A coordinate projection in complex three-space

For $v=(x,y,z)$, $v^{\mathrm{\prime }}=(x^{\mathrm{\prime }},y^{\mathrm{\prime }},z^{\mathrm{\prime }})$ and $a,b\in \mathbb{C}$, the formula gives $P(av+b{v}^{\mathrm{\prime }})=(ax+b{x}^{\mathrm{\prime }},a(x+z)+b(x^{\mathrm{\prime }}+z^{\mathrm{\prime }}),az+b{z}^{\mathrm{\prime }})=aP(v)+bP(v^{\mathrm{\prime }})$. Moreover $P^2(x,y,z)=P(x,x+z,z)=(x,x+z,z)=P(x,y,z)$.

For every $(x,y,z)$, $P(x,y,z)=x(1,1,0)+z(0,1,1)\in U$. Conversely any $u=s(1,1,0)+t(0,1,1)=(s,s+t,t)$ satisfies $P(u)=(s,s+t,t)=u$, so $\mathrm{im}P=U$ and $P{\mathrm{\mid }}_U={\mathrm{id}}_U$. This explicitly realizes the projection supplied by F1.

The equation $P(x,y,z)=0$ is equivalent to $x=0$ and $z=0$, with $y$ arbitrary. Hence $\ker P=\{(0,y,0):y\in \mathbb{C}\}=\mathrm{span}\{(0,1,0)\}$. Every vector decomposes as $(x,y,z)=(x,x+z,z)+(0,y-x-z,0)$ with the first summand in $U$ and the second in $\ker P$. If $(s,s+t,t)=(0,y,0)$, then $s=t=y=0$, so the intersection is zero and the decomposition is unique. Thus the stated kernel is exactly the direction along which $P$ projects.