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I know that all positive semidefinite matrices has a rank-1 decomposition. (Equivalently, all quadratic nonnegative polynomial is sum of squares of linear function.)

$$A = \sum_{i=1}^r x_i x_i^T = X X^T$$.

I am wondering if there is an entrywise nonnegative X satisfying $A=X X^T$ if $A$ is a entrywise nonnegative positive semidefinite matrix. The converse is obviously true..

I think an important tool is that solution set of $A=X X^T$ is closed with the operation $X \rightarrow XP$ where the $P$ is an orthonormal matrix. For example, $$a^T \begin{pmatrix} 2 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 2 \\ \end{pmatrix}a$$

can be represented as $(a_1+a_2+a_3)^2 + (a_1 - a_3)^2$, but with multiplying by

$$\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \\ \end{pmatrix}, $$

we get an 'positive representation', $\frac{1}{2}(2a_1+a_2)^2 + \frac{1}{2}(a_2 + 2a_3)^2$.

Seho Yim
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    J.E. Maxfield, H. Minc, On the equation X'X = A, Proc. EdinburghMath. Soc. 13 (1962) 125–129.

    I found this is false. There is an counterexample in this paper with $n = 5$. Entrywise nonnegative positive semidefinite matrix is called doubly nonnegative(DNN) and a matrix that can be represented as $XX^T$ with nonnegative X is called completely positive matrix(CP).

     – Seho Yim May 23 '22 at 12:33

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