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“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, ncmathsadist Jan 29 '13 at 2:03

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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

1 Answer 1

up vote 2 down vote accepted

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 $X$.

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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

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