# Can a unitary matrix $U$ be factorized $U=\overline{R}R^{-1}$ for a non-singular matrix $R$?

Suppose $$U$$ is a unitary matrix. Does there exist a non-singular matrix $$R$$ such that $$U=\bar{R}R^{-1}?$$ The bar denotes the complex conjugation of each entry of the matrix $$R$$.

Remarks: 1) It is obviously true for $$1\times 1$$. 2) If $$U$$ admits the decomposition $$U=VDV^{T}$$, where $$V$$ is orthonormal and $$D$$ diagonal with diagonal entries of unit modulus, then $$R:=VD^{-1/2}V^{T}$$.

The required $$R$$ exists if and only if $$U$$ is complex symmetric.
If $$U=\overline{R}R^{-1}$$, then $$UR=\overline{R}$$. Hence $$U\overline{UR}=U\overline{\overline{R}}=UR=\overline{R}$$. In turn, $$U\overline{U}=I$$ and $$\overline{U}=U^\ast$$. Hence $$U=U^T$$.
Conversely, if $$U$$ is complex symmetric, it admits a special form of singular value decomposition $$V\Sigma V^T$$, known as Takagi factorisation. Since $$U$$ is unitary, its singular values are all equal to $$1$$. Hence $$\Sigma=I$$ and $$U=VV^T$$. Now take $$R=\overline{V}$$ and we are done.