---
tag: M76S68K6YD
kind: proof
name: Proof of the polar decomposition from the singular value decomposition
msc: [15A23]
proves: 0J2C7BDJQA
uses: [000NPTMX6J, 0D6RMJ10M9, 0EX045E2BP, 0EKZCX4K3P, 0880F9EQCP, 0AM050KCAW, 0N5GP3YT8T, 0753Y2X5K6, 09GK5PS824, 04CPJ6AF00, FBSAPTP1PR]
sources:
  - url: https://en.wikipedia.org/wiki/Polar_decomposition
    locator: Relation to the SVD
---

## Proof

Here $^*$ is the [[conjugate transpose|0EX045E2BP]], and $(XY)^* = Y^* X^*$ by the [[transpose of a product|09GK5PS824]] with conjugated entries.

1. If $X$ and $Y$ are [[unitary|0D6RMJ10M9]] $n \times n$ matrices, then $(XY)^* XY = Y^* X^* X Y = I$. So $XY$ is unitary.
2. If $X$ is unitary, then $X X^* = I$ by the [[one-sided inverse theorem|0753Y2X5K6]]. So $X^*$ is unitary too.
3. 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 $\sigma_1, \dots, \sigma_n \ge 0$.
4. Let $U = W V^*$ and $P = V \Sigma V^*$. By steps 1 and 2, $U$ is unitary.
5. Since $\Sigma$ is real and diagonal, $\Sigma^* = \Sigma$. So $P^* = V \Sigma^* V^* = P$, and $P$ is [[Hermitian|0EKZCX4K3P]].
6. 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]].
7. Since $V^* V = I$, we get $UP = W V^* V \Sigma V^* = W \Sigma V^* = A$. This proves existence.
8. For uniqueness of $P$, let $A = UP$ with $U$ unitary and $P$ positive semidefinite Hermitian. Then $A^* A = P^* U^* U P = P^2$.
9. By [[the Gram matrix result|0AM050KCAW]], $A^* A$ is positive semidefinite. By step 8, $P$ is a positive semidefinite matrix whose square is $A^* A$.
10. By the definition of the [[square root|0N5GP3YT8T]], there is exactly one such matrix. So $P = (A^* A)^{1/2}$.
11. Now let $A$ be invertible, and let $A = UP$ as in step 8. Since the [[determinant is multiplicative|04CPJ6AF00]], $\det U \det P = \det A \ne 0$.
12. So $\det P \ne 0$, and $P$ is [[invertible|FBSAPTP1PR]]. Then $U = A P^{-1}$.
13. By step 10, $P$ is determined by $A$. So $U = A P^{-1}$ is determined by $A$ as well.

## Idea

The 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.