= Solution
\b[The <spectral norm> is the induced Euclidean <operator norm>]:
$$
\boxed{\|A\|=\sup_{\|v\|_2=1}\|Av\|_2
=\sqrt{\lambda_{\max}(A^\dagger A)}.}
$$
It is the largest <singular value>. From the definition, multiplying on either side by a <unitary operator> leaves the norm unchanged: right multiplication permutes the unit sphere of possible inputs, and left multiplication preserves output lengths. In particular every <unitary operator> has norm one.
For <unitary product telescoping>, replace the factors one at a time. With empty products interpreted as the identity,
$$
U_m\cdots U_1-V_m\cdots V_1
=\sum_{j=1}^m U_m\cdots U_{j+1}(U_j-V_j)V_{j-1}\cdots V_1.
$$
All cross terms cancel. By the triangle inequality and unitary invariance of the <spectral norm>,
$$
\boxed{\|U_m\cdots U_1-V_m\cdots V_1\|
\leq\sum_{j=1}^m\|U_j-V_j\|<m\epsilon.}
$$
This bound is independent of the dimension and does not assume that any of the factors commute.
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