The Gramian of a complex matrix $A$ is $A^\ast A$. The equation $$ \langle \boldsymbol{x}, A^\ast A\boldsymbol{x}\rangle = \langle A\boldsymbol{x}, A\boldsymbol{x}\rangle=\lVert A\boldsymbol{x}\rVert^2\geq 0 $$ tells us that Gramians are positive semidefinite.
The converse is also true. The spectral theorem implies that every positive semidefinite matrix $H$ factors as $$ H=UDU^\ast $$ where $U$ is unitary and $D$ is diagonal with nonnegative entries. Defining $A=\sqrt{D}U^\ast$ then gives $H=A^\ast A$.
For example, consider the factorization $$ \overset{H}{\left[\begin{array}{rrr} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 2 & -2 & 8 \end{array}\right]} =\overset{U}{ \left[\begin{array}{rrr} \frac{1}{\sqrt{18}} & \frac{1}{\sqrt{2}} & \frac{2}{3} \\ -\frac{1}{\sqrt{18}} & \frac{1}{\sqrt{2}} & -\frac{2}{3} \\ \frac{4}{\sqrt{18}} & 0 & -\frac{1}{3} \end{array}\right] } \overset{D}{\left[\begin{array}{rrr} 9 & & \\ & 1 & \\ & & 0 \end{array}\right]} \overset{U^\ast}{\left[\begin{array}{rrr} \frac{1}{\sqrt{18}} & -\frac{1}{\sqrt{18}} & \frac{4}{\sqrt{18}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{2}{3} & -\frac{2}{3} & -\frac{1}{3} \end{array}\right]} $$ It follows that $H=A^\ast A$ where $A$ is given by $$ \overset{A}{ \left[ \begin{array}{rrr} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & \frac{4}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 0 \end{array} \right] } = \overset{\sqrt{D}}{\left[\begin{array}{rrr} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right]} \overset{U^\ast}{\left[\begin{array}{rrr} \frac{1}{\sqrt{18}} & -\frac{1}{\sqrt{18}} & \frac{4}{\sqrt{18}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{2}{3} & -\frac{2}{3} & -\frac{1}{3} \end{array}\right]} $$ It's worth observing that our original $H$ had entries in $\Bbb Z$ but this $A$ does not have integer entries.
However, in this example it is possible to write $H=A^\ast A$ where $A$ does have integer entries $$ \overset{H}{\left[\begin{array}{rrr} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 2 & -2 & 8 \end{array}\right]}=\overset{A^\ast}{\left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \\ 2 & -2 \end{array}\right]}\overset{A}{\left[\begin{array}{rrr} 1 & 0 & 2 \\ 0 & 1 & -2 \end{array}\right]} $$ With this context in mind, here are my questions.
- Suppose $H$ is $n\times n$ positive semidefinite with entries in a subring $R$ of $\Bbb C$. Can we write $H=A^\ast A$ where $A$ is $m\times n$ with entries in $R$? If so, is there a way to construct $A$?
- More precisely, let $R$ be a subring of $\Bbb C$ and let $\mathcal{P}_n(R)$ be the collection of $n\times n$ positive semidefinite matrices with entries in $R$. Is there an equality $$ \mathcal{P}_n(R)=\bigcup_{i=1}^k\{A^\ast A\mid A\in\mathcal{M}_{i\times n}(R)\} $$ for some $n\leq k\leq\infty$?
- If this factorization is not possible, are there conditions on $R$ where the factorization is possible?