# Which vectors have unique representation when it isn't a direct sum?

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."

• if some vector $v$ has more than one representation, then $0$ has more than one representation. Then $u + w = u + w + 0 = \ldots$ Commented May 14, 2020 at 15:37

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.

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 $$x=(x-v)+v=(u'+w')+(u+w)=(u'+u)+(w'+w)$$.

• 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. Commented May 14, 2020 at 5:53
• 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 $j$, $u_j+u_j^*\neq u_j+u_j^{**}$. Commented May 17, 2020 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. Commented May 17, 2020 at 19:40

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 $$v=0+v$$.

• That's true, but that's just a sufficient condition, not a necessary one. Commented May 14, 2020 at 5:31
• @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. Commented May 14, 2020 at 5:37
• @bowlofpetunia: $U\cap W$ is not empty, since $0 \in U\cap W.$
– Fred
Commented May 14, 2020 at 5:42
• @Fred Yes you're right I should have been writing non-trivial thank you. I don't believe this changes anything though. Commented May 14, 2020 at 5:43
• @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. Commented May 14, 2020 at 5:47