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  "body": "\n## Proof\n\nHere $^*$ is the [[conjugate transpose|0EX045E2BP]], and $(XY)^* = Y^* X^*$ by the [[transpose of a product|09GK5PS824]] with conjugated entries.\n\n1. If $X$ and $Y$ are [[unitary|0D6RMJ10M9]] $n \\times n$ matrices, then $(XY)^* XY = Y^* X^* X Y = I$. So $XY$ is unitary.\n2. If $X$ is unitary, then $X X^* = I$ by the [[one-sided inverse theorem|0753Y2X5K6]]. So $X^*$ is unitary too.\n3. By the [[singular value decomposition|000NPTMX6J]], $A = W \\Sigma V^*$ with $W$ and $V$ unitary $n \\times n$ matrices. Here $\\Sigma$ is diagonal with real entries $\\sigma_1, \\dots, \\sigma_n \\ge 0$.\n4. Let $U = W V^*$ and $P = V \\Sigma V^*$. By steps 1 and 2, $U$ is unitary.\n5. Since $\\Sigma$ is real and diagonal, $\\Sigma^* = \\Sigma$. So $P^* = V \\Sigma^* V^* = P$, and $P$ is [[Hermitian|0EKZCX4K3P]].\n6. Let $x \\in \\C^n$ and $y = V^* x$. Then $x^* P x = y^* \\Sigma y = \\sum_i \\sigma_i \\abs{y_i}^2 \\ge 0$. So $P$ is [[positive semidefinite|0880F9EQCP]].\n7. Since $V^* V = I$, we get $UP = W V^* V \\Sigma V^* = W \\Sigma V^* = A$. This proves existence.\n8. For uniqueness of $P$, let $A = UP$ with $U$ unitary and $P$ positive semidefinite Hermitian. Then $A^* A = P^* U^* U P = P^2$.\n9. By [[the Gram matrix result|0AM050KCAW]], $A^* A$ is positive semidefinite. By step 8, $P$ is a positive semidefinite matrix whose square is $A^* A$.\n10. By the definition of the [[square root|0N5GP3YT8T]], there is exactly one such matrix. So $P = (A^* A)^{1/2}$.\n11. Now let $A$ be invertible, and let $A = UP$ as in step 8. Since the [[determinant is multiplicative|04CPJ6AF00]], $\\det U \\det P = \\det A \\ne 0$.\n12. So $\\det P \\ne 0$, and $P$ is [[invertible|FBSAPTP1PR]]. Then $U = A P^{-1}$.\n13. By step 10, $P$ is determined by $A$. So $U = A P^{-1}$ is determined by $A$ as well.\n\n## Idea\n\nThe singular value decomposition $W \\Sigma V^*$ already contains a rotation and a stretch. Regrouping it as $(W V^*)(V \\Sigma V^*)$ gives a unitary factor times a positive semidefinite factor. Any such factorisation has $P^2 = A^* A$, and square roots of positive semidefinite matrices are unique.\n",
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