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linear transition -> linear transformation

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Tom Krüger 1 year ago
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@ -239,7 +239,7 @@ The defining property of a superposition is the probability distribution of its
\end{proof}
\subsection{Transition Matrices}
It follows directly from \cref{eq:exp_state_single_bit} that $\ptrans : \spanspace\parens{\mathbf{B}} \to \spanspace\parens{\mathbf{B}}$ is a linear transition on the space spanned by state basis $\mathbf{B}$ and \cref{thm:superpositionsClosedUnderProbabilisticTransition} even states that $\ptrans : \mathbf{B}^n \to \mathbf{B}^n$ and $\mathbf{B}^n$ is closed under $\ptrans$. It is well known, that the space of all linear maps $\Hom_{\R}\parens{V,W}$ between two finite-dimensional real vector spaces $V$ and $W$ is isomorphic to $\R^{\parens{\dim(W), \dim(V)}}$. So, there must exist an isomorphism between transition functions $\ptrans$ and $\R^{\parens{N,N}}$.
It follows directly from \cref{eq:exp_state_single_bit} that $\ptrans : \spanspace\parens{\mathbf{B}} \to \spanspace\parens{\mathbf{B}}$ is a linear transformation on the space spanned by state basis $\mathbf{B}$ and \cref{thm:superpositionsClosedUnderProbabilisticTransition} even states that $\ptrans : \mathbf{B}^n \to \mathbf{B}^n$ and $\mathbf{B}^n$ is closed under $\ptrans$. It is well known, that the space of all linear maps $\Hom_{\R}\parens{V,W}$ between two finite-dimensional real vector spaces $V$ and $W$ is isomorphic to $\R^{\parens{\dim(W), \dim(V)}}$. So, there must exist an isomorphism between transition functions $\ptrans$ and $\R^{\parens{N,N}}$.
\begin{theorem}%[see \cite{Knabner}]
\label{thm:probabilistic_matrix_representation}
Let $\mathbf{B} = \parensc{\mathbf{b}_i}_{i=1}^N$ be a $n$-bit state basis and $\mathscr{B} = \parensc{\mathbf{v}_j}_{j=1}^N$ a basis of $\R^N$, then there exists a matrix $A = (a_{ij}) \in \R^{\parens{N,N}}$ such that


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