TITLE: Invariant Length Function on $n \times n$ Unitary Group
QUESTION [2 upvotes]: Let $1$ denote the identity of $U(n)$, the group of all $n \times n$ complex unitary matrices. Given $x \in M_n(\Bbb{C})$, define $||x|| := \sqrt{\frac{1}{n}tr(x^*x)}$; i.e., the normalized Hilbert-Schmidt norm. 
If $x,y$ are $n \times n$ unitaries, why is it true that $||xy-1|| \le ||x-1|| + ||y-1||$? I'm having trouble verifying this inequality. I need to show this in order to show that $x \mapsto \frac{1}{2} ||x-1||$ is a invariant length function on the $n \times n$ unitary group.

REPLY [1 votes]: Note that if $x\in U(n)$, then for any $z\in M_n(\mathbb C)$ we have 
$$\|xz\|=\sqrt{\frac1n\operatorname{Tr}(z^*x^*xz)}=\sqrt{\frac1n\operatorname{Tr}(z^*z)}=\|z\|.$$
Thus, for $x,y\in U(n)$, we have
$$\|xy-1\|=\|x(y-1)+(x-1)\|\leq\|x(y-1)\|+\|x-1\|=\|y-1\|+\|x-1\|.$$
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