Upper bound on the distance of orthogonal matrices 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!
 A: 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 $X=\left[ \begin{array}{cc} 1 & 2 \\ 1 & 2 \end{array} \right]$.
A: With some algebra, we get two key facts based on the orthogonality of $O.$
First, 
$$
||X - OX||_F^2 = {{\rm Tr}} (X-OX)(X-OX)^\top ={{\rm Tr}} (XX^\top - XX^\top O^\top - OXX^\top + OXX^\top O^\top)=
2 ||X||_F^2 {{\rm Tr}}(I-O).
$$
And second, 
$$||I-O||_F^2 = {{\rm Tr}} (I-O)(I-O)^\top = {{\rm Tr}} (I - O - O^\top + OO^\top)
=2{{\rm Tr}}(I-O) .$$
Putting it all together, we have that
$$
||I-O||_F^2  =2{{\rm Tr}}(I-O) = \frac{||X - OX||_F^2}{||X||^2_F}
\le \frac{\epsilon^2}{||X||^2_F}.
$$
Taking the square root, $||I-O||_F \le \frac{\epsilon}{||X||_F}.$
If $||X||^2_F=0,$ then the Cauchy Schwartz inequality says
$$
||X - OX||_F \leq ||X||_F + || OX||_F = ||X||_F + || X||_F =0 .
$$
Therefore, I don't think we can say anything about $||I-O||_F$ because $\{O \mid ||X - OX||_F \leq \epsilon\}=O(n)$.
