The definition of probability simplex in $\mathbb{S}^n$:
$$\{X\in \mathbb{S}^n: X\geq 0, \text{tr}(X) = 1\},$$ where $\mathbb{S}^n$ is the vector space of symmetric $n\times n$ matrices.
A symmetric matrix $A\in \mathbb{S}^n$ has a diagonalization:
$$A = U\Lambda U^T$$
By the property of "trace of a matrix", we have:
$$\text{tr}(A) = \text{tr}(\Lambda)$$
So I think the projection of $A$ onto the probability simplex is the following:
- $\pi(A) = U\Lambda^+U^T$, where $\Lambda^T$ is defined as:
$$\lambda_i'=\lambda_i, \text{ for } \lambda_i\geq 0 \\ \lambda_i' = -\lambda_i, \text{ for } \lambda_i< 0$$
Note: $\pi()$ is the projection onto the probability simplex
My question is how to satisfy the condition of the unity trace?