Dear math stackexchange users,
I have a question on orthogonal matrices: suppose I have a matrix $X\in\mathbb{R}^{n\times n}$ and I consider the orbit of the orthogonal group $O(n)$ acting from the left on $X$: $$Orb= \{Y\in\mathbb{R}^{n\times n}\ |\ Y=OX, O\in O(n)\}.$$ Now let $\epsilon>0$ and suppose I have $OX\in Orb$ satisfying $\|X-OX\|_F\leq \epsilon$.
If $I$ is the identity matrix, is there any way to give an upper bound on the distance $\|I-O\|_F$ in terms of $\|X\|_F$ and $\epsilon$?
Of course, if I have an upper bound on $\|I-O\|_F$, then by submultiplicativity of the Frobenius norm this gives an upper bound on $\|X-OX\|$. However, is there anything for the other direction? The orthogonal group is a compact set and this gives a trivial upper bound. This one however, is not applicable in my problem.
Thank you very much!
If $X$ is invertible, then $I-O=(X-OX)X^{-1}$ and: $$\|I-O\| \leq \|X-OX\| \|X^{-1}\| \leq \epsilon\|X^{-1}\|$$
If $X$ is not invertible, take $\|X^{-1}\|=\infty$. I don't think it's possible to get a bound in terms of $\|X\|$. Consider the $2\times 2$ case with $O=\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right]$ and $X=\left[ \begin{array}{cc} 1 & 2 \\ 1 & 2 \end{array} \right]$.