# Finite dimensional normed vector spaces

Due to a deplorable oversight in the preparation of my Functional Analysis course, I had forgotten to present in the first chapter the elementary properties of finite-dimensional normed vector spaces. Since I didn’t want to stop the flow of the course to come back to this artificially (and since I didn’t really need those facts yet), I delayed until a better moment came. In the end, I am going to prove these things in the chapter containing the Banach Isomorphism Theorem, by an argument that may be called “cheating”, but which I find amusing.

There are three properties, which are closely related, which I wanted to state for finite-dimensional normed vector spaces (over C, although everything also holds for real vector spaces):

(Property 1) Any finite-dimensional normed vector space is a Banach space (i.e., is complete);

(Property 2) Any two norms on a finite-dimensional space are equivalent: there exist constants c>0, C such that

$c||v||_1\leq ||v||_2\leq C||v||_1$

for all vectors v;

(Property 3) Any finite-dimensional subspace of an arbitrary normed vector space is closed with the induced norm.

Since Cn is a Banach space with the norm

$(*)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||(x_1,\ldots,x_n)||=\max |x_i|,$

we see that Property (2) clearly implies Property (1). Moreover, the latter gives Property (3), since if W is finite-dimensional in V, the induced norm will make it a complete subset of V by Property (1), and so it must be closed in W by basic topology.

The second property is typically proved with a little compactness argument in the first chapter of textbooks in functional analysis, and so the others follow. Here is the alternate argument I will present.

We prove Property (1) first, by induction on the dimension of the finite dimensional normed vector space V. For dimension 1, with V spanned by some vector e, homogeneity of the norm gives

$||te||=c|t|\text{ with } c=||e||$

which implies that V is homeomorphic to C, hence complete. Now if we assume that this Property (1) holds for spaces of dimension n-1, and V has dimension n, we consider a basis e1,…,en of V, and for every i=1,…,n, look at the i-th coordinate functional

$\lambda_i : V\rightarrow \mathbf{C}.$

We do not know (in our setting!) that these are continuous, but the kernel Vi is of dimension n-1, so with the induced norm, it is a Banach space by the induction hypothesis, and so in particular is must be closed in V. Since the i-th basis vector is not in Vi, a corollary of the Hahn-Banach Theorem (that is in fact typically proved directly in that particular case) states that λi does have a continuous extension to V, with the property that

$\tilde{\lambda}_i(e_i)\not=0.$

But this extension must then be a non-zero multiple of λi, and so those coordinate functions themselves are continuous.

Now, having done this, the linear map

$T : V\rightarrow \mathbf{C}^n$

mapping v to

$T(v)=(\lambda_1(v),\ldots,\lambda_n(v))$

is then clearly bijective, and it is continuous (where the target has any of the standard norms, for instance the maximum norm (*) above). The inverse is also continuous since it is given by

$T^{-1}(\alpha_1,\ldots,\alpha_n)=\sum_{i=1}^n{\alpha_i e_i}$

and

$||\sum_{i=1}^n{\alpha_i e_i}||\leq D \max|\alpha_i|,\ \text{ where }\ D=n\max_i ||e_i||.$

In other words, T is a homeomorphism, and hence V is complete since Cn is. [Note : as was pointed out in a comment, general homeomorphisms do not preserve completeness, but here T and its inverse are also linear, hence both are Lipschitz, and then completeness is preserved.]

Having proved this first property of finite-dimensional normed vector spaces, we obtain (2) as a corollary of the Banach Isomorphism Theorem: given an arbitrary norm on a finite-dimensional vector space, we can compare it in one direction to the norm

$||\sum_{i=1}^n{\alpha_i e_i}||_2=\max|\alpha_i|$

defined in terms of a basis, as above. This means the identity bijective linear map

$(V,||\cdot||_2)\rightarrow (V,||\cdot ||)$

is a continuous bijection between Banach spaces (by Property (1), both norms are Banach norms, which is required to apply the Isomorphism Theorem), and hence its inverse is continuous by the Banach Isomorphism Theorem, which implies the two norms are in fact equivalent.

And as we already described, the Property (3) follows directly from Property (1).

### Kowalski

I am a professor of mathematics at ETH Zürich since 2008.

## 3 thoughts on “Finite dimensional normed vector spaces”

1. Attila Smith says:

“In other words, T is a homeomorphism, and hence V is complete since Cn is.”

Completeness needn’t be preserved under homeomorphisms.

2. Yes, very true, thanks for the correction! Here of course we have a linear bijective homeomorphism, hence it is (bi)-Lipschitz, so completeness is preserved. I’ll correct the text.

3. Guero says:

This may not help too much, but topological completeness is preserved by homeomorphisms.