Hilbert-Schmidt distance positive semi-definite

Question

Given two density matrices $\rho_1,\rho_2$ (let them be finite dimensional). I try to show that $\mathrm{Tr}((\rho_1-\rho_2)^2) = 0 \implies \rho_1 = \rho_2$. What I tried is the following.

I realised that $(\rho_1 - \rho_2)^2$ is semi positive-definite since $\rho_1,\rho_2$ are Hermitian and semi positive-definite. Therefore the eigenvalues are real and non-negative. Since the sum of these eigenvalues is zero all the eigenvalues of $(\rho_1 - \rho_2)^2$ must be zero. However I don't see how to connect this to the conclusion that $\rho_1 = \rho_2$?

It is true that we have $(\rho_1 - \rho_2)^2 v = 0$ for all $v \in \mathbb{C}^n$ (since you can express $v$ in terms of complete eigenbasis of $(\rho_1 - \rho_2)^2$) and hence $(\rho_1 - \rho_2)^2 = 0$. But this does not imply $\rho_1 = \rho_2$ right?

Answer 1

If $A$ is hermitian and $A^2=0$, then $A=0$. The proof is quite simple, suppose $A\neq 0$

$$ \langle i|A^2|i\rangle=\sum_j \langle i|A|j\rangle\langle j|A|i\rangle=\sum_j |\langle i|A|j\rangle|^2\neq 0$$

at least for one of the $i$s, because I assumed $A\neq 0$. Hence if $A\neq 0$, $A^2\neq 0$. This clearly implies that if $A^2=0$, then $A=0$.

Answer 2

Note that $\rho\equiv (\rho_1-\rho_2)$ is hermitian. If $\rho^2 =0$, then for an element of the underlying Hilbert space $ \psi\in \mathscr{H}$ it holds that: $$ (\psi,\rho^2 \psi) = (\rho \psi,\rho \psi) =((\rho_1 - \rho_2)\psi,(\rho_1 - \rho_2)\psi) = ||(\rho_1-\rho_2)\psi||^2 =0 \quad ,$$

where $(\cdot,\cdot)$ denotes the scalar product and $|| \cdot||$ the norm. The properties of the norm then imply that $$ (\rho_1-\rho_2) \psi = 0 \quad . $$ Since $\psi \in \mathscr{H}$ was chosen arbitrarily and an operator is defined by its action on the elements of the Hilbert space, we conclude that $\rho_1 = \rho_2$.