Let $A$ be a square complex matrix. Its polar decomposition is $A=UH,$ where $U$ is a unitary and $H$ is a positive-semidefinite Hermitian matrix (i.e. all its eigenvalues are real not negative).
Literature says that the matrix $H$ is always unique positive-semidefinite and given by $H=\sqrt{A^{*}A},$ where $A^{*}$ is the transpose conjugate of $A.$ This is because $L=A^{*}A$ is a positive-semidefinite Hermitian matrix, and therefore has a unique positive-semidefinite Hermitian square root.
I'm interested in the following question: Why is it always assumed the condition of positive-semidefiniteness for the Hermitian matrix $H$?
I mean, for example if $A$ is a $2\times2$ matrix then there are four square roots of the matrix $L=A^{*}A$. All four roots can be obtained from the eigenvalues $\left( \lambda_{1},\lambda_{2}\right) $ and eigenvectors $\left( v_{1},v_{2}\right) $ of $L$. If we define the matrix $$ S=% \begin{bmatrix} \pm\sqrt{\lambda_{1}} & 0\\ 0 & \pm\sqrt{\lambda_{2}}% \end{bmatrix}, $$ then the four square roots read as $H=\sqrt{A^{*}A}=VSV^{-1},$ where $V$ is the matrix whose columns are the eigenvectors $\left( v_{1},v_{2}\right)$, and $S$ is any of the sign combinations of above matrix.
Here an example with real numbers: The matrix $$ A = \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}, $$ can be decomposed as $$ A = U_{1}H_{1}=% \begin{bmatrix} -0.5145 & 0.8575\\ 0.8575 & 0.5145 \end{bmatrix}% \begin{bmatrix} 2.0580 & 2.4010\\ 2.4010 & 3.7730 \end{bmatrix} , $$ $$ A =U_{2}H_{2}=% \begin{bmatrix} -0.9806 & 0.1961\\ -0.1961 & -0.9806 \end{bmatrix}% \begin{bmatrix} -1.5689 & -2.7456\\ -2.7456 & -3.5301 \end{bmatrix} . $$ Eigenvalues of $H_{1}$ are $\left( 0.366,5.465\right) $. Eigenvalues of $H_{2}$ are $\left( 0.366,-5.465\right) $. The first option $H_{1}$ is the expected positive-semidefinite Hermitian matrix (both eigenvalues positive). However, it is a fact that the second option represents a different but valid polar decomposition of the matrix $A$. Why it is ignored in the literature?