The Exercise

A square matrix $A$ has a polar decomposition: a unitary $W$ and two positive matrices $J$ and $K$ with $$A = WJ = KW, \qquad J = \sqrt{A^\dagger A}, \qquad K = \sqrt{A A^\dagger}.$$(1) The exercise asks what this decomposition is for a positive matrix $P$, for a unitary matrix $U$ and for a Hermitian matrix $H$.

The Square Root and the Sign

The positive square root $\sqrt{B}$ of a positive matrix $B$ is the positive matrix whose square is $B$. If $B = \sum_i \mu_i |i\rangle\langle i|$ in an orthonormal eigenbasis, with eigenvalues $\mu_i \ge 0$, then $$\sqrt{B} = \sum_i \sqrt{\mu_i}\,|i\rangle\langle i|.$$(2)

The sign $\mathrm{sgn}(H)$ of a Hermitian matrix $H$ is the matrix with the eigenvectors of $H$ whose eigenvalues are $+1$ where those of $H$ are positive or zero and $-1$ where they are negative.^✦ In an orthonormal eigenbasis, where $H = \sum_i \lambda_i |i\rangle\langle i|$, $$\mathrm{sgn}(H) = \sum_i s(\lambda_i)\,|i\rangle\langle i|, \qquad s(\lambda) = \begin{cases} +1, & \lambda \ge 0, \\ -1, & \lambda < 0. \end{cases}$$(3)^✦

The Claim

What Is Checked

The claim is stated over the positive square root (2) and over the sign of a Hermitian matrix (3). Of the sign, the library proves that it is a unitary, (10).