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Assume that $U$ is a real $2 \times 2$ matrix arising from a singluar value decomposition $$ M = USV' $$

As a part of a bigger calculation, it suggested in this paper (see text right beneath equation (44)) that we can in this case use the properties of $U$ to create a so-called orthomorphic transformation, i.e. a matrix $X$ such that $$ X = (I_2 + U)^{-1}(I_2-U) $$ and $$ U = (I+X)^{-1}(I-X) $$ Now, here is my problem: It seems to me that the matrix $I_2 + U$ is almost never invertible. If $U$ is a real matrix, then by nature of the SVD it is typically a rotation matrix with $U_{11}=-U_{22}$. Then we have $$ |I_2+U| = |U|+U_{11}+U_{22}+1 $$ with $|U| = \pm1$ since $U$ is orthogonal. So in that case, any time $|U|=-1$, the said matrix is singular.

Am I right in that it is a somewhat exceptional case that $(I_2+U)$ is in fact invertible? Or am I missing something? I would be truly grateful if anyone could shed some light on this problem.

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  • $\begingroup$ Experimentally this does seem true (almost) always for random real unitary 2x2 matrices, sometimes for such 3x3 matrices, and (almost) never for 4x4 and larger. When considering complex unitary matrices of any dimension, it seems to (almost) never happen. Example matlab/octave code to test this, drawing matrices uniformly at random from the manifold of orthonormal matrices with the induced measure is as follows: n=2; [U,~,~] = svd(randn(n,n)); eig(U+eye(n)) $\endgroup$ – Nick Alger Jul 29 '16 at 19:56
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$I+U$ is singular if and only if $-1$ is an eigenvalue of $U$. So, you are right that the sum is singular when $\det U=-1$. As $U$ is a $2\times2$ real orthogonal matrix, it has an eigenvalue $-1$ only when $\det U=-1$ or $U=-I_2$. However, you may do the following:

  • if $U$ is equal to $-I_2$ or $D=\operatorname{diag}(1,-1)$, replace $(U,V^T)$ by $(I,UV^T)$;
  • if $\det U=-1$ and $U\notin\{-I_2,D\}$, replace $(U,V^T)$ by $(UD,DV^T)$.

In both cases, the SVD is preserved and the new $U$ will give you a valid $X$.

By the way, when $-1$ is not an eigenvalue of $U$, the mapping $U\mapsto(I+U)^{-1}(I-U)$ is more commonly known as the Cayley transform.

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  • $\begingroup$ What if the eigenvalue of $U$ is $-1$ with multiplicity $2$? Then the determinant is $1$, but $I_2 + U$ will be singular. $\endgroup$ – Rodrigo de Azevedo Jul 30 '16 at 17:43
  • $\begingroup$ @RodrigodeAzevedo Thanks for catching that. Now fixed. $\endgroup$ – user1551 Jul 30 '16 at 18:04

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