Projection onto a linear transformation of the non-negative orthant

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Given a vector $b \in \mathbb{R}^n$ and a symmetric positive-definite matrix $A \in \mathbb{R}^{n \times n}$.

The question: How to find the projection of the vector $b$ onto the convex set $\{ x \in \mathbb{R}^n | Ax \ge 0 \}$ ? (the inequality means that each component of LHS is non-negative)

I suspect that the solution might use the function $\max(\; \cdot \; , 0)$ and/or the spectral decomposition of $A$. However, I have not solved the problem yet.

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Your problem has no closed-form solution because it is equivalent to non-negative least square problem.

Let $y=Ax$, your problem is equivalent to: $$\arg\min_{y\geq0}\|A^{-1}y-b\|^2$$ which is the standard form of NNLS, a problem which is known that it has no closed-form solution.