# 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 '13 at 23:54
Fixed, thank you. –  user35359 Feb 2 '13 at 19:31

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 $\text{Ran}(P)$.
I think it should be $Px\mapsto Ax$ on $ran(P)$. Is that correct? Thanks for the answer. –  user35359 Feb 2 '13 at 19:32
@Robert Israel: In other notations the matrix P is considered as an absolute value operator for A, i.e. $P=|A|=(A^TA)^{1/2}$ where $(\cdot)^{1/2}$ is the principle square root operator. Can we apply the triangular inequality with $|\cdot|$ for matrices, i.e., $|X-Y|\leq |X|+|Y|$. –  user2987 Apr 16 '14 at 20:49
@Bel: try some $2 \times 2$ examples with $X = I$. –  Robert Israel Apr 17 '14 at 5:29