[Math] Assume that $V$ and $W$ are isomorphic vector spaces. What does it imply if $V$ and $W$ are not finite dimensional.

linear algebravector-spaces

All of the theorems in my textbook in regards to isomorphisms and dual spaces speak specifically about relationships between finite vector spaces.

For example, "Two finite dimensional vector spaces over $\mathbb{F}$ are isomorphic if and only if they have the same dimension."

And "Suppose $V$ is finite dimensional, then $V'$ is also finite dimensional and dim$V'$=dim$V$.

I have encountered a problem that seems to conspicuously leave out the normal preface of "assume $V$ and $W$ are finite dimensional…". I'm wondering how far I can extend the rules of finite dimensional vector spaces, dual spaces and isomorphisms, and their relationships.

EDIT: For example, I believe the following proof would be correct if we are only considering finite vector spaces:

Assume $V'$ and $W'$ are isomorphic, then $V$ and $W$ are isomorphic.

  • dim$V'$ = dim$W'$ = dim$V$ = dim$W$ $\longrightarrow$ $V$ is isomorphic to $W$.

What happens if finite dimension is not mentioned?

Best Answer

Yes, two infinite vector spaces are isomorphic if and only if they have the same dimension - but you have to appropriately define 'dimension'. To do this, you'll have to understand cardinal numbers, which are a way of measuring the relative sizes of (possibly infinite) sets.

Every set $S$ has an associated "size", or cardinality, a cardinal number denoted $|S|$ or $\#S$. A cardinal number may be finite or infinite. Without going into too much detail, the idea is that if two sets can be put in bijection with each other then they must have the same 'size'. That is, if $S,T$ are sets and $f: S \rightarrow T$ is a bijection, then $|S| = |T|$. In fact, some infinite cardinals are larger than others. For example, the cardinality of $\mathbb{R}$ is strictly larger than the cardinality of $\mathbb{N}$: $|\mathbb{R}| > |\mathbb{N}|$.

Every vector space has a basis, including infinite-dimensional ones. In the case of an infinite-dimensional vector space, the dimension can still be defined - we set $\dim(V) = |B|$ for any basis $B$ of $V$. (To prove this definition is well defined, you need to first show that $|B| = |B'|$ for any bases $B, B'$ of $V$.) If there is an isomorphism $T: V \rightarrow V'$ between two vector spaces $V$ and $V'$, choose any basis $B$ of $V$; then $B' = T(B)$ is a basis for $V'$ and $T$ is a bijection between $B'$ and $B$. This shows that $B$ and $B'$ are the same size in the sense of cardinality.

Since some infinite cardinals are larger than others, it is not true that all infinite-dimensional vector spaces are isomorphic.

Note, however, that many things that hold for finite-dimensional vector spaces and linear maps do not hold for infinite-dimensional ones. For example, linear transformations between isomorphic finite-dimensional vector spaces are injective if and only if they are surjective. This is not true of infinite-dimensional vector spaces. Infinite-dimensional spaces have isomorphic proper subspaces, etc.