Question

Can someone explain the polar decomposition of a square matrix A. It is defined as A = Q*S where Q is orthogonal and S is symmetric positive definite. Why can't you reverse Q*S and still obtain any matrix A?

Answer 1

Hi,
I would suggest you to look at Gilbert Strang's book: Linear Algebra, Geodesy and GPS, chapter 7. I will just outline what I have read from that book. Basically, the SVD of a matrix A can be written as\[A=U\Sigma V ^{T}\]
Any complex number (a real number is just a special case of a complex number) can be written as a magnitude 'r' that multiplies a phase \[e ^{j\theta}\]
This concept is extended to matrices in polar decomposition (and hence the name). That is done by a "trick" where V^T V (which is identity!)is inserted before the diagonal matrix. Hence it looks like this:
\[A=U(V^TV)\Sigma V^T\]
Re-writing this, we obtain:
\[A=(UV^T)(V\Sigma V^T)\]
The first term UV^T is a product of two orthogonal matrices and hence it is orthogonal too. It's called Q and the adjacent product is called H. So finally, we obtain
\[A = QH\]

For the "reverse" product, instead of V^T V, use U^T U

Like I said, perhaps, the best thing would be to read about it in a book :-) Good luck!