Principal square root of a positive semidefinite matrix ($\sqrt{A}$)

If $A=Q\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)Q^T$ is real symmetric and positive semidefinite, its principal square root isIt is symmetric and positive semidefinite, and is positive definite exactly when $A$ is.

Every invertible real matrix has the polar decompositionwhere $P$ is symmetric positive definite and $R$ is orthogonal.

Every invertible complex matrix has the unique polar decompositionwhere $P$ is a positive-definite Hermitian matrix and $U$ is a unitary matrix. Writing $P=e^B$ gives $B=\frac{1}{2}\log (M{M}^{\ast })$, which is Hermitian.