Trace distance of two classical-quantum states

Question

I have these two classical-quantum states:

$$\rho = \sum_{a} \lvert a\rangle \langle a\lvert \otimes q^a \\ \mu = \sum_{a} \lvert a\rangle \langle a\lvert \otimes r^a $$

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:

$$T(\rho, \mu) = \frac{1}{2} ||\rho - \mu||_1 \\ = \frac{1}{2} || \sum_a \lvert a\rangle \langle a\lvert \otimes(q^a - r^a)||_1$$

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.