Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

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.

share|improve this answer

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 $\text{Ran}(P)$.

share|improve this answer
    
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 at 20:49
    
@Bel: try some $2 \times 2$ examples with $X = I$. –  Robert Israel Apr 17 at 5:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.