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).

  • $\begingroup$ i.e. the claim is $\left\|\mathbf{Q}_1-\mathbf{Q}_2\right\|\stackrel{?}{=}\sqrt{n}$ where the $\mathbf{Q}_i$ are orthogonal matrices. $\endgroup$ – J. M. isn't a mathematician Sep 23 '10 at 9:20
  • $\begingroup$ The difference between two orthogonal matrices is not necessarily orthogonal: see alext87's examples. $\endgroup$ – Agustí Roig Sep 23 '10 at 10:10
  • $\begingroup$ If $A$ is an $n \times n$ orthogonal matrix, then $\|A\| = \sqrt{n}$. Maybe that was the claim. $\endgroup$ – bubba Jul 26 '17 at 0:47

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.


$$ \begin{align} \|A - B\|^2 &= \mathrm{tr} \left( (A-B)^t (A-B) \right) \\ &= \mathrm{tr} \left( (A^t -B^t) (A-B) \right) \\ &= \mathrm{tr} (A^tA -A^tB - B^tA + B^tB) \\ &= \mathrm{tr} (2I) - \mathrm{tr}(2A^tB) \\ &= 2n - 2\mathrm{tr}(A^tB) \end{align} $$

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 $\sqrt{8}$.

| cite | improve this answer | |
  • 1
    $\begingroup$ Probably the original claim of $||A-B||^2 \propto n$ was for the case when $tr(A^tB)=0$. If $tr(A^tB)=0$, then we can call $A$ and $B$ to be orthogonal to each other using definition of Frobenius inner product. $\endgroup$ – Mohit Pandey Jan 4 '19 at 20:35

I don't think this is true. Let $A=I_2$ and $B$ be 2 by 2 matrix with $b_{11}=b_{22}=0$ and $b_{12}=b_{21}=1$. Then both these matrices are orthogonal but

and $||A-B||_F = 2\neq \sqrt{2}$

An even easier counterexample:

Let $A=1$ and $B=-1$ the 1 by 1 orthogonal matrices. Then $A-B=2$ so $||A-B||_F=2\neq 1$.

| cite | improve this answer | |

If the distance is defined as $$d(X,Y) = \sqrt{\textrm{trace}(X^{\top}Y)}$$ then it follows from the fact you can write the Frobenius norm as $$\Vert A \Vert_{F}^{2} = \sum_{i,j} |a_{ij}|^{2}$$ since $X^{\top}Y$ is orthogonal if X,Y are and an orthogonal matrix has as its columns (or rows) 3 orthonormal vectors and so their squared norms are each 1 and so their sum is n and so their Frobenius norm is $\sqrt{n}$.

The difference between elements of a group wouldn't be defined by treating them as if they belong to a vector space (or have addition/subtraction defined on them at least). Orthogonal matrices form a group so their combination must be under that binary operation. This is why commutators for abelian groups (well all groups but the abelian ones have the appropriate right hand side for the point I'm making) are written as $$[X,Y] = XYX^{-1}Y^{-1} = e$$ and not $XY - YX = 0$. If X,Y belong to both a group and a separate ring then you can reshuffle $XY-YX=0$ into $XY=YX$ then $XYX^{-1}Y^{-1} = 1 \cong e$ but that's meaningless if X,Y do not have addition/subtraction defined on them. Sure, a matrix representation easily does so but that isn't forming a representation of the underlying group then.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.