# polar decomposition of a square A when A is not full rank/invertible

I am on the proof of the polar decomposition w/o using SVD:

Let $$A$$ be a real $$n \times n$$ matrix. Show that there exists $$Q$$: orthogonal, $$P$$: positive semi-definite symmetric matrix such that $$A = QP$$.

Consider the product $$A^*A$$, it is a self-adjoint and positive semi-definite (as we are in $$\mathbb{R}$$) symmetric matrix, so it has a decomposition $$EDE^{-1}$$ where $$D$$ is diagonal and $$E$$ is orthonormal and contains eigenvectors of $$A^*A$$. Now let $$P = \sqrt{A^*A} = E\sqrt{D}E^{-1}$$ where $$\sqrt{D}$$ has diagonal elements being the squareroot of $$D$$. Notice that $$||Px|| = ||Ax||,$$ so we can expect the product $$AP^{-1}$$ to be orthogonal. Provided that $$P^{-1}$$ exists (i.e. $$P$$ is positive definite), we have $$||AP^{-1}x||=||x||$$, so we do find our desired decomposition by letting $$Q = AP^{-1}$$.

The problem is that if $$A$$ is not invertible, then $$P$$ will also not invertible, regarding the whole proof useless. How should I fix it in this case?

A standard workaround for singularities is to use analytic techniques. The below is an existence argument, not a uniqueness argument. Note Polar Decomposition is not unique when $$\det\big(A\big)=0$$.

$$A_k := A + \frac{\delta}{k}I$$ for some small enough $$\delta \gt 0$$. Then for all $$k=1,2,3,...$$
$$A_kP_k^{-1}=Q_k \in O_n\big(\mathbb R\big)$$

by compactness of $$O_n\big(\mathbb R\big)$$, the $$Q_k$$ form a bounded sequence and hence there is (at least one) convergent subsequence $$k_1\lt k_2\lt...$$ whose limit is given by $$\lim_{i\to\infty}Q_{k_i}=Q \in O_n\big(\mathbb R\big)$$ .

Similarly
$$P_{k_i} = Q_{k_i}^TA_{k_i}$$
whose limit exists because each term on the RHS does. And for avoidance of doubt:
1.) $$P_{k_i}$$ is symmetric for all $$i$$ hence the limit is too.
2.) by topological continuity of eigenvalues, since $$P_{k_i}$$ has all positive eigenvalues for all $$i$$, its limit has all real non-negative eigenvalues, i.e. it is PSD.

• I would like to take this as the answer, however is there a more simple way to show this? Nov 2, 2020 at 11:47
• the simple way is to get the result via another path entirely-- i.e. via SVD. Nov 2, 2020 at 16:05