Invertibility of $U+I$ where $U$ comes from a SVD $M=USV'$

Question

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.

Answer 1

$I+U$ 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.