Consider a Haar random quantum state $|\psi\rangle$. Note that
$$\rho =\mathbb{E}[|\psi\rangle\langle \psi|] = \frac{\mathbb{I}_{n}}{2^{n}}.$$
$\mathbb{I}_n$ is the identity operator on $n$ qubits. Now consider what happens when we trace out the last (or the $n^{\text{th}}$) qubit of $|\psi\rangle$. Denote that density matrix by $\rho^{[n-1]}$. Note that the partial trace is a linear operator. Hence,
$$ \rho^{[n-1]} = \mathbb{E}\left[\text{Tr}_n( |\psi\rangle\langle \psi|)\right] = \text{Tr}_n\left(\mathbb{E}\left[ |\psi\rangle\langle \psi|\right]\right) = \frac{\mathbb{I}_{n-1}}{2^{n-1}}$$
Clearly, the rank of this matrix is $2^{n-1}$ and the eigenvalues are each equal to $\frac{1}{2^{n-1}}$.
However, we could also write $|\psi\rangle$ in terms of its Schmidt decomposition:
$$|\psi\rangle = \alpha_1 |\phi_1 \rangle |\chi_1\rangle + \alpha_2 |\phi_2 \rangle |\chi_2\rangle,$$
where $|\phi_1 \rangle$ and $|\phi_2 \rangle$ are orthonormal vectors over $n-1$ qubits and $|\chi_1\rangle$ and $|\chi_2\rangle$ are orthonormal vectors over $1$ qubit. Clearly,
$$\text{Tr}_n( |\psi\rangle\langle \psi|) = |\alpha_1|^{2} |\phi_1 \rangle \langle \phi_1 | + |\alpha_2|^{2} |\phi_2 \rangle \langle \phi_2 |. $$
Clearly, the rank of this density matrix is $2$ and the eigenvalues are exactly $|\alpha_1|^{2} $ and $|\alpha_2|^{2}$. Since they have to sum to $1$, the eigenvalues are certainly large, and at least one of them is much larger than $\frac{1}{2^{n-1}}$.
How can both of these statements be simultaneously true? That is, how can the probabilistic average of a bunch of rank $2$ matrices suddenly give rise to a matrix of rank $2^{n-1}$? Rank of a sum of matrices is sub-additive, so it is certainly mathematically possible, but I am missing the qualitative intuition.
Furthermore, for a typical Haar random quantum state $|\psi\rangle$, is there anything we can say about how large or small $|\alpha_1|^{2} $ and $|\alpha_2|^{2}$ are and what the eigenvectors look like?
