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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.

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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.

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

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