Given \(A \in M_{m,n}\) with \(m \geq n\), 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 \(A = UH\). (State in detail any auxiliary results used without proof.)
Solution
1. The uniqueness assertion for \(U\) is false as printed. If \(A=0\), then \(H=0\), while every \(m\times n\) matrix with orthonormal columns satisfies \(A=UH\).
Since \(m\ge n\), such matrices exist and are not unique.
2. The correct statement is that \(H=(A^*A)^{1/2}\) is uniquely determined, there exists an isometry \(U:\mathbb C^n\to\mathbb C^m\) with \(A=UH\), and \(U\) is unique exactly when \(A\) has full column rank.
We use the standard fact that a Hermitian positive semidefinite matrix has a unique Hermitian positive semidefinite square root.
3. Put \(H=(A^*A)^{1/2}\) and \(K=\ker H\). Then \(K=\ker A\) and \(\operatorname{im}H=K^\perp\).
For every \(x\), \[ \|Ax\|^2=\langle A^*Ax,x\rangle=\langle H^2x,x\rangle=\|Hx\|^2, \] so \(Ax=0\) iff \(Hx=0\). Since \(H\) is Hermitian, \(\operatorname{im}H=(\ker H)^\perp\).
4. Define \(U_0:K^\perp\to\mathbb C^m\) by \(U_0(Hx)=Ax\). This is well-defined and isometric.
If \(Hx=Hy\), then \(x-y\in K=\ker A\), so \(Ax=Ay\). Also \(\|U_0(Hx)\|=\|Ax\|=\|Hx\|\).
5. Extend \(U_0\) to an isometry \(U:\mathbb C^n\to\mathbb C^m\). Then \(U\) has orthonormal columns and \(A=UH\).
Extend an orthonormal basis of \(K^\perp\) to one of \(\mathbb C^n\), and extend its \(U_0\)-image to an orthonormal \(n\)-frame in \(\mathbb C^m\); this is possible because \(m\ge n\). Then \(UH x=U_0(Hx)=Ax\).
6. The factor \(H\) is unique.
If \(A=U_1H_1\) with \(U_1^*U_1=I\) and \(H_1\ge0\), then \[ A^*A=H_1U_1^*U_1H_1=H_1^2. \] Hence \(H_1\) is the positive semidefinite square root of \(A^*A\), so \(H_1=H\).
7. If \(A\) has full column rank, then \(H\) is invertible and \(U=AH^{-1}\), so \(U\) is unique.
Full column rank is equivalent to \(\ker H=0\) by 3.
8. If \(A\) is rank-deficient, then \(U\) is not unique.
Now \(K\ne0\). The map \(U\) is fixed on \(K^\perp\) but may be changed on \(K\) by choosing a different orthonormal extension; since \(H\) vanishes on \(K\), all such extensions still satisfy \(UH=A\).