# Counterexample to uniqueness of complement of a vector subspace

I am currently working through Axler's Linear Algebra Done Right. From what I understand, given an infinite dimensional vector space $V$ and subspaces $U$, $W_1$, and $W_2$ of $V$ such that $V=U\oplus W_1=U\oplus W_2$, it is not necessarily the case that $W_1$ and $W_2$ are isomorphic.

Intuitively, I would think that given a basis of $U$, one could extend this to a basis of $V$, and that "basis extension" would uniquely determine the "direct-sum complement" of $U$ (up to isomorphism). Thus, I am wondering if someone could provide an example of a vector space $V$ and a subspace $U$ of $V$ such that the "direct-sum complement" of $U$ is not unique, and provide some insight into why my line of reasoning fails?

• Take $U=...\oplus\mathbb{K}\oplus\mathbb{K}$ (infinite sum of the field of scalars), $M_1=\mathbb{K}$ and $M_2=\mathbb{K}\oplus\mathbb{K}$. – orole Jan 4 '18 at 22:31
• Your approach was trying a different problem. Assume $U\subset V$ is given (Observe how in the example above $U$ doesn't sit in the same way inside $U\oplus M_1$ and inside $U\oplus M_2$). Assume that $M_1, M_2\subset V$ are subspaces that complement $U$ in $V$. Can $M_1$ and $M_2$ have bases of different cardinalities? That is what you were trying to answer. – orole Jan 4 '18 at 22:42
• By the way, in your approach you mentioned 'uniquely determine'. Not quite, it is still true that there are many different complements, even in finite dimension. Look in the plane the $Y$ axis complement the $X$ axis, but the line $y=x$ also complement the $X$ axis. There are many ways to extend the basis of a subspace to a basis of the whole space. – orole Jan 4 '18 at 22:47
• I guess he means unique "up to isomorphisms" – Tommaso Seneci Jan 4 '18 at 22:48
• @orole: In their problem, they are given $U$ as a subspace of $V$. So their problem is the problem of your second comment, not your first comment. – Eric Wofsey Jan 4 '18 at 23:05

Your understanding is incorrect. If $V=U\oplus W_1=U\oplus W_2$, then $W_1$ and $W_2$ are always isomorphic. Indeed, given that $V=U\oplus W_1$, if you take the projection map $V\to V/U$ and restrict it to $W_1$, you get an isomorphism $W_1\to V/U$. So $W_1\cong V/U$, and similarly $W_2\cong V/U$, and so $W_1\cong W_2$.
(For an explicit direct isomorphism between $W_1$ and $W_2$, note that any element $x\in W_1$ can be written uniquely as $u+w$ where $u\in U$ and $w\in W_2$. The map taking $x$ to $w$ is then an isomorphism $W_1\to W_2$.)
• Do we need $W_1$ and $W_2$ to be closed? – Filburt Jan 18 '18 at 13:23
• No, there need not even be any topology on $V$. – Eric Wofsey Jan 18 '18 at 16:05