Finite vs infinite dimensional vector spaces

Solution 1:

For simplicity, all the vector spaces in the following are over $\mathbb{C}$, or some complete field.

All norms on a finite dimensional vector space are equivalent. This is not true for infinite dimensional vector spaces over (consider $L^p$ norms). I believe this comes from the fact that the unit ball is compact for a finite dimensional normed linear spaces (NLS), but not in infinite dimensional NLS.

The weak topology on a finite dimensional vector space is equivalent to the norm topology. This is always false for infinite dimensional vector spaces. More generally, there are many topologies of interest on an infinite dimensional vector space, but just one of interest on a finite dimensional space (from a linear algebra/functional analysis perspective).

There is a nontrivial translation invariant measure for finite dimensional vector spaces (say over $\mathbb{C} $ or $ \mathbb{R}$, the Lebesgue measure). This is not true for an infinite dimensional Hilbert space (the unit ball has infinitely many disjoint translates of a ball of radius $\sqrt{2}/4$).

This is a property of bounded operators on a NLS: The spectrum of a linear map $T:V \to V$ (the set of $\lambda\in \mathbb{C}$ such that $T-\lambda I$ is not invertible) consists precisely of the eigenvalues of $T$. However, if $V$ is infinite dimensional, then $T-\lambda I$ may not be invertible even if $\lambda$ is not an eigenvalue.

Solution 2:

An infinite dimensional vector space can be isomorphic to one of its proper subspaces ($\mathbb{R}\subset\mathbb{C}$ as vector spaces over $\mathbb{Q}$).

Solution 3:

Also if $V$ is finite dimensional vector space, then the algebra $\operatorname{End}(V)$ has no non-trivial $2$-sided ideals. For $V$ infinite-dimensional, $\operatorname{End}(V)$ has proper $2$-sided ideal which consists of endomorphisms of finite rank.