$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 10.8}$

*The Schmidt Decomposition.*

Every $m\times n$ matrix $M$, with $m \leq n$, has a singular value decomposition $M = UDV$ where $D$ is an $m\times n$ diagonal matrix with non-negative real entries, and $U$ and $V$ are $m\times m$ and $n\times n$ unitary matrices.

Let $\ket\psi \in A\otimes B$, where $A$ has dimension $m$ and $B$ has dimension $n$, with $m \leq n$. Let $\{ \ket i \}$ be a basis for $A$ and $\{ \ket j \}$ be a basis for $B$, then for some choice of $m_{ij}\in\bf C$

(1)Let $M$ be the $m\times n$ matrix with entries $a_{ij}$. Use the singular value decomposition (SVD) for $M$ to find sets of orthonormal unit vectors $\{ \ket{\alpha_i}\} \in A$ and $\{ \ket{\beta_j}\} \in B$ such that

(2)where $\lambda_i$ is non-negative. The $\lambda_i$ are called the {\it Schmidt coefficients}\index{Schmidt coefficients}, and $K$, the number of $\lambda_i$, is called the {\it Schmidt rank}\index{Schmidt rank} or {\it Schmidt number}\index{Schmidt number} of $\ket\psi$.