= Solution
The first three output components equal $x_1,x_2,x_3$, while the fourth is $x_4-\sum_{k=1}^4x_k=-x_1-x_2-x_3$. Hence
$$
\boxed{P=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\-1&-1&-1&0\end{pmatrix}.}
$$
Its outputs have component sum zero, and applying $P$ to any such output leaves it fixed. Thus
$$
\boxed{P^2=P,\qquad\det P=0,\qquad\operatorname{tr}P=3.}
$$
The first three columns give a <basis> of the <image of a linear map>:
$$
v_1=(1,0,0,-1)^T,\quad v_2=(0,1,0,-1)^T,\quad v_3=(0,0,1,-1)^T.
$$
They are <linearly independent> because their first three coordinates are the standard <basis>. Their <span> is precisely the three-dimensional hyperplane $y_1+y_2+y_3+y_4=0$.
Apply the <Gram-Schmidt process>. First $q_1=v_1/\sqrt2$. Subtracting the $q_1$ projection from $v_2$ gives $(-1,2,0,-1)^T/2$, of length $\sqrt{3/2}$. Subtracting both preceding projections from $v_3$ gives $(-1,-1,3,-1)^T/3$, of length $2/\sqrt3$. Therefore an <orthonormal basis> is
$$
\boxed{q_1=\frac{(1,0,0,-1)^T}{\sqrt2},\qquad q_2=\frac{(-1,2,0,-1)^T}{\sqrt6},\qquad q_3=\frac{(-1,-1,3,-1)^T}{\sqrt{12}}.}
$$
Their pairwise <dot products> vanish and each has length one. Although this image admits an <orthonormal basis>, $P$ itself is an <oblique projection>: its <kernel> is the fourth coordinate axis, which is not perpendicular to the image, and $P\ne P^T$.
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