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If two vector spaces $V$ and $W$ are isomorphic, do they always have the same dimension?

In other words: $dim V = dimW $ is a sufficient condition for isomorphism, is it also necessary ?

Ehab Shoubaki
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    for finite dimensional vector spaces, having same dimension is equivalent to being isomorphic. (also in the title, the word you should use is "converse" not "inverse") – peek-a-boo Jan 12 '20 at 18:03
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    @peek-a-boo It's true for infinite dimensional spaces too (assuming AC of course). – Noah Schweber Jan 12 '20 at 18:03
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    To see this, simply note that a surjective linear map always takes a spanning set to a spanning set and an injective linear map always takes a linearly independent set to a linearly independent set. – WoolierThanThou Jan 12 '20 at 18:04
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    It bothers me that a yes/no question has not received an explicit yes/no answer. With that said: the answer is yes. – Ben Grossmann Jan 12 '20 at 18:08
  • @NoahSchweber that's good to know; I always assumed there would be be some issues in the infinite dimensional case, so I never really looked into that – peek-a-boo Jan 12 '20 at 18:11
  • @NoahSchweber What is AC ? Is that the "Axiom of Choice" ? – Ehab Shoubaki Jan 12 '20 at 18:14
  • @EhabShoubaki Yup: Basically, it's a statement about vector spaces with bases (since the notion of dimensions depends on a basis), and without the axiom of choice, you can't ensure that every vector space has a basis. – WoolierThanThou Jan 12 '20 at 18:16
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    @peek-a-boo Well, the key fact is that injective linear maps preserve linear independence, and surjective linear maps preserve spanning. That holds in all vector spaces, without using choice. So if $B$ is a basis in $V$ and $f:V\rightarrow W$ is an isomorphism, then $f[B]$ is a basis in $W$ - and clearly $f$ restricts to a bijection between $B$ and $f[B]$. This means that isomorphic vector spaces have the same dimension provided that dimension makes sense: we need AC to prove "Every vector space has a basis, and any two bases for the same vector space have the same cardinality." – Noah Schweber Jan 12 '20 at 18:52
  • Essentially the issue is whether the dimension of a vector space is well-defined. Without AC, a vector space doesn't necessarily have a basis, but you could define "$X$ has dimension "$\alpha$" to mean "$X$ has a basis of cardinality $\alpha$", and then we don't need AC to say: if $X$ and $Y$.are isomorphic and one has dimension $\alpha$, then so does the other. – Robert Israel Jan 12 '20 at 18:54
  • If you defined "$X$ has dimension $\alpha$" as I suggested, the fact that there could be a vector space with bases of different cardinalities would mean that dimension is not unique. My statement would be true, it just wouldn't preclude the possibility that $X$ and $Y$ have other dimensions as well. – Robert Israel Jan 13 '20 at 15:25
  • @RobertIsrael You're right, I'm sorry, I misread your definition. – Noah Schweber Jan 13 '20 at 15:31

1 Answers1

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The answer is yes.

The key point is the following:

Lemma: if $f:V\rightarrow W$ is a linear map and $B$ spans $V$, then $f[B]$ spans $f[V]$. Here $f[B]=\{f(b): b\in B\}$ and $f[V]=\{f(v): v\in V\}$; in particular, $f[V]$ is the image of $f$.

Proof: Fix $w\in f[V]$. Since $w\in f[V]$ there is some $v\in V$ with $f(v)=w$ (there may be more than one, since $f$ might not be injective, but that's fine). Since $B$ spans $V$ there are scalars $s_1,..., s_n$ and vectors $b_1,...,b_n\in B$ such that $$s_1b_1 + ... + s_nb_n=v.$$ But then by linearity of $f$ we have $$s_1f(b_1)+...+s_nf(b_n)=f(v)=w.$$ So $w$ is a linear combination of elements of $f[B]$. $\quad\Box$

Now we can prove the desired result. If $f:V\rightarrow W$ is an isomorphism, let $B$ be a basis for $V$. Then:

  • $f[B]=\{f(b): b\in B\}$ is a spanning set for $W$, by the above, and since $f$ is injective we have $\vert f[B]\vert=\vert B\vert$.

  • If $f[B]$ is linearly independent in $W$, then $f[B]$ is a basis for $W$, and so $dim(V)=dim(W)$ since they have bases of the same cardinality.

  • So suppose $f[B]$ is not linearly independent in $W$. Then we can find a $C\subsetneq B$ which spans $W$.

  • But then - since $f^{-1}$ is also a bijective linear map - we have that $f^{-1}[C]=\{f^{-1}(c): c\in C\}$ spans $V$ and is a proper subset of $B$. Since no proper superset of a spanning set can be linearly independent, this contradicts the fact that $B$ was a basis for $V$.

There's a subtle issue in the above, around the existence of bases. It's easy to show that if $V$ has a finite spanning set then $V$ has a basis and any two bases of $V$ have the same cardinality, but the situation is more complicated with respect to general vector spaces: it's consistent with ZF (= set theory without the axiom of choice) that $(1)$ there are vector spaces which have no bases at all and $(2)$ there are vector spaces with bases of different cardinalities! So without AC, we might have a vector space with no dimension or multiple dimensions.

When we look carefully at the above argument, the "choice-free" theorem is the following:

Suppose $f:V\rightarrow W$ is an isomorphism of vector spaces. Then letting "$Bases(U)$" be the set of bases of $U$, the map $$F: Bases(V)\rightarrow Bases(W): B\mapsto f[B]$$ is a bijection and for all $B\in Bases(V)$ we have $\vert B\vert=\vert F(B)\vert$.

Noah Schweber
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