Yes, at least for finite dimensional vector spaces. First, I'll show you for two vector spaces $H_1$ and $H_2$ and you can proceed by induction. Use the fact that any linear operator can be represented as a sum of tensor product of linear operators, and if $H$ is Hermitian then $\frac{H + H^\dagger}{2} = H$. That is,
$$H = \frac{\sum_{i = 1}^m A_i \otimes B_i + A_i^\dagger \otimes B_i^\dagger}{2}$$
for linear operators $A_i :H_1 \to H_1$ and $B_i:H_2 \to H_2$. Now
$$ H = \sum_{i = 1}^m \left(\frac{A_i + A_i^\dagger}{2}\right) \otimes \left(\frac{B_i + B_i^\dagger}{2}\right) - \sum_{i = 1}^m \left(\frac{A_i - A_i^\dagger}{2i}\right) \otimes \left(\frac{B_i - B_i^\dagger}{2i}\right),$$
and each term is a tensor product of Hermitian operators.
I use my notations.
i) Let $A\in M_n,B\in M_m$ be diagonalizable matrices over $\mathbb{C}$. Let $(u_i)_{i\leq n},(v_j)_{j\leq m})$ be bases of eigenvectors associated to the eigenvalues $spectrum(A)=(\lambda_i),spectrum(B)=(\mu_j)$. Then $(u_i\otimes v_j)_{i,j}$ is a basis of eigenvector of $A\otimes B$ because $(A\otimes B)(u_i\otimes v_j)=A(u_i)\otimes B(v_j)=\lambda_i\mu_j u_i\otimes v_j$.
ii) Assume that $\tau=\lambda_i\mu_j=\lambda_k\mu_l$ where $(i,j)\not= (k,l)$.
Then $(A\otimes B)(u_i\otimes v_j+u_k\otimes v_l)=\tau(u_i\otimes v_j+u_k\otimes v_l)$. Assume that $u_i\otimes v_j+u_k\otimes v_l$ is a tensor product $a\otimes b$; then, $ab^T$, the associated $n\times m$ matrix has rank $1$ and is the sum of two matrices of rank $1$: $u_i{v_j}^T+u_k{v_l}^T$ (one has the same result for the transposes). This can only happen if the images are the same, that is $span(u_i)=span(u_k)$ or, in the same way, if $span(v_j)=span(v_l)$; in other words, $u_i\otimes v_j+u_k\otimes v_l$ is a tensor product iff $i=k,j\not=l$ or $j=l,i\not= k$ iff $\lambda_i$ is a multiple eigenvalue of $A$ or $\mu_j$ is a multiple eigenvalue of $B$.
iii) Let $z$ be an eigenvector of $A\otimes B$ that is not a tensor product. Then $z=\sum_{(i,j)\in K}z_{i,j}u_i\otimes v_j$ where $z_{i,j}\not= 0$. There is $\tau$ s.t.
$(A\otimes B)(z)=\sum_{i,j} z_{i,j}\lambda_i\mu_j u_i\otimes v_j=\sum_{i,j}\tau z_{i,j}u_i\otimes v_j$, that implies
for every $(i,j)\in K$, $\tau=\lambda_i\mu_j$, and $A\otimes B$ has a multiple eigenvalue.
$\textbf{Conclusion.}$. Finally, your equivalence is valid when $A,B$ have simple eigenvalues.
For example, for $A=diag(1,2),B=I_2$, then all the eigenvectors of $A\otimes B$ are tensor products.
Best Answer
It's basically the same, it's mainly a matter of implicit identifications.
Namely, in a Hilbert space $H$ there is a natural isomorphism $H\to H^*$ (which is the bra-ket duality), so $H\otimes H\simeq H\otimes H^*$.
Now there is a map $H\otimes H^*\to L(H)$ which corresponds to what you call the outer product.
So for two kets $|\psi\rangle$ and $|\phi\rangle$ in $H$, you can either see $|\psi\rangle\otimes |\phi\rangle$ as an element of $H\otimes H$ and interpret that as a kind of two-particles state, or you can look at its canonical image in $H\otimes H^*$, which the physicists note $|\psi\rangle \langle\phi|$ (which makes sense since $\langle\phi|$ is the element of $H^*$ corresponding to $|\phi\rangle$), and then interpret it as an operator in $L(H)$.