# Linear function between finite-dimensional vectorspaces is continuous [duplicate]

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.

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## marked as duplicate by Gerry Myerson, Asaf Karagila, 5PM, Brandon Carter, ncmathsadistJan 29 '13 at 2:03

You don't need $Y$ to be finite-dimensional. But you need $X,Y$ to be normed spaces for this to make sense. Or at least topological vector spaces. –  1015 Jan 28 '13 at 22:36
The link you gave is too difficult for me, sorry. Can one show it more elementary? By $\epsilon - \delta$ - criterion maybe? –  math12 Jan 28 '13 at 22:48
The root of all this is the equivalence of norms in finite dimension. Are you aware of this? –  1015 Jan 28 '13 at 22:50
I guess no. Sorry! :( –  math12 Jan 28 '13 at 22:53

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 $X$.
Let $\left\{a_1,...,a_n\right\}$ be a basis of the n-dimensional normed vectorspace $X$. Then I tried this but could not continue it: $\lVert f(x)\rVert_Y=\lVert\sum\limits_{i=1}^{n}\alpha_i f(a_i)\rVert_Y\leq\sum\limits_{i=1}^n\lVert\alpha_i f(a_i)\rVert_Y=\sum\limits_{i=1}^n\lvert\alpha_i\rvert\lVert f(a_i)\rVert_Y\leq\max\limits_{1\leq i\leq n}\left\{\lvert\alpha_i\rvert\right\}\cdot\sum\limits_{i=1}^n\lVert f(a_i)\rVert_Y$ –  math12 Jan 29 '13 at 10:27
@math12 : $\sum\limits_{i=1}^n\lVert f(a_i)\rVert_Y\le n\max\limits_{1\le i\le n}\lVert f(a_i)\rVert_Y$ –  xavierm02 Feb 13 '13 at 14:59