I'm trying to solve the following exercise:
Let $V,W$ be finite dimensional inner product spaces over $\mathbb{C}$. Show that for every $\psi \in V\otimes W$ with $\langle \psi, \psi \rangle_{V\otimes W} = 1$, the following are equivalent:
(i) There exist $v \in V$, $w\in W$ with $\langle v,v\rangle_V = \langle w,w\rangle_W = 1$ and $\psi = v \otimes w$.
(ii) $\rho = \rho_V \otimes \rho_W$, where $\rho = |\psi \rangle \langle \psi |$, $\rho_V = \text{Tr}_W (\rho)$, $\rho_W = \text{Tr}_V (\rho)$.
I am allowed to use without proof, that $\rho, \rho_V, \rho_W$ are density operators. Tr$_X(\cdot)$ denotes the partial trace in $X$.
My proof goes as follows:
(i) $\Longrightarrow$ (ii): By assumption we have $v\in V, w\in W$ with $v^\dagger v = w^\dagger w = 1$ and $\psi = v \otimes w$. Define $\rho := \psi \psi^\dagger = vv^\dagger \otimes ww^\dagger.$ Now, compute $$\rho_V = \text{Tr}_W \rho = \text{Tr}(ww^\dagger)vv^\dagger = (w^\dagger w)vv^\dagger = vv^\dagger,$$ $$\rho_W = \text{Tr}_V \rho = \text{Tr}(vv^\dagger)ww^\dagger = (v^\dagger v)ww^\dagger = ww^\dagger.$$ Thus indeed, $\rho = \rho_V \otimes \rho_W$, as desired.
(i) $\Longleftarrow$ (ii): By assumption we have $\psi \in V\otimes W$ with $\psi^\dagger \psi = 1$. Thus we know that $\exists v\in V, w\in W$ such that $\psi = v\otimes w$. We define $$\rho := \psi \psi^\dagger = (v\otimes w)(v^\dagger \otimes w^\dagger) = vv^\dagger \otimes ww^\dagger.$$ It remains to show that $v^\dagger v = w^\dagger w = 1.$ We also know that $\rho = \rho_V \otimes \rho_W$. Comparing with the above definition of $\rho$, we see that $vv^\dagger = \rho_V$ and $ww^\dagger = \rho_W$. Since these are density operators, we know that $1 = \text{Tr}(\rho_V) = v^\dagger v$, and $1 = \text{Tr}(\rho_W) = w^\dagger w$, as desired. Done.
I have a bad feeling about the proof, feels dodgy somehow. Is it even true, that $A \otimes B = C\otimes D \quad \Longrightarrow \quad A=C, B = D$? Any feedback?