Take U,W to be subspaces of $\mathbb{R}^{3}$
$U = \operatorname{Lin}\left\{\left(\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right),\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right)\right\}$
and
$V=\operatorname{Lin}\left\{\left(\begin{array}{l} 1 \\ -2 \\ 1 \end{array}\right)\right\}$.
Also : $U \oplus V = \mathbb{R}^{3}$.
I need to find vectors $u \in U$ and $x \in U^{\perp}$, such that
$\left(\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right)$ = $u +x$
That is, find a choice of $u$ and $x$ that is unique and that gives me this vector.
Now, my strategy is to find a orthogonal projection matrix that sends $\mathbb{R}^{3}$ to $U$.
This should give me a vector in $U$ such that when added to a vector $x \in V$, say $\left(\begin{array}{l} 1 \\ -2 \\ 1 \end{array}\right)$ , I can obtain the required vector (in a unique way).
But I if a build a projection matrix P = $U(U^{T}U)^{-1}U^{T}$, and then take $u' = P \left(\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right)$ , where $U = \left[\begin{array}{cc} -1 & 1 \\ 0 & 1 \\ 1 & 1 \end{array}\right]$ ... it doesn't quite give me the vector that I need.
This is is my inital strategy. What am I missing ? What is the right way to approach this ?