Every Positive Semidefinite Matrix Has Exactly One Positive Semidefinite Square Root

In the setting of Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices, let $n\in\mathbb{N}$ and let $A\in\mathcal{S}(n)$, the set of symmetric matrices, be positive semidefinite, with positive definite matrices, the set $\mathrm{O}(n)$ of orthogonal matrices, the matrix product, the transpose, the diagonal matrix $\operatorname{diag}(\lambda)$ of $\lambda\in\mathbb{R}^{n}$, and square roots of nonnegative real numbers.

¶ There is exactly one $B\in\mathcal{S}(n)$ that is positive semidefinite and satisfies $BB=A$.

¶ If $Q\in\mathrm{O}(n)$ and $\lambda\in\mathbb{R}^{n}$ satisfy $A=(Q\operatorname{diag}(\lambda))Q^{\top}$, then $\lambda_{k}\ge0$ for every $k\in[n]$, and the matrix $B$ of clause square-root equals $(Q\operatorname{diag}(\sqrt{\lambda}))Q^{\top}$, where $\sqrt{\lambda}=(\sqrt{\lambda_{1}},\dots,\sqrt{\lambda_{n}})\in\mathbb{R}^{n}$.

¶ The matrix $B$ of clause square-root satisfies $AB=BA$.

¶ If $A$ is positive definite, then so is the matrix $B$ of clause square-root.