Suppose $V$ is vector space and let $U,W$ are subspaces of $V$.
Define $U+W:=\{u+w: u\in U,w\in W\}$. We say that the sum of subspaces $U$ and $W$ is direct if any vector $x\in U+W$ has a unique representation $x=u+w$ with $u\in U$ and $w\in W$.
Question 1: if the vector $x$ can be written as the sum $x=u+w$ and we also know that $x=w+u$. Could I conclude that $u=w$? This uniqueness from definition confuses me a bit!
Question 2: I want to show the following: if the sum of $U$ and $W$ is direct then $U\cap W=\{0\}$.
Let $z\in U\cap W$ and $z\neq 0$ then one can write $z=z+0=0+z$. Since the sum is direct hence there is a unique representation could I conclude that $z=0$?
I would be very grateful if anyone can in detail explain that confusing moment with uniqueness.
Best Answer
Perhaps an example is helpful. Let $V = \mathbb{R}^2$, let $U$ be the $x$-axis, and let $W$ be the $y$-axis. Then consider the element $x := (2, 1) \in V$. The unique representation of $x$ as a sum of elements from $U$ and $W$ is $$ (2, 1) = (2, 0) + (0, 1), $$ where $u = (2, 0) \in U$ and $w = (0, 1) \in V$. Now, you could also write the sum in the order: $$ (2, 1) = (0, 1) + (2, 0). $$ But what you can't do is use any other element from $U$ than $(2, 0)$. Like, you couldn't use $(7, 0)$ instead; no matter how hard you look, you'll never find a $w \in W$ such that $$ (2, 1) = (7, 0) + w. $$ The element that comes from $U$, whether you write it first or second in the sum, must be $(2, 0)$. And similarly, the element from $V$ must be $(0, 1)$. That's what uniqueness of the representation as $u + w$ means.
A bit more formally, what uniqueness means is that if $u + w = u' + w'$, where $u, u' \in U$ and $w, w' \in W$, then it must be the case that $u = u'$ and $w = w'$.
For the second question, since $z$ and $0$ are in both $U$ and $W$, we can write $z$ as $$ z + 0 \qquad\text{(with $u = z$ and $w = 0$)} $$ or as $$ 0 + z \qquad\text{(with $u' = 0$ and $w' = z$)}. $$ But uniqueness means that $u = u'$, which is to say $z = 0$ (and also $w = w'$, which again tells us that $z = 0$).