Trace distance of two classical-quantum states

Question

I have these two classical-quantum states: \begin{align*}\rho &= \sum_{a} | a\rangle \langle a| \otimes q^a \\ \mu &= \sum_{a} | a\rangle \langle a| \otimes r^a \end{align*} where $a$ are the classical basis vectors, $q^a, r^a$ are arbitrary matrices dependent on $a$. Now, we can take the trace distance of these two classical-quantum states, which would be: $$T(\rho, \mu) = \frac{1}{2} \|\rho - \mu\|_1 = \frac{1}{2} \| \sum_a | a\rangle \langle a| \otimes(q^a - r^a)\|_1$$ Now, my question is, can we rewrite the above expression in the following way: $$T(\rho, \mu) = \frac{1}{2} \sum_a \|q^a - r^a\|_1$$ i.e., just pulling the summation out of the norm?

Answer 1

Yes, since the trace norm is the sum of the absolute value of the singular values, and the singular values can be found for each of the $a$ blocks independently.