Let $R$ be a subring of $\Bbb C$ closed under complex conjugation and let $P$ be an $n\times n$ positive semidefinite matrix with entries in $R$. I'm curious if it is always possible to factor $P$ as $P=A^\ast A$ where $A$ is $m\times n$ with entries in $R$. 


If $P$ is $1\times 1$ and $R=\Bbb Z$, then Lagrange's four-square theorem says $P=A^\ast A$ where $A$ is $4\times 1$. Is there anything known beyond this observation?



Do matrices of the form $A^\ast A$ where $A$ has entries in $R\subset\Bbb C$ account for all positive semidefinite matrices with entries in $R$?
$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}$I did some searching and thinking, and this is what I have come up with. Disclaimer: I couldn't actually find full copies of the papers I'm going to cite by Mordell and Ko, so this is based on reading summaries of them in other papers.




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The case $R = \ZZ$ was considered by Mordell in "A new Waring’s problem with squares of linear forms", *Quarterly Journal of Mathematics* **1** (1930), 276–288 and "On the representation of a binary quadratic form as a sum of squares of linear forms", *Mathematische Zeitschrift* **35** (1932), 1–15. To see the connection, note that, if the $B = A^T A$ and the columns of $A$ are $a\_1$, $a\_2$, ..., $a\_m$, then $x^T B x = \sum \langle a\_i, x\_i \rangle^2$. So Mordell thought about this problem as expressing a given quadratic form as a sum of squares of linear forms. Mordell showed that this is always possible for quadratic forms in $2$ variables, and that $5$ squares suffice. Without the bound $5$, this appeared on the 1995 IMO shortlist. ([solution](https://prase.cz/kalva/short/soln/sh95a2.html)) 


Chao Ko showed in "On the representation of a quadratic form as a sum of squares of linear forms" *The Quarterly Journal of Mathematics*, Volume os-8, Issue 1, (1937), Pages 81–98 that, for $3$, $4$ or $5$ variables, $6$, $7$ or $8$ squares suffice.


However, it was already noted by Mordell that the $E\_6$ Cartan matrix (below) is not a sum of any number of squares!
$$E\_6 = \left[ \begin{smallmatrix} 
2&-1&0&0&0 &0\\
-1&2&-1&0&0 &0 \\
0&-1&2&-1&0 &-1 \\
0&0&-1&2&-1 &0 \\
0&0&0&-1&2 &0 \\
0&0&-1&0&0 &2 \\
\end{smallmatrix} \right].$$
To see that this can be proved by a finite search, note that $\mathrm{Tr}(B) = \sum A\_{ij}^2$, so in this case we would have $\sum A\_{ij}^2 = 12$ and so the matrix $A$ can have at most $12$ nonzero entries. If you think a bit harder, it isn't too hard to get this down to a number of cases that can be done by a pencil and paper search; Hint: First show that, after reordering columns and switching their signs, we can assume that the first five rows of $A$ are $(e\_2-e\_1, e\_3-e\_2, e\_4-e\_3, e\_5-e\_4, e\_6-e\_5)$.


Since then, the main people thinking about this seem to be Myung-Hwan Kim and Byeong-Kweon Oh. They've produced a series of papers on this sort of question over the last 20 years. Their paper "[Bounds for quadratic Waring's problem](http://www.math.snu.ac.kr/~mhkim/thesis/thesis_29.pdf)", *Acta Arith.* **104** (2002), no. 2, 155–164 is pretty readable and seemed like a good starting point.


I couldn't find many papers on the Hermitian variant, but I did find "[On a Waring's problem for integral quadratic and hermitian forms](https://arxiv.org/abs/1702.08854)" which also has many good references in it.


Based on this, I would suspect that there is no simple answer for $\ZZ$, and thus no simple answer for subrings of $\CC$ more complicated than $\ZZ$. What that leaves are the subfields of $\CC$, so I'll turn to those next.




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If $K$ is a field of characteristic not $2$, and $B$ is a symmetric matrix with entries in $K$, then there is an invertible matrix $S$ such that $SBS^T$ is diagonal (see, for example, [here](https://www.encyclopediaofmath.org/index.php/Lagrange_method).) I claim that a diagonal matrix $D$ is a sum of squares if and only if each diagonal entry of $D$ is a sum of squares. If each entry of $D$ is a sum of squares, the claim is clear; conversely, if $D = A^T A$ then $D\_{jj} = \sum\_i A\_{ij}^2$. So we are reduced to the question of which elements of $K$ are sums of squares.


Similarly, let $L$ be a field of characteristic not $2$ and let $\sigma : L \to L$ be an involution. For a matrix $A$, define $A^{\dagger}$ to be the matrix with $A^{\dagger}\_{ij}= \sigma(A\_{ji})$. A very similar argument shows that, if $B$ is a matrix with $B^{\dagger} = B$, then there is an invertible matrix $S$ such that $SBS^T$ is diagonal. So, as above, we are reduced to the case of diagonal matrices. Let $K$ be the fixed field of $L$, so a diagonal matrix obeys $D^{\dagger} = D$ if and only if its entries are in $K$. An argument as above shows that a diagonal matrix is of the form $A^{\dagger} A$ if and only if its diagonal entries are sums of norms from $L$ to $K$. The extension $L/K$ is quadratic; if $L = K(\sqrt{-1})$ then we can simply say that a diagonal matrix is of the form $A^{\dagger} A$ if and only if each entry is a sum of squares.


Now, let $L$ be a subfield of $\CC$ and let $\sigma$ be complex conjugation, so $K = L \cap \RR$ and $L = K(\sqrt{-1})$. So the above argument reduces the question of which Hermitian matrices are of the form $A^{\dagger} A$ to the question of which elements of $K$ are sums of squares. 


In a number field $K$, an element is a sum of squares if and only if it is nonnegative for every embedding $K \to \RR$. I think I could also say something for finitely generated subfields of $\CC$, but this seems like it might be pretty far from the OP's interest.


