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Polar decomposition

July 30, 2009

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

<Ae_i,Ae_j> = <e_i, A^*Ae_j> = \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 <Ae_i,Ae_i> \neq 0, then w_i = \frac{Ae_i}{\sqrt{<Ae_i,Ae_i>}}.

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.

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