TITLE: If $\dim(V) $ is infinite, show that $V\oplus V$ is isomorphic to $V$
QUESTION [7 upvotes]: 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 $V\oplus V \cong V$?

REPLY [3 votes]: 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.