Singular points of a matrix when the entries are restriced to a Lie Group

Question

Let $\mathsf{SO}(3)$ be the set of $3 \times 3$ rotation matrices. Let $R\in\mathsf{SO}(3)$ and $r_{ij}$ represent the entry of $R$ sitting at the $i^{th}$ row and $j^{th}$ column, i.e., $$ R \in\ \mathsf{SO}(3) =\left(\begin{array}{ccc} r_{11} & r_{12} & r_{13}\\ r_{21} & r_{22} & r_{23} \\ r_{31} & r_{32} & r_{33} \end{array}\right). $$ For the matrix $D$ given below $$ D =\left(\begin{array}{ccc} -r_{22}-r_{33} & r_{21} & r_{31}\\ r_{12} & -r_{11}-r_{33} & r_{32} \\ r_{13} & r_{23} & -r_{11}-r_{22} \end{array}\right), $$ I want to find the following:

1. Find the set of all point for which $D$ loses rank? In other words I want to find all values of $r_{ij}\in\mathbb{R}$ for $i,j=\{ 1,2,3\}$ such that the matrix $D$ becomes singular. Although each $r_{ij}$ take values only on a closed subset $ [-1,1]\subset \mathbb{R}$ (subject to more conditions), but here we assume they can take any value on $\mathbb{R}$ just to make things simple. Since $r_{ij}$ is allowed to take any value on $\mathbb{R}$, I denote this singular set by $S_{\mathbb{R}} = \{ r_{ij}\in\mathbb{R}: det(D) = 0\}$. At least one way (maybe a naive one) is to compute an expression of the determinant of $D$, which in this case would be a nonlinear expression of $9$ variable and find its roots. However, this is a non-trivial task because the determiant of $D$ is $$ det(D) = r_{11}r_{12}r_{21} - r_{11}^{2} r_{22} - r_{11}r_{33}^{2} - r_{11}^{2}r_{33} - r_{22}r_{33}^{2} - r_{22}^{2}r_{33} - r_{11}r_{22}^{2} + r_{11}r_{13}r_{31} + r_{12}r_{21}r_{22} - 2 r_{11}r_{22}r_{33} + r_{12}r_{23}r_{31} + r_{13}r_{21}r_{32} + r_{13}r_{31}r_{33} + r_{22}r_{23}r_{32} + r_{23}r_{32}r_{33}. $$ How to find roots of this equation? Or, is there a better way to approah this problem?

2. Find all the values of $r_{ij}$ for $i,j=\{ 1,2,3\}$ such that when each $r_{ij}$ is the entry of a rotation matrix $R \in \mathsf{SO}(3)$ the matrix $D$ loses rank. I think this second problem is even more challenging as it requires to find conditions under which $D$ loses rank subject to the condition that the problem is restricted to a Lie group. Again, my question is how it can be solved? What tools are needed to study this kind of problems?

Thanks a lot, specially if you have read all the way to the end!! :)

Answer 1

Ok, typing my last comment I think I see a partial answer:

Let $v \in \mathbb{R}^3$ be such that $Dv = 0$. From $D = R^{-1} - tr(R)I$ we see that

$$R^{-1}v = tr(R)v$$

Now since $R^{-1}$ is a rotation matrix an the last equation states that $v$ is an eigenvector of $R^{-1}$ we must have that $v$ lies on the rotation axis of $R$ and $R^{-1}$ and moreover that $tr(R) = 1$ or $tr(R) = -1$: rotations cannot make vectors longer or shorter so $\pm 1$ are the only eigenvalues.

This tells you something: we either have

$$r_{11} + r_{22} + r_{33} = 1$$ OR $$r_{11} + r_{22} + r_{33} = -1$$.

Conversely, whenever $R$ is a rotation matrix satisfying the FIRST of the last two equations we do know that $D$ is singular because $R$, being a rotation, necessarily HAS a rotation axis and the vectors on that axis hence ARE eigenvalues with eigenvalue 1.

The more interesting question is what happens when $r_{11} + r_{22} + r_{33} = -1$. The only rotations with eigenvalue $-1$ are those over $180$ degrees and so you get some additional constraints on the $r_{ij}$ to incorporate that.

In summary:

The solutions are:

• Any rotation matrix with trace equal to 1 (note that I didn't check if they actually exist, I leave that to you)
• Rotation matrices with trace $-1$ encoding a rotation over $180$ degrees

There are no others. I leave it to you to work out the corresponding values of the $r_{ij}$.