# polar decomposition for non square matrices

I know that for a square complex matrix $A\in\mathbb{C}^{n\times n}$ there exist matrices $O$ and $P$ in $\mathbb{C}^{n\times n}$ with $O$ unitary and $P$ hermitian positive semidefinite such that $A=OP$. The proof starts with applying the spectral theorem to the matrix $AA^*$.

Now I read that similarly for $A\in \mathbb{C}^{n\times m}$, $A=OP$ with $O\in\mathbb{C}^{n\times m}$, $P\in\mathbb{C}^{m\times m}$, $P$ hermitian positive semidefinite and $O$ isometric, if $m<n$. How can this be proved?

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In English it's called a "square matrix", not "quadratic". – Robert Israel Jan 30 at 23:54
Fixed, thank you. – user35359 Feb 2 at 19:31

## 2 Answers

The usual polar decomposition for complex matrices is $A = O P$ where $O$ is a partial isometry and $P$ is (hermitian) positive semidefinite, not necessarily symmetric. Namely $P$ is the positive semidefinite square root of $A^* A$. Since $\|P x\|^2 = x^* A^* A x = \|A x\|^2$ for any $x$, the map $Ax \mapsto Px$ on $\text{Ran}(A)$ is an isometry (you have to show that this is well-defined and linear, but that's not hard). To complete the definition of the partial isometry $O$, you take any partial isometry from the orthogonal complement of $\text{Ran}(A)$ to the orthogonal complement of $\text{Ran}(P)$.

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I think it should be $Px\mapsto Ax$ on $ran(P)$. Is that correct? Thanks for the answer. – user35359 Feb 2 at 19:32

Do you know singular value decomposition? http://en.wikipedia.org/wiki/Singular_value_decomposition

Polar decomposition is trivial consequence of singular value decomposition, which is defined for any matrix.

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