8
$\begingroup$

Let $A, B$ be $n$-square (Hermitian) positive definite matrices. Let $AB=U|AB|$ be the polar decomposition of $AB$. So $U$ is unitary (called the unitary factor of $AB$). What is the optimal constant $c$ such that $\|I-U\|\le c$, where the norm is the usual spectral norm?

I want to have some understanding on the behaviour of the unitary factor for certain classes of matrices (e.g. matrices with real eigenvalues). Perhaps this is well known, any pointer to the existing papers is welcome.

$\endgroup$
  • 2
    $\begingroup$ Numerical experiment suggests that the real part of the eigenvalues of $U$ may be negative. That is, $c\ge \sqrt{2}$. $\endgroup$ – M. Lin Nov 23 '14 at 3:29
5
$\begingroup$

Here I consider only the case when the matrices are real. Let $C_n$ be the best bound $c$ in dimension $n$. I did not write the entire proof, but it is clear (for me) that $C_2=\sqrt{2}$. Morover $||I-V||=||I-U||$ if $U$ is the orthogonal matrix associated to $AB$ and $V$ to $diag(A,1)diag(B,1)$. Then $(C_n)_n$ is non-decreasing. Numerical experiments show that $C_3>1.549,C_4>1.564,C_5>1.678,C_6>1.721$.

The interesting question is: Is $\lim_n C_n=2$ true ?

$\endgroup$
  • 1
    $\begingroup$ For some finite $n$, can $U$ have an eigenvalue $-1$? $\endgroup$ – M. Lin Nov 26 '14 at 0:48
  • 1
    $\begingroup$ @ Lin , I don't think so ; yet even if $-1$ is never an eigenvalue, that does not imply that $C_n<2$ because $C_n$ may be a LimitSup. $\endgroup$ – loup blanc Nov 26 '14 at 11:13
2
$\begingroup$

I think that the part $(a)$ of proposition $2.4$ of this paper shows that for $n$ sufficiently large one can construct a $n \times n$ unitary matrix $U=-I_{2}\oplus U'$, which can be decomposed as the product of three positive matrices. For such $U$ we have $\parallel U-I\parallel=2$. In fact one can take a unitary matrix $U'$ with $Det \;U'=1$ such that $0$ lies in the interior of the convex hull of the eigenvalues of $U$. This convex hull is equal to $W(U)$, the numerical range of $U$.

In fact if $U=ABC$ for positive $A,B,C$ then $UC^{-1}=AB$ so $C^{-1}=|AB|$ is the positive factor of $U $ in its polar decomposition. Of course such $U$ has $-1$ as an eigenvalue.

An explicit example is the following

$$\begin{pmatrix}-1& 0&0&0\\0&-1&0&0\\0&0&\sqrt{3}/2&-1/2\\0&0& 1/2&\sqrt{3}/2 \end{pmatrix} $$ According to the aboved linked paper this matrix is a product of three positive matrices. This matrix, with distance $2$ from the identity matrix $I_{4}$, is the unitary factor of polar decomposition of $AB$ for two positive matrices $A,B$. So according to the notation $C_{n}$ in the answer by Loup Blanc, we have $C_{n}=2, \; \forall n\geq 4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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