An orthogonal matrix is a matrix $A$ over the reals such that $A^t=A^{-1}$ (its transpose is its inverse). The Frobenius norm over $n\times n$ real matrices is given by $\|A\| = \sqrt{trace(A^tA)}$.
I have come across the following claim: The distance (induced by the Frobenius norm) between any two (non equal) orthogonal matrices is $\sqrt{n}$. I can't find a proof for this claim, but no refutation either (of course, if the difference between two orthogonal matrices is itself an orthogonal matrix the claim is clear, but I don't know if that's true either).
Best Answer
I don't know where you found your claim, but it seems that the distance induced by the Frobenius norm between any two orthogonal matrices can be any real number.
Because:
$$ \begin{align} \|A - B\|^2 &= \mathrm{tr} \left( (A-B)^t (A-B) \right) \\ &= \mathrm{tr} \left( (A^t -B^t) (A-B) \right) \\ &= \mathrm{tr} (A^tA -A^tB - B^tA + B^tB) \\ &= \mathrm{tr} (2I) - \mathrm{tr}(2A^tB) \\ &= 2n - 2\mathrm{tr}(A^tB) \end{align} $$
The last but one equality is due to the fact that $A^tA = B^tB = I$ and $(B^tA)^t = A^tB$ and a matrix and its transpose have the same trace.
Now, take $A = I$ and we've got
$$ \| I - B\|^2 = 2n - 2\mathrm{tr}(B) $$
for any orthogonal matrix $B$. So, if you take as $B$ the family of orthogonal matrices
$$ \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \ , $$
$\theta \in [0, 2\pi]$, their traces $\mathrm{tr} (B) = 2\cos\theta$ can be any real number between $-2$ and $2$. So, their Frobenius distances to the unit matrix $I$ can be any real number from $0$ to $\sqrt{8}$.