# Polar decomposition for real rectangular matrices

I'm reading a book where polar decomposition is given for linear operator. This implies a decomposition for real square matrix into $$SP$$ where $$S$$ is orthogonal matrix and $$P$$ is positive semidefinite symmetric matrix.

As an exercise, I tried to prove a version of polar decomposition for real rectangular matrix:

Theorem. Let $$A$$ be $$n$$ by $$p$$ real matrix with $$n\ge p$$. $$A=SP$$ where $$S$$ is $$n$$ by $$p$$ matrix with orthonormal columns and $$P$$ is $$p$$ by $$p$$ positive semi-definite symmetric matrix.

Since I made this up myself so I'm not sure this is actually true. But the attempt is this:

Attempt. Let $$P=\sqrt{A^TA}$$ where $$P$$ is positive semidefinite symmetric matrix as $$A^TA$$ is positive operator (positive semidefinite symmetric matrix). $$\|Pv\|=\|Av\|, \forall v$$ as one can easily check. This fact implies that null space of $$P$$ and $$A$$ is identical. Using this fact, we may define $$S'$$ from range of $$P$$ to range of $$A$$ by $$S'(Pv)=Av$$ such that $$S'$$ is well defined. One can check $$S'$$ is linear, bijective, and preserves norm.

Now number of columns in $$S'$$ is equal or less than $$p$$. We consider extending $$S'$$ to $$p$$ columns by appending columns $$c_i$$. Let $$\mathbb{R}^p=\text{range}S' \oplus \text{range}S'^\perp$$. We may choose $$c_i$$ such that they are orthonormal basis of $$\text{range}S'^\perp$$. Call the extended matrix $$S$$. We may verify easily that $$\|Sv\|=\|v\|, \forall v \in \mathbb{R}^p$$, using Pythagoras theorem as well as $$S'$$ preserves norm and that $$c_i$$'s are orthonormal. Now $$v^T v=v^TS^TSv, \forall v \iff S^TS=I \iff$$ all columns of $$S$$ are orthonormal.

• Which book????? Jul 12, 2023 at 14:35

## 1 Answer

Yes, the proof is correct. I think that this version should be given in the books as it can immediately lead to what is known as the Singular Value Decomposition for a rectangular matrix, which has many applications in applied mathematics and engineering. Furthermore, one can show that the polar decomposition of a rectangular matrix $$A$$ is unique if and only if $$A$$ is of full rank.

• There is no need to capitalise the title. It sounds like the OP is unhinged and screaming into the void Jul 12, 2023 at 14:36