What is the implication of Schmit decomposition? According to schmidt decomposition if I have pure state $|\psi\rangle$ in the composite hilbert space $AB$ ( both $A$ and $B$ are hilbert spaces of dimension $n$ )
then it can be writen as
$$|\psi\rangle = \sum_i \lambda_i |i_A\rangle |i_B\rangle$$
here $\{i_A\}$ and  $\{i_B\}$ are orthonormal basis for hilbert spaces $A$ and $B$ respectively.
The above holds true for any  vector $|\psi\rangle$ ( if I am not wrong ) even if its not normalized ( if not normalized then $\sum_i \lambda_i^2$ wont be equal to 1 ). Thus any vector of of a space of dimension $n$ x $n$ is being written in linear combination of only $n$ orthonormal vectors ( $|i_A\rangle |i_B\rangle$ for $1 <= i <= n$ ), which should not be possible. Am I missing something or interpreting schmidt decomposition incorrectly ?
 A: Denote $|\psi\rangle = \sum\limits_{i = 1}^m \sum\limits_{j = 1}^n h_{ij} |ij\rangle$ as $|\psi\rangle \rightarrow H = (h_{ij})_{m \times n}$. Then we have the following lemma:

Lemma: Define matrix $U$ (in the original basis) as a new setting for Alice and $V$ for Bob, then a state $|\psi\rangle$ in the basis of the new settings is $U^* H V^\dagger$.
Proof:
  Denote the original bases of Alice and Bob both as $|0\rangle, |1\rangle$, the new setting as $|0_a\rangle, |1_a\rangle$ and $|0_b\rangle, |1_b\rangle$ for Alice and Bob, respectively. and so
  $$
\begin{bmatrix}
|0_a\rangle\\
|1_a\rangle
\end{bmatrix}
=
U
\begin{bmatrix}
|0\rangle\\
|1\rangle
\end{bmatrix},
\begin{bmatrix}
|0_b\rangle\\
|1_b\rangle
\end{bmatrix}
=
V
\begin{bmatrix}
|0\rangle\\
|1\rangle
\end{bmatrix}.
$$
  In this way, the state is expressed as
  $$
|\psi\rangle = [|0\rangle, |1\rangle] H \begin{bmatrix}
|0\rangle\\
|1\rangle
\end{bmatrix} =
[|0_a\rangle, |1_a\rangle] (U^\dagger)^T  H V^\dagger\begin{bmatrix}
|0_b\rangle\\
|1_b\rangle
\end{bmatrix},
$$
  Thus, in the basis of new setting, the state is expressed as $U^* H V^\dagger$.

Linear algebra tells us that: Given a matrix $H$, there are some unitary matrices $U,V$ such that $U^* H V^\dagger$ is a diagonal matrix. So, given any state $|\psi\rangle$, there are some basses $|0_a\rangle, |0_b\rangle$ such that $|\psi\rangle$ can be expressed as a diagonal matrix ${\rm Diag}(\lambda_1,\lambda_2,\cdots,\lambda_n)$, that is, 
$$
|\psi\rangle = \sum_i \lambda_i |0_a\rangle |0_b\rangle.
$$
Obviously, $|0_a\rangle, |0_b\rangle$ may be different for different $|\psi\rangle$.
A: As suggested by WSA aka RV, I copy my comments into a (partial) answer.
The key point is that the theorem says "for any given $\psi$ there exist two bases $\{i_A\}$ and $\{i_B\}$ such that...". This means that the choice of the bases depends on the vector $\psi$ we are considering.
So there is not an $n$ dimensional common basis that span the whole $n\times n$ space; and if we would be very precise we may write the formula as
$$\psi= \sum_{i=1}^n \lambda_i(\psi)i_A(\psi)\otimes i_B(\psi) \; ,$$
i.e. clarifying the $\psi$-dependence of the bases (and coefficients obviously).
