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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.

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    $\begingroup$ All norms are equivalent. Unit ball is relatively compact. Subspaces are 'automatically' closed. $\endgroup$ – copper.hat Oct 9 '13 at 22:02
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(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 continuos.

(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.

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  • $\begingroup$ Is 'injective endomorphism $T$' different from endomorphism $T$ with $\ker(T)=\{0\}$? $\endgroup$ – John Smith Kyon Jan 22 '20 at 9:29
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A finite dimensional vector space is always isomorphic to its dual, but this is false for an infinite dimensional vector space.

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  • $\begingroup$ 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... :-)) $\endgroup$ – Asaf Karagila♦ Oct 10 '13 at 12:40
  • $\begingroup$ What's the counterexample? $\endgroup$ – ante.ceperic Oct 11 '13 at 7:00
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For example, not every (infinite) matrix corresponds to a linear operator.

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