Question

Question 11: Prove that any \(n \times n\) matrix A admits a polar decomposition of the form \(A = PQ\), where P is a \(n \times n\) positive semidefinite matrix with the same rank as A and where Q is an \(n \times n\) orthogonal matrix. (Hint: Use a singular value decomposition, \(A = U\sum {V^T}\), and observe that \(A = \left( {U\sum {U^T}} \right)\left( {U{V^T}} \right)\).) This decomposition is used, for instance, in mechanical engineering to model the deformation of a material. The matrix P describe the stretching or compression of the material (in the directions of the eigenvectors of P), and Q describes the rotation of the material in space.

Short answer

It is proved that any \(n \times n\) matrix A admits a polar decomposition of the for \(A = PQ\).

Step-by-step solution

The Singular Value Decomposition

Consider \(m \times n\) matrix with rank \(r\) as \(A\). There is a \(m \times n\) matrix \(\sum \) as seen in (3) wherein the diagonal entriesin \(D\) are the first \(r\) singular valuesof A, \({\sigma _1} \ge \cdots \ge {\sigma _r} > 0,\) and there arises an \(m \times m\) orthogonal matrix \(U\) and an \(n \times n\) orthogonal matrix \(V\) such that \(A = U\sum {V^T}\).

Show that any \(n \times n\) matrix A admits a polar decomposition of the form \(A = PQ\)

Consider that \(A = U\sum {V^T}\) as a singular value decomposition of A. \({U^T}U = I\) and \(A = U\sum {U^T}U{V^T} = PQ\) wherein \(P = U\sum {U^T} = U\sum {U^{ - 1}}\) and \(Q = U{V^T}\) because \(U\) is orthogonal. \(P\) contains nonnegative eigenvalues since \(P\) are the same as \(\sum \) that is diagonal with nonnegative diagonal entries because \(\sum \) are symmetric. Therefore, \(P\) is positive semidefinite. \(Q\) is orthogonal since \(Q\) is the product of orthogonal matrices.

Hence, it is proved that any \(n \times n\) matrix A admits a polar decomposition of the form \(A = PQ\).