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Suppose I'm free to chose a basis $P$ for my input matrix $A$ before applying polar decomposition, how do I find a basis $P$ such that the rotation factor of polar decomposition becomes as close to identity as possible?

We know that $1/\sqrt{2}$ of random 2D matrices admit a basis where $U$ is the identity matrix, I'm curious about the distribution of angles in the remaining 30%

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Yaroslav Bulatov
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  • Note that having real eigenvalues doesn't allow a change of basis in which $U$ is the identity matrix; for the standard polar decomposition, we would also require the eignvalues of the matrix to be non-negative. – Ben Grossmann Jul 31 '23 at 13:47
  • @BenGrossmann doesn't real eigenvalue allow similarity transformation into matrix that is positive semi-definite? So the rotation factor of that matrix would be the identity – Yaroslav Bulatov Jul 31 '23 at 14:50
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    No: similarity transformations preserve eigenvalues, and positive semidefinite matrices have non-negative eigenvalues. A matrix with a negative, real eigenvalue cannot be similar to a positive semidefinite matrix. – Ben Grossmann Jul 31 '23 at 14:52
  • You might be able to use this ideas of angles in this context, however, if you modify the polar decomposition such that $P$ is required only to be symmetric. I suspect (based completely on conjecture) that the relaxation on $P$ can be accounted for by a restriction on $U$ that its eigenvalues should have non-negative real part. – Ben Grossmann Jul 31 '23 at 14:57
  • @BenGrossmann hm, I was confused by this answer where you show that matrix has real eigenvalues by showing it's similar to positive definite. I guess such existence is sufficient condition for real eigenvalues but not necessary? – Yaroslav Bulatov Jul 31 '23 at 15:39
  • Exactly right. As it so happened in that case, the eigenvalues of the matrices $D_i$ were necessarily positive – Ben Grossmann Jul 31 '23 at 19:19
  • Another trick that might be noteworthy here: if $A$ has real eigenvalues, then there exists a constant $k$ (for example, $k = |A|$ for any typical matrix norm $|\cdot|$) such that $A + kI$ will have non-negative eigenvalues, from which it follows that $A + kI$ is similar to a positive semidefinite matrix. – Ben Grossmann Jul 31 '23 at 19:21

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