# An identity involving matrix norm and vector norm in relation to a density matrix in physics

This question is self-contained. For those who are interested, it arises from my study of the paper:

K Lendi (1987), Evolution matrix in a coherence vector formulation for quantum Markovian master equations of N-level systems, J. Phys. A: Math. Gen., 20(1): 15-23.

The question is about the squared Frobenius norm $\|\rho\|_F^2=\operatorname{Tr}(\rho^2)$ (see formula 2.32 in the paper) of a self-adjoint "density matrix" in physics. This density matrix has a representation $$\rho = \frac{1}{N} \operatorname{id} + \sum_{k=1}^{N^2-1} v_k F_k \tag{2.16}$$ where the matrices $F_i$ form an orthonormal basis of the space of traceless Hermitian matrices. It follows that $v_i=\operatorname{Tr}(\rho F_i)$.

Now consider the squared Euclidean norm $\|v\|_2^2 = \sum_{k=1}^{N^2-1} v_k^2$ (2.33).

I want to know why $\|\rho\|_F^2 = \frac{1}{N} + \|v\|_2^2$ holds.

• In $(2.33)$, instead of summing up $v_k^2$, one should sum up $|v_k|^2$ instead, because $v_k$ is not necessarily real. – user1551 Sep 19 '15 at 7:11

Just calculate $$\rho^2=\rho\rho^*=\left(\frac{1}{N}\text{id}+\sum_{k=1}^{N^2-1}\nu_kF_k\right) \left(\frac{1}{N}\text{id}+\sum_{m=1}^{N^2-1}\nu_mF_m\right)^*=\\ =\frac{1}{N^2}\text{id}+\sum_{k,m=1}^{N^2-1}\nu_k\bar\nu_mF_kF_m^*+\text{terms with only F_k or F_m^*}$$ and use the fact that $\{\text{id},F_k\}$ is an ON basis (w.r.t. the Frobenius norm) $$\|\rho\|_F^2=\text{tr}\,(\rho\rho^*)=\frac{1}{N^2}\text{tr(id)}+ \sum_{k,m=1}^{N^2-1}\nu_k\bar\nu_m\underbrace{\text{tr}\,(F_kF_m^*)}_{=\delta_{k,m}}+0=\frac{1}{N^2}N+ \sum_{k=1}^{N^2-1}|\nu_k|^2.$$