Given \(A\in M_{m,n}\) with \(m\ge n\), prove that there exists a unique \(U\in M_{m,n}\) with orthonormal columns, and a unique Hermitian positive semidefinite \(H\in M_n\) such that \(A=UH\).
Solution
The uniqueness assertion for \(U\) is false as printed when \(A\) is rank-deficient. For example, if \(A=0\), then necessarily \(H=0\), but every \(m\times n\) matrix with orthonormal columns satisfies \(A=UH\). Thus \(U\) is not unique as soon as there is more than one such isometry.
The correct polar-factorization statement is the following: for every \(A\in M_{m,n}\) with \(m\ge n\), there is a unique Hermitian positive semidefinite matrix \[ H=(A^*A)^{1/2} \] and at least one \(U\in M_{m,n}\) with orthonormal columns such that \(A=UH\). Moreover, \(U\) is unique if and only if \(A\) has full column rank \(n\).
Indeed, let \[ A=W\Sigma V^* \] be a singular-value decomposition in which \(W\in M_{m,n}\) has orthonormal columns, \(V\in M_n\) is unitary, and \(\Sigma=\operatorname{diag}(\sigma_1,\ldots,\sigma_n)\) with every \(\sigma_i\ge0\). Set \[ U=WV^*,\qquad H=V\Sigma V^*. \] Then \[ U^*U=VW^*WV^*=I_n, \] so \(U\) has orthonormal columns, while \(H\) is Hermitian positive semidefinite and \[ UH=WV^*V\Sigma V^*=W\Sigma V^*=A. \] Furthermore, \[ A^*A=V\Sigma^2V^*=H^2. \] A Hermitian positive semidefinite matrix has a unique Hermitian positive semidefinite square root, so every factorization \(A=U'H'\) with \((U')^*U'=I_n\) and \(H'\ge0\) satisfies \[ (H')^2=A^*A, \] and hence \(H'=H\). Thus \(H\) is always unique.
Suppose now that \(A\) has full column rank. Then \(A^*A\) is positive definite, so \(H\) is invertible. Any factorization \(A=UH\) therefore satisfies \[ U=AH^{-1}, \] which proves uniqueness of \(U\).
Conversely, suppose \(\operatorname{rank}A<n\). Then \(\ker H=\ker A\) is nonzero. In the singular-value decomposition above, after reordering the singular values we may write \[ \Sigma=\operatorname{diag}(\sigma_1,\ldots,\sigma_r,0,\ldots,0), \qquad r<n. \] Choose a unitary diagonal matrix \[ D=\operatorname{diag}(1,\ldots,1,-1,1,\ldots,1) \] whose \(-1\) occurs in one of the zero-singular-value positions. Then \(D\Sigma=\Sigma\), so \[ U'=WDV^* \] also has orthonormal columns and \[ U'H=WDV^*V\Sigma V^*=WD\Sigma V^*=W\Sigma V^*=A. \] But \(U'\ne U\). Hence \(U\) is not unique when \(A\) fails to have full column rank.