Let \(A\) be a nonsingular real \(n\times n\) matrix. Prove that there exist a unique orthogonal matrix \(Q\) and a unique positive-definite symmetric matrix \(B\) such that \[ A=QB. \]
Solution
Set \[ S=A^TA. \]
1. The matrix \(S\) is symmetric positive definite.
Symmetry is immediate: \[ S^T=(A^TA)^T=A^TA=S. \] For every nonzero vector \(v\in\RR^n\), \[ v^TSv = v^TA^TAv = \norm{Av}^2 > 0, \] because \(A\) is nonsingular and hence \(Av\neq0\).
2. There exists a positive-definite symmetric matrix \(B\) suchthat\[ B^2=S. \]
By the real spectral theorem, there is an orthogonal matrix \(U\) and positive real numbers \(\lambda_1,\ldots,\lambda_n\) such that \[ S = U \operatorname{diag}(\lambda_1,\ldots,\lambda_n) U^T. \] Define \[ B = U \operatorname{diag} (\sqrt{\lambda_1},\ldots,\sqrt{\lambda_n}) U^T. \] Then \(B\) is symmetric, all of its eigenvalues are positive, and direct multiplication gives \(B^2=S\).
3. The positive-definite symmetric square root of \(S\) is unique.
Suppose \(C\) is positive definite and symmetric with \[ C^2=S. \] Then \(C\) commutes with \(S\), because \[ CS = CC^2 = C^3 = C^2C = SC. \] Therefore \(C\) preserves every eigenspace \(E_\lambda\) of \(S\).
On \(E_\lambda\) one has \[ C^2=\lambda I. \] The restriction of \(C\) to \(E_\lambda\) is again symmetric and positive definite, so it has an orthonormal eigenbasis with positive eigenvalues \(\mu\). Each such eigenvalue satisfies \[ \mu^2=\lambda, \] hence \(\mu=\sqrt\lambda\). Thus \[ C|_{E_\lambda}=\sqrt{\lambda} I. \] Step 2 shows that \(B\) acts in exactly the same way on every eigenspace of \(S\). Since those eigenspaces span \(\RR^n\), one has \(C=B\).
4. Define\[ Q=AB^{-1}. \]Then \(Q\) is orthogonal.
The matrix \(B\) is positive definite and therefore invertible. Since \(B\) is symmetric and \(B^2=A^TA\), \[ \begin{aligned} Q^TQ &= B^{-1}A^TAB^{-1}\\ &= B^{-1}B^2B^{-1}\\ &= I. \end{aligned} \] Hence \(Q\) is orthogonal.
5. One has\[ \boxed{A=QB} \]with \(Q\) orthogonal and \(B\) positive-definite symmetric.
By step 4, \[ QB = AB^{-1}B = A. \] The required properties of \(Q\) and \(B\) were established in steps 2 and 4.
6. This factorization is unique.
Suppose also that \[ A=Q_1B_1, \] where \(Q_1\) is orthogonal and \(B_1\) is positive definite and symmetric. Then \[ A^TA = B_1Q_1^TQ_1B_1 = B_1^2. \] Thus \(B_1\) is a positive-definite symmetric square root of \(S=A^TA\). By step 3, \[ B_1=B. \] Since \(B\) is invertible, \[ Q_1 = AB^{-1} = Q. \] So both factors are unique.
7. Q.E.D.
Step 5 proves existence, and step 6 proves uniqueness.