The least purification environment dimension is the density rank

Given: A density operator $\rho$ on $H$ and a finite-dimensional purification space $K$ whenever a purification in $H\otimes K$ is under consideration.

Ensemble and purification: a purification is a unit vector whose auxiliary partial trace is $\rho$.

Schmidt decomposition: a bipartite vector has a Schmidt decomposition with positive coefficients and orthonormal families in both factors.

Purifications exist and are unitarily equivalent after padding: $\rho$ has a purification on an auxiliary space of dimension $\mathrm{rank}(\rho )$.

Ordered tensor registers, their inner product, and wire permutations, Dirac Notation Conversion and Partial trace over a finite tensor factor: the product basis uses the fixed factor order, $\mathrm{\mid }x⟩⟨y\mathrm{\mid }$ acts by $z\mapsto ⟨z,y⟩x$, and partial trace contracts equal auxiliary basis indices.

The partial trace is characterized by the trace pairing and is basis-independent: the partial-trace formula may use the auxiliary Schmidt basis, independently of the originally fixed basis.

Rank and nullity of a linear map with finite-dimensional domain: the rank of a finite-dimensional endomorphism is the dimension of its image.

Linear combination of a finite list, and the span $\mathrm{span}(S)$ as the smallest linear subspace containing $S$, The orthogonal complement $W^{\perp }=\{v:⟨v,w⟩=0\text{ for all }w\in W\}$, For a subspace $W$ of a finite-dimensional inner product space, $V=W\oplus W^{\perp }$, If ${\dim }_{F}V=n$ and $U$ is a linear subspace of $V$, then $U$ is finite-dimensional, ${\dim }_{F}U\le n$, and ${\dim }_{F}U=n$ if and only if $U=V$ and Every finite-dimensional real or complex inner product space has an orthonormal basis: the span $L$ of the finite orthonormal family is a subspace, its orthogonal complement is finite-dimensional and has an orthonormal basis, and adjoining that basis completes the family to an orthonormal basis of $K$.

Finite-dimensional vector space, and its dimension ${\dim }_{F}V$; infinite-dimensional means having no finite basis, Every finite orthogonal list of nonzero vectors is linearly independent and If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $\mathbb{N}$: an orthonormal list is linearly independent, and in a finite-dimensional $K$ every linearly independent set has size at most $\dim K$.