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I know that $V = U \oplus W$ means that every $v \in V$ can be written uniquely as $v = u + w$ for some $u \in U, w \in W$. However, what happens if $V = U + W$ is not direct? Then this means that some vector $v$ does not have a unique representation. But can we say exactly which vectors have a unique representation, and which don't?

Is this even a useful question? Intuitively, I am thinking "sometimes we don't have a direct sum, but perhaps we can still work with what we've got. In particular, let's see which vectors we can write uniquely, and maybe we can work with those."

twosigma
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    if some vector $v$ has more than one representation, then $0$ has more than one representation. Then $u + w = u + w + 0 = \ldots$ – preferred_anon May 14 '20 at 15:37

3 Answers3

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If $V=U+W$ is not direct, then there is $z \in U \cap W$ with $z \ne 0.$

Let $v \in V,$ then there are $u \in U$ and $w \in W$ such that

$$v=u+w.$$

We also have

$$v=(u-z)+(w+z).$$

Observe that $u-z \in U, w+z \in W, u-z \ne u$ and $w+z \ne w.$

Consequence: each $v \in V$ does not have a unique representation.

Fred
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If one vector can be written in more than one way, then all can. Let $v=u+w$ for different pairs $\{u,w\}$ and write $x=(x-v)+v=(u'+w')+(u+w)=(u'+u)+(w'+w)$.

joriki
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    To clarify: $u' + w'$ is not a representation of $v$. It is a representation of $x - v$. So at the end when we write $x = (u' + u) + (w' + w)$, we can substitute different values for $u$ and $w$ because $v$ doesn't have a unique representation. For example if $v = u_1 + w_1$ and $v = u_2 + w_2$ then $x = (u' + u_1) + (w' + w_1)$ and $x = (u' + u_2) + (w' + w_2)$, yes? So $x$ has non-unique representation too. – twosigma May 14 '20 at 05:53
  • Note: this answer generalizes to any finite sum. (The other answers that use the condition $U\cap W=0$ works for two subspaces but can be generalized; see this post). Specifically, suppose $V=\sum U_i$ isn’t direct, say $v=\sum u_i^=\sum u_i^{}$. Then for any $x$, if $x-v=\sum u_i$, we have $x=(x-v)+v=\sum u_i+\sum u_i^=\sum (u_i+u_i^)$ and similarly $x=\sum (u_i+u_i^{})$. Since $u_j^\neq u_j^{}$ for some $j$, $u_j+u_j^*\neq u_j+u_j^{}$. – twosigma May 17 '20 at 19:40
  • Alternatively, we can reduce the general question of the finite case to the case of two subspaces. The idea is that we can break down the non-direct finite sum into a non-direct sum of two subspaces: if $V = \sum U_i$ isn’t direct, say there is $v \in V$ such that $v = \sum u_i^* = \sum u_i^{}$ with say $u_j^* \neq u_j^{}$ for some $j$, then write $V=\sum U_i=U_j + (\sum_{i\neq j} U_i)$, which is a non-direct sum of two subspaces. – twosigma May 17 '20 at 19:40
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First note that the difference between $U\oplus W$ and $U+W$ is that in the latter case we allow $U\cap W$ to be non-trivial. So there exists a $v\neq 0$ such that $v\in U\cap W$ so we can write $v=v+0$ and $v=0+v$.

bowlofpetunias
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  • That's true, but that's just a sufficient condition, not a necessary one. – joriki May 14 '20 at 05:31
  • @joriki Is it not true that if $V=U+W$ and $V\neq U\oplus W$ then $U\cap W$ is non-empty? And so based on the wording of the question if $V=U+W$ and this is not direct then this directly implies $U\cap W$ is non-empty. – bowlofpetunias May 14 '20 at 05:37
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    @bowlofpetunia: $U\cap W$ is not empty, since $0 \in U\cap W.$ – Fred May 14 '20 at 05:42
  • @Fred Yes you're right I should have been writing non-trivial thank you. I don't believe this changes anything though. – bowlofpetunias May 14 '20 at 05:43
  • @bowlofpetunias: I don't understand why you're asking whether it's not true, since I started my comment by saying that it's true (irrespective of the detail of the distinction between "non-empty" and "non-trivial"). – joriki May 14 '20 at 05:44
  • @joriki sorry perhaps I'm misunderstanding but what do you mean that it is not a necessary condition? – bowlofpetunias May 14 '20 at 05:46
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    @bowlofpetunias: The question asks "exactly which vectors have a unique representation". You show that particular vectors don't have a unique representation, but that doesn't answer the question, since it doesn't follow that all others do. – joriki May 14 '20 at 05:47
  • @joriki yes you are right. So I suppose I've shown existence of one and with your proof the rest follows quite nicely. Thank you for pointing that out. – bowlofpetunias May 14 '20 at 05:48