0

I have not done math for a long time and I need a little help.

I need to know if it's true that if $X$ is not isomorphic to $Y$ then $ X \times X$ is not isomorphic to $Y \times Y$ being $X$ and $Y$ vector spaces. I tried trying that if $ X \times X$ is isomorphic to $Y \times Y$ then $X$ is isomorphic to $Y$ but I do not know how to continue.

If it is true can you help me with the proof? And if it is false can you give me an example of vector spaces $X$ and $Y$ such that $X$ is not isomorphic to $Y$ but however $ X \times X$ is isomorphic to $Y \times Y$.

Thank you to all.

Catacroker
  • 93

1 Answers1

1

It is indeed true. Recall that two vector spaces are isomorphic iff the cardinality of their basis coincide.

Let's first treat the finite-dimensional case. Let $n:=dim(X)$ and $m:=dim(Y)$. If $X\times X$ is isomorphic to $Y\times Y$, then we have (using $dim(V\times V)= 2 dim(V)$ for a vector space $V$)

$$ 2n=dim(X\times X) = dim(Y\times Y) = 2m.$$

We get $n=m$ and hence $X$ and $Y$ are isomorphic.

The infinite-dimensional case is essentialy the same. We need that the Cardinality of the Cartesian Product of Two Equinumerous Infinite Sets is the same as the infinite set itself. And redo the argument above.

Edit: As pointed out in the comment section the argument is not clear in the infinite-dimensional case. Let $E$ be a basis of an infinite-dimensional vector space $V$. Note that we have

$$\vert E \vert \leq \vert basis \ V\times V\vert= \vert E \ \sqcup E\vert \leq \vert E \times E\vert = \vert E \vert,$$

where the last equality is due to Cardinality of the Cartesian Product of Two Equinumerous Infinite Sets. Hence, the dimension of $V$ and $V\times V$ coincide. Thus, if $X\times X$ and $Y\times Y$ have the same dimension, so do $X$ and $Y$.

Severin Schraven
  • 19,179
  • 1
    Are you absolutely sure that $\dim(E\times F)=\dim(E)\times\dim(F)$? What is the dimension of $\mathbb{R}^n$ over $\mathbb{R}$? – C. Falcon May 12 '17 at 16:34
  • 2
    @C.Falcon Thanks for pointing out my stupid mistake. – Severin Schraven May 12 '17 at 16:37
  • Note that there is nothing natural about the isomorphism! – JJR May 12 '17 at 16:49
  • @JJR Is it possible to obtain a natural isomorphism? My feeling is telling my that in the infinite-dimensional case there is no such thing, but what about the finite-dimensional case? – Severin Schraven May 12 '17 at 16:52
  • @SeverinSchraven I also don't think so and additionally I am not sure that the argument with the cardinality in infinite dimensional case works since $\mathbb{R}$ and $\mathbb{C}$ have same cardinality as sets but are not isomorphic over $\mathbb{R}$. – JJR May 12 '17 at 16:58
  • @JJR I'm not applying the cardinality argument to the the spaces, but to their respective basis. – Severin Schraven May 12 '17 at 16:59
  • You probably mean $|\text{Basis of }X| = |\text{Basis of }X\times X|$ – JJR May 12 '17 at 17:04
  • is it really so trivial to prove that $|A\amalg A| = |A|$ for any cardinal $|A|$ – JJR May 12 '17 at 18:06
  • @JJR Do you see any problems with the inequalities above? – Severin Schraven May 12 '17 at 18:17
  • They are definitely correct just not sure if trivial https://math.stackexchange.com/questions/198904/cardinality-of-infinite-sets – JJR May 12 '17 at 19:06