I'd be much obliged if someone could give a proof for this, preferably an elementary one. Thanks!

• $T$ has to be continuous. Because continuous linear transformation on a normed space is bounded: $\| Tx\| \leq M\| x\|$ for some $M>0$ and all vectors $x$. Hence if $\{ a_n\}$ is a Cauchy sequence, then $\| Ta_m-Ta_n\| \leq M\| a_m-a_n\|$ gives that $\{ Ta_n\}$ is a Cauchy sequence as well. – Janko Bracic Jan 17 '15 at 8:54

Given $n,m\in \mathbb N; \exists p\in \mathbb N$ such that $||a_n-a_m||<\epsilon \forall m,n\geq p$
Now $||T(a_n)-T(a_m)||=||T(a_n-a_m)||<\epsilon$