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Let $H=\mathbb{R}^3$, $P_2$ be a projection of $H$ on the $XY$ plane and $P_1$ a projection of $H$ on the line $L: y = x$ in the $XY$ plane. Find $Y_1^\bot$, $Y_2^\bot$ (where $Y_1 = P_1 (H)$, $Y_2 = P_2 (H)$) and the orthogonal complement of $Y_1$ in $Y_2$, also determine the coordinates of $(P_2-P_1)\overline{x}$, where $\overline{x}=(x,y,z)\in\mathbb{R^3}$. Is $P_1 + P_2$ a projection?

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$P_2 (x,y,z) = (x,y,0)$ is fairly clear. The orthogonal projection of $(x,y,z)$ onto the line $\ell = \{ (\alpha,\alpha,0) : \alpha \in \mathbb{R} \}$ must be the unique closest point on $\ell$ to $(x,y,z)$ because orthogonal projection and closest point projection onto a subspace are the same. So you can find that point by minimizing $$ f(\alpha)=(x-\alpha)^2+(y-\alpha)^2+(z-0)^2 \\ 0 = f'(\alpha) = 2(x-\alpha)+2(y-\alpha) = 2\left[\frac{x+y}{2}-\alpha\right] $$ Therefore, $$ P_1(x,y,z) = \frac{x+y}{2}(1,1,0)=(\frac{x+y}{2},\frac{x+y}{2},0). $$ This point is on $\ell$. The orthogonal complement of $\ell$ contains $(0,0,\beta)$ and $(\alpha,-\alpha,0)$; these project to $0$ under $P_2$. You can directly check whether or not the following defines a projection: $$ (P_1+P_2)(x,y,z)=(x+\frac{x+y}{2},y+\frac{x+y}{2},0). $$

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