This post proves the existence of polar decomposition.

The polar decomposition is a generalization of the polar form of complex numbers $z = re^{i\theta}$.

Theorem 1 (Existence of polar decomposition) Let $A$ be an $n \times n$ complex matrix. Then $A = SU$, where $S$ is a positive semi-definite Hermitian matrix, and $U$ is unitary. Furthermore, $S$ is uniquely determined by $A$.

Lemma 1 There exists a unique positive semi-definite square root for a positive semi-definite Hermitian matrix $B$.

Proof

Existence: By spectral theorem, there exists unitary $P$ such that $PBP^{-1}$ is diagonal. Take the non-negative square root of each diagonal entry (remember that a Hermitian matrix admits real eigenvalues only), we get a diagonal matrix $D$, such that $D^2 = PBP^{-1}$. Then $(P^{-1}DP)^2 = B$.

Uniqueness: Suppose that $C^2 = B$. BY spectral theorem, there exists an orthonormal eigenbasis $\{e_1,\cdots,e_n\}$ for the left multiplication map by $C$, i.e. $Ce_i = \lambda_i e_i$ for $i = 1, 2, \cdots ,n$. Squaring, we get $Be_i = \lambda_i^2 e_i$ for all $i$ Since $\lambda_i^2$ is uniquely determined by $B$ (its eigenvalues), $\lambda_i$ are also uniquelydetermined by $B$, asserting the uniqueness of positive semi-definite square root.

Proof of Theorem 1

Notice that $A^*A$ is a Hermitian matrix. By spectral theorem, there exists an orthonomal eigenbasis $\{e_1,\cdots,c_n\}$ for the left multiplication map by $A^*A$. Let $A^*Ae_i = \lambda_ie_i$. If $<,>$ is the complex inner product, we have

$ = = \lambda_j \delta_{ij}$

This means that $\{Ae_i\}$ is an orthogonal set. Extend this to an orthogonal basis, and make it orthonomal: $\{w_1,\cdots, w_n\}$, i.e. if $ \neq 0$, then $w_i = \frac{Ae_i}{\sqrt{}}$.

Then it is trivial to see that the map $e_i \rightarrow Ae_i$ can be decomposed to $e_i \rightarrow w_i \rightarrow Ae_i$. The first operation is a unitary one because it is a change of orthonomal bases. The second operation is a mere scaling up, so it is positive semi-definite. $\Box$

Remark

If we suppose that such a decomposition exists, then $AA^* = SUU^*S^* = S^2$, i.e. $S$ is the positive semi-definite Hermitian square root of $AA^*$, thus is unique.