“Every linear mapping on a finite dimensional space is continuous”

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Every linear function on a finite-dimensional space is continuous.

I was wondering what the domain and codomain of such linear function are?

Are they any two topological vector spaces (not necessarily the same), as along as the domain is finite-dimensional? Can the codomain be a different normed space (and may not be finite-dimensional)?

I asked this because I saw elsewhere the same statement except the domain is a finite-dimensional normed space, and am also not sure if the codomain can be a different normed space (and may not be finite-dimensional).

Thanks and regards!

Answer

The result we can show is the following:

Let E and F two topological vector spaces, and T:EF a linear map. If E is finite dimensional, then T is continuous.

First, if (e1,,en) is a basis of E, then any set of n+1 vectors of T(E) is linearly dependent, so T(E) has a dimension . Let k be the dimension of T(E), and (v_1,\ldots,v_k) a basis of this space. We can write for any x\in E: T(x)=\sum_{i=1}^ka_i(x)v_i and since v_i is a basis each a_i is linear.
We have to show that each map T_i\colon E\to F, T_i(x)=:a_i(x)v_i is continuous.


Added: the map x\mapsto a_i(x) is well-defined because (v_1,\ldots,v_k) is a basis. In particular, it takes finite values.


By definition of a topology on a topological vector space we only have to show that the map x\mapsto a_i(x) is continuous. To do that, we use the fact that a finite dimensional topological vector space can be equipped with a norm which gives the same topology (in fact it’s the unique one), namely put N\left(\sum_{j=1}^n\alpha_jx_j\right):=\sum_{j=1}^n|\alpha_j|.
Now the continuity is easy to check: denoting x=\sum_{j=1}^nx_je_j and y=\sum_{j=1}^ny_je_j
|a_i(x)-a_i(y)|\leqslant \sum_{j=1}^n|a_i((x_j-y_j)e_j)|=\sum_{j=1}^n|x_j-y_j|\cdot |a_i(e_j)|\leqslant N(x-y)\sum_{j=1}^n|a_i(e_j),
since |x_j-y_j|\leqslant N(x-y) for all 1\leqslant j\leqslant n.

Attribution
Source : Link , Question Author : Tim , Answer Author : Davide Giraudo

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