How can eigenvectors of a Hermitian matrix be entangled? This question comes from physics, but I will reduce the amount of physics jargon.
You have a tensor product space $H_1 \otimes H_2$. Any vector $w$ in this space has a Schmidt decomposition:
$$
\mathbf{w} = \sum_{i} \alpha_i \mathbf{u_i}\otimes\mathbf{v_i}
$$
Vector $w$ is not entangled if the Schmidt rank is 1, i.e., only one possible $\alpha_i$.
Now, in Physics you consider the eigenvectors of a Hermitian matrix (the Hamiltonian). According to the Spectral Theorem, this matrix is always diagonalizable.
So given the eigenvectors, you can go to a basis where these eigenvectors are just the trivial basis for $\mathbb{C}^n$ (by trivial basis I mean the columns of the identity matrix $\mathbb{1}_n$).
We see that in this basis the eigenvectors are clearly not entangled...
But entanglement is basis independent! And I know as a fact that the eigenvectors of some Hamiltonians are entangled. So whats going on here? Where have I forgotten?
 A: 
We see that in this basis the eigenvectors are clearly not entangled

It doesn't make sense to say that a vector is "entangled" with respect to an arbtirary basis for $H_1 \otimes H_2$.  Entanglement ultimately has to do with how the operator relates to the underlying spaces $H_1,H_2$.  When we change bases, the relation to the underlying spaces is not altered.  Howevever, one might say that the vector $(1,0,0,\cdots,0)$ "looks entangled" because, under a typical basis for $H_1 \otimes H_2$, it would be.  Let's resolve that paradox:
Here's an example: take $H_1 = H_2 = \Bbb C^2$, and the operator $A$ over $\Bbb C^2 \otimes \Bbb C^2 \cong \Bbb C^4$ given by
$$
A = A|_{\mathcal A}= \left[ 
\begin{array}{cc|cc}
1&0&0&1\\0&0&0&0\\
\hline
0&0&0&0\\
1&0&0&1
\end{array}
\right]
$$
We have taken the standard bases $\Bbb C^2$, i.e. $e_1 = (1,0)$ and $e_2 = (0,1)$. We have taken the standard basis over $\Bbb C^2 \otimes \Bbb C^2$ to be $\mathcal A = \{e_i \otimes e_j:i,j \in \{1,2\}\}$ where the pairs $(i,j)$ are taken in lexicographical order (i.e. $(1,1),(1,2),(2,1),(2,2)$).  Note that this basis for $\Bbb C^2 \otimes \Bbb C^2$ depends on the basis that we've chosen for the underlying spaces.
For our basis of eigenvectors, we can take $v_1 = e_1 \otimes e_1 + e_2 \otimes e_2 = (1,0,0,1)$ and any three vectors $v_j$ orthonormal to this one.  If we were to take such a basis $\mathcal B = \{v_1,v_2,v_3,v_4\}$, the matrix of $A$ with respect to this basis would be
$$
A|_{\mathcal B} = 
\begin{bmatrix}
2&0&0&0\\
0&0&0&0\\
0&0&0&0\\
0&0&0&0
\end{bmatrix}
$$
and within this new basis, $v_1$ would like $(1,0,0,0)$.  If we were to consider $(1,0,0,0)$ as a coordinate vector relative to our original basis $\mathcal A$, then this would correspond to the vector $e_1 \otimes e_1$, which is clearly separable (not entangled).  However, because we applied a change of basis to the overall space $H_1 \otimes H_2$, this apparent separability does not correspond to our definition of separability because our basis is no longer tied to the bases for the underlying spaces.  $v_1 = e_1 \otimes e_1 + e_2 \otimes e_2$ cannot be expressed in the form $u \otimes w$ (with $u \in H_1$ and $w \in H_2$), and this property has nothing to do with the basis that we've chosen.

Note that if our change in basis in $H_1 \otimes H_2$ comes from a change in basis over our underlying spaces $H_1$ or $H_2$, then this property of apparent separability will be preserved.  For instance, consider the basis $\{f_1,f_2\}$ of $H_2$ where
$$
f_1 = \frac 1{\sqrt{2}} (e_1 + e_2) \qquad
f_2 = \frac 1{\sqrt{2}} (e_1 - e_2)
$$
The eigenvector can be written as
$$
v_1 = e_1 \otimes (f_1 + f_2) + e_2 \otimes (f_1 - f_2) = \\
e_1 \otimes f_1 + e_1 \otimes f_2 + e_2 \otimes f_1 - e_2 \otimes f_2
$$
So, relative to the associated basis $\{e_i \otimes f_j\}$ taken in lexicographical order, our eigenvector has coordinate vector $(1,1,1,-1)$, which is indeed "apparently entangled".
