# Solving a quadratic equation for a hermitian matrix

I am looking for a procedure to find solution(s) for a square matrix equation

$H^T H = S$

where $H = H^\dagger$ is a hermitian ($n\times n$) matrix and $S$ is a given symmetric complex matrix. Due to hermiticity of $H$, $S$ should satisfy $n$ conditions $(\operatorname{im}(\operatorname{tr}(S^i))=0,\quad i = 1,...,n)$.

I am interested in simple solutions for small $n=3$ matrices. For $n=2$, this can be solved by an explicit parametrization of $H$ which leads to a quadratic equation giving four solutions. Since the problem is similar to taking a square root of a matrix, presumably there are $2^n$ solutions for this problem, too.

note: This question is perhaps similar to this one. Here, the equation is simpler but applies to complex, not real matrices.

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This is a very partial answer. Assume that $S$ is invertible ; then $S$ and $I$ are congruent and there is $K$ s.t. $K^TK=S$. Here a solution $H$ must be a hermitian matrix ; then $HH^T=\overline{S}$. Since $H$ is invertible, $HH^T$ and $H^TH$ are orthogonally similar. Consequently $S$ and $\overline{S}$ are orthogonally similar. In particular, $S$ must have the spectrum of a real matrix.

For the construction of $K$, have a look at page 130 of: Turnbull, Aitken. An introduction to the theory of canonical matrices.

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Any Hermitian matrix is normal, i.e., we may write $H = UDU^*$ where $D$ is a real diagonal matrix, and $U$ is unitary. Thus, $S = (U^* D U)(U^*DU) = U^* D^2 U$ so we may diagonalize $S$ with the unitary matrix $U$. Hence, all eigenvalues of $S$ must by necessity be non-negative and real. Hence, $S$ must be positive semi-definite for it to be a solution. Furthermore, $S$ can be diagonalized with the unitary matrix $U$, so $S$ is normal.

Hence, we can find solutions iff $S^*S=SS^*$ and all eigenvalues are non-negative and real.

From here, we can easily see ways to find all solutions with the following method:

1) Diagonalize $S$, $S = U^*D'U$, where $D'$ have non-negative entries.

2) We can now choose $H = U D U^*$ where $D$ is any root of $D'$. There are be $2^n$ choices here.

(Note, $D^2=D'$).

EDIT: Ah, I misread it as $H^*H=S$. Maybe the above method can be used in some special case?

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