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For a vector space $V$ of infinite dimension, to show that $V\oplus V$ is isomorphic to $V$ is to show that there exists an invertible linear transformation between $V \oplus V $ and $V$.

Every vector space have a basis. If $B$ is the infinite set of basis for $V$, then the set $B\oplus0 \cup 0\oplus B$ is the basis for $V \oplus V$. Using axiom of choice the cardinality $|B\oplus0 \cup 0\oplus B|= |B\oplus0|+|0 \oplus B|=|B|+ |B| = \max\{|B|,|B|\}= |B|$ and then I was trying to argue that this implies that there is an isomorphism between the bases of $V\oplus V$ and $V$, so they are isomorphic. But I'm not sure how $|B|+|B|=\max\{|B|,|B|\}= |B|$ is true by the axiom of choice. How could we show that if $V$ is infinite dimensional vector space, then $V\oplus V \cong V$?

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    $\begingroup$ The basis for the sum is the union, not the product of the bases. $\endgroup$ – Tobias Kildetoft May 10 '16 at 11:15
  • $\begingroup$ I edited the post. $\endgroup$ – MathsMy May 10 '16 at 11:23
  • $\begingroup$ I think this is rather misleading now, $B \cup B$ makes me think it's just $B$. Wouldn't $B \oplus 0 \cup 0 \oplus B$ be better? $\endgroup$ – TastyRomeo May 10 '16 at 11:23
  • $\begingroup$ yes, thanks for the correction. I will re-edit the post $\endgroup$ – MathsMy May 10 '16 at 11:24
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    $\begingroup$ Re: But I'm not sure how $|B|+|B|=\max\{|B|,|B|\}= |B|$ is true by the axiom of choice. Assuming AC, for any two infinite cardinals you have $a+b=\max(a,b)$. In particular, $b+b=b$. See, for example, this post. $\endgroup$ – Martin Sleziak May 10 '16 at 12:28
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Your argument is good, you just need to finish it up.

Assuming choice, you can consider a basis $\mathcal{B}$ of $V$. Then the set $\mathcal{C}=\{(v,0):v\in\mathcal{B}\}\cup\{(0,v):v\in\mathcal{B}\}$ is a basis for $V\oplus V$.

Since $\mathcal{B}$ is infinite, there is a bijection $f\colon\mathcal{C}\to\mathcal{B}$, because $\mathcal{C}$ is just the disjoint union of two copies of $\mathcal{B}$. Then the linear map $T\colon V\oplus V\to V$, defined on $\mathcal{C}$ by $T(w)=f(v)$ and extended by linearity, is an isomorphism, because it sends a basis onto a basis.

Note that choice is needed in two places: to obtain a basis for $V$ and to use that $|X|+|X|=|X|$ when $X$ is an infinite set.

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Since your vector space $V$ is infinite-dimensional, let $b_1, b_2, \dots, b_n, \dots$ be a basis of $V$. Then the set $$(B \oplus 0) \cup (0 \oplus B) = \{b_1 \oplus 0, b_2 \oplus 0, \dots, 0 \oplus b_1, 0 \oplus b_2, \dots\}$$

forms a basis of of $V \oplus V$.

Assuming the dimension of $V$ is countable, the dimension of $V \oplus V$ is countable too. Can you come up with an invertible linear transformation yourself, by thinking of how you show that $|\mathbb{N}| = |\mathbb{Z}|$?

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    $\begingroup$ What if there is no countable basis (functional space for instance)? $\endgroup$ – pmichel31415 May 10 '16 at 11:36
  • $\begingroup$ I hadn't really considered that, but you could think of a similar argument as above, no? After all, $|\mathbb{R}^+| = |\mathbb{R}|$. $\endgroup$ – TastyRomeo May 10 '16 at 11:43
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I think there's a clearest way of stating this :

The Axiom of choice gives you a basis $B=(b_i)_{i\in I}$ where $I$ (and incidentally $B$) is infinite and may or may not be countable.

Now given $J_1,J_2\subset I$ such that $J_1\cup J_2=I$ and $J_1\cap J_2=\emptyset$, it is clear that $X_1=(b_j)_{j\in J_1}$ and $X_2=(b_j)_{j\in J_2}$ are such that $span(X_1)\oplus span(X_2)=V$.

Now is it possible to find $J_1,J_2$ such that $span(X_1)$ and $span(X_2)$ are each isomorphic to $V$? (In which case $V\oplus V\cong span(X_1)\oplus span(X_2)=V$)

This boils down to the following question :

  • Given an infinite set $I$, is it possible to find $J_1,J_2\subset I$ such that $J_1\cup J_2=I$ , $J_1\cap J_2=\emptyset$ and $J_1\sim I$ and $J_2\sim I$ ($\sim$ being here used as a (non-standard) way of saying "there exists a bijection between the two")

Now if $I$ is infinite, it is in bijection with $I\times\{0,1\}$ (using AC).

Now if $f$ is your bijection just set $J_1=f^{-1}(I\times\{0\}),J_2=f^{-1}(I\times\{1\})$.

Qed

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