Linear function between finite-dimensional vectorspaces is continuous 
Possible Duplicate:
“Every linear mapping on a finite dimensional space is continuous” 


I would like to show that 
  $$
f\colon X\to Y, f\mbox{ linear}, X, Y\mbox{ vectorspaces }. \mbox{ dim }X=n, \mbox{ dim }Y=m,
$$
  is continuous.


How can I show that?
At first I thought of sequence theorem of continuity. But it seems to me that this is very circuitous.
 A: You'll need a little more information to solve this problem, but I'm guessing that wherever you're getting this problem from has that information as implied; in particular, we need to work with some notion of distance, a.k.a. a norm on the vector space.
As you correctly mentioned, we can work with the sequential form of continuity, since this suffices for vector spaces. So what we want to show is that if $x_n \rightarrow x$, for $x_n, x \in X$, then $f(x_n) \rightarrow f(x)$.
The key problem here is: What does "$x_n \rightarrow x$" mean? The way to sort this out is to use a norm on the vector space $X$; then the above symbols is equivalent to $\|x_n - x\|_{X} \rightarrow 0$ as $n \rightarrow \infty$. Similarly, we need a norm on $Y$, and then we are trying to show that $\|f(x_n) - f(x)\|_{Y} \rightarrow 0$. 
So, in summary, see if you can show that, if $\|x_n - x\|_{X} \rightarrow 0$, then $\|f(x_n) - f(x)\|_{Y} \rightarrow 0$. You'll need to show that there exists some number $C > 0$, such that $\|f(z)\|_Y \le C \|z\|_X$ for any $z \in X$. This relies on the finite-dimensionality of $X$.
