# Prove that $\|UVU^{-1}V^{-1}-I\|\leq 2\|U-I\|\|V-I\|$

$U,V$ are unitary $n\times n$ matrices, and the norm is the operator norm (so we can use $\|UV\|\leq\|U\|\|V\|$).

I've noticed that \begin{align} \|UVU^{-1}V^{-1}-I\|&= \|(UV-VU)U^{-1}V^{-1}\|\\ &\leq \|UV-VU\|\|U^{-1}V^{-1}\| \end{align}

I can bound the first term by $\|UV\|+\|VU\|$, but I don't think this is useful.

Hints (rather than complete answers) would be appreciated.

The question comes from here (exercise 1)

Let $[A,B]=AB-BA$ denote the commutator of two $n \times n$ matrices $A,B$.
Hint: For $U,V$ unitary $n\times n$ matrices, one has the identity $$\lVert UVU^{-1}V^{-1}-1 \rVert = \lVert [U,V]U^{-1}V^{-1} \rVert = \lVert [U,V] \rVert = \lVert [U-1,V-1] \rVert.$$
• Why is $\|[U,V]U^{-1}V^{-1}\| = \|[U,V]\|$? I can see how we can get an inequality: $\|[U,V]U^{-1}V^{-1}\| \leq \|[U,V]\| \|U^{-1}V^{-1}\| = \|[U,V]\|$ (where the last equality follows because $U^{-1}V^{-1}$ is unitary and hence has norm equal to $1$). – man_in_green_shirt May 25 '17 at 8:49
• Either way, that is enough to prove the claim, as then we have $\|[U-I,V-I]\| = \|(U-I)(V-I)-(V-I)(U-I)\| \leq \|(U-I)(V-I)\|+\|(V-I)(U-I)\|\leq 2\|U-I\|\|V-I\|$ – man_in_green_shirt May 25 '17 at 8:55
• @man_in_green_shirt For the inequality in the other direction, we have $\lVert [U,V] \rVert \leqslant \lVert [U,V] U^{-1}V^{-1}\rVert \lVert VU \rVert =\lVert [U,V] U^{-1}V^{-1}\rVert$ – Jonas Dahlbæk May 25 '17 at 9:42
Note that $\|UVU^{-1}V^{-1}\|=1$ for all unitaries $U,V$. We would like to show that $$\tag{1} \|UVU^{-1}V^{-1}-I\|\leq 2\|U-I\|\,\|V-I\|.$$ The left-hand-side, as hinted in the exercise, is $\|UV-VU\|$. Now $$\tag{2} \|UV-VU\|\leq\|UV-V\|+\|V-VU\|=2\|U-I\|.$$ As the roles of $U$ and $V$ are symmetric, we can get, by multiplying $(2)$ and the corresponding inequality for $V$, $$\|UV-VU\|\leq 2\|U-I\|^{1/2}\|V-I\|^{1/2},$$ which is sharper than $(1)$ when $\|U-I\|>1$ and $\|V-I\|>1$.