Differences between infinite-dimensional and finite-dimensional vector spaces

Question

I've just started a course in Representation Theory, and in solving our first homework I've used a couple of theorems about finite-dimensional vector spaces (for an example, rank-nullity theorem). My colleagues pointed out to me that we are working in general vector spaces, so I've patched up places where I've used those theorems with more general arguments.

So, why did I make that mistake? Well, in my previous linear algebra courses we mostly worked with finite-dimensional vector spaces, so in my mind I started to consider all vector spaces finite-dimensional.

To fix that, and to prevent future mishaps, I would like to see a few differences between finite-dim. and infinite-dim. vector spaces. More 'obvious' fact is in finite-dimensional space, the better.

All norms are equivalent. Unit ball is relatively compact. Subspaces are 'automatically' closed. — copper.hat, Oct 09 '13 at 22:02

score 14 · Accepted Answer · edited Mar 15 '24 at 19:41

14

(1). There are endomorphisms $T$ with $\ker(T)=\{0\}$ which are not surjective.

(2). Not in every case a linear form $\phi$ is representable by a vector $v$ in presence of a scalar product, i.e., there doesn't exist a vector $v$ that $\phi(.)=\langle v,.\rangle$.

(3). Not all linear mappings are continuous.

(4). You can equip a vector space with two different norms such that the unit ball in respect to the first norm in unbounded in respect to the second.

It's just a brand new world.

edited Mar 15 '24 at 19:41

rschwieb

153,510

answered Oct 10 '13 at 12:32

Michael Hoppe

18,103
3
32
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Is 'injective endomorphism $T$' different from endomorphism $T$ with $\ker(T)={0}$? – BCLC Jan 22 '20 at 09:29

score 5 · Answer 2 · answered Oct 09 '13 at 22:15

5

A finite dimensional vector space is always isomorphic to its dual, but this is false for an infinite dimensional vector space.

answered Oct 09 '13 at 22:15

Jeff

2,594

It should be mentioned that if we considered the algebraic dual, then this is always false for infinite dimensional spaces. (At least under the assumption of choice... :-)) – Asaf Karagila Oct 10 '13 at 12:40
What's the counterexample? – ante.ceperic Oct 11 '13 at 07:00

score 4 · Answer 3 · answered Oct 09 '13 at 22:07

4

For example, not every (infinite) matrix corresponds to a linear operator.

answered Oct 09 '13 at 22:07

Boris Novikov

17,470

score 0 · Answer 4 · answered Mar 15 '24 at 19:24

Here is a paraphrase of the first item in Michael Hoppe's answer: Proper linear subspaces of finite dimensional vector spaces have strictly smaller dimension. (compare e.g. Prove that a set is infinite if and only if it is equipotent to a proper subset.)

As an example consider the vector space of (one-sided) sequences of real numbers and its subspace consisting of all those sequences whose zeroth entry is zero. (see e.g. Spectrum of right shift operator contradiction)

Note that this distinguishes the finite dimensional paradigm and the infinite dimensional paradigm without reference to topology.

Differences between infinite-dimensional and finite-dimensional vector spaces

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