# Prove that $V=U \oplus W$, given a particular $T: V \rightarrow V$

The projection on $U$ along $W$ is the function $T:V\rightarrow V$ defined by $T(u+w)=u$, where $u \in U$, $w \in W$. Let $V=\mathbb{R}^2$, and $U=\{(x,-x): x \in \mathbb{R}\}$, and $W=\{(x,0): x\in \mathbb{R}\}$. Prove that $V=U \oplus W$, and give formulas for $T$, the projection on $U$ along $W$, and $S$, the projection on $W$ along $U$.

Is it enough to show that T is linear to show that $V=U \oplus W$? I'm uncertain as to whether that is enough to show what I want. Thanks in advance.

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And where is the difficulty? Instead of just dumping a problem here, it's better if you show us what you've done, what you know, where you get stuck, and so on - makes it easier to write useful replies. –  Gerry Myerson Sep 25 '12 at 5:45
Hint: $(x,y) = (x+y, 0) + (-y, y)$ –  martini Sep 25 '12 at 5:52
Sorry @GerryMyerson I typed hastily and forgot to mention that important information. –  tk2 Sep 25 '12 at 6:12

To show that $V = U \bigoplus W$ we need to show that every element $v\in V$ can be written in the form $v = u + w$ where $u \in U$ and $w \in W$, and moreover we need to show that $U\cap W = 0$.
First let $v\in V$. Then we can write $v = (x,y)$ since $V \subset \mathbb{R}^2$ and hence $$v = (x + y, 0) + (-y, y) \in W + V$$ so we have showed the first part, and now we need to show that the intersection is the $0$ space.
Suppose $a\in U\cap V$ then write $a = (x,y)$; Since $a \in W$ that $y = 0$ so $a = (x,0)$ but then since $a\in U$ this implies that $x = -0 = 0$ and hence $$a = (0,0) \in 0$$ So we can now see that $V = U\bigoplus W$.
Then we need to define the projection onto $U$ and $W$, but these are defined exactly by the formula you cited in the fist line, and are well defined linear operators since $V = U \bigoplus W$. One can see this is well defined because, by the direct sum, we can uniquely write $v = u + w$ for all $v$, and hence we only have one possible value for the projection onto either $U$ or $W$ (namely $u$ or $w$ respectively)