Can finitely many hermitian (positive-semidefinite) operators always be embedded into a small dimensional space preserving inner product? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T02:06:41Z http://mathoverflow.net/feeds/question/68354 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68354/can-finitely-many-hermitian-positive-semidefinite-operators-always-be-embedded Can finitely many hermitian (positive-semidefinite) operators always be embedded into a small dimensional space preserving inner product? Penghui Yao 2011-06-21T09:13:12Z 2011-09-13T18:22:12Z <p>Given $n$ hermitian (positive-semidefinite) operators $Q_1,\cdots,Q_n$ in finite dimensional Hilbert space (the dimension can be very large), is there a mapping $\phi$ maps $Q_i$ to $P_i$, which preserves inner product, i.e. $\langle P_i, P_j\rangle =\langle Q_i,Q_j\rangle$, and all $P_i$'s are hermitian (positive-semidefinite) operators staying in a smaller dimensional space, say $poly(n)$ ?</p> <p>Further question, given $n^2$ real numbers $c_{ij}$ $1\leq i,j\leq n$, how to decide if there exist $n$ hermitian (positive-semidefinite) operators $P_1,\cdots,P_n$ satisfying $\langle P_i,P_j\rangle=c_{ij}$? If exists, what is the minimum dimension of operators?</p> <p>For vectors, the questions above are trivial. I wonder if there are any known results for operators? Thanks.</p> http://mathoverflow.net/questions/68354/can-finitely-many-hermitian-positive-semidefinite-operators-always-be-embedded/68369#68369 Answer by Chris Godsil for Can finitely many hermitian (positive-semidefinite) operators always be embedded into a small dimensional space preserving inner product? Chris Godsil 2011-06-21T12:33:17Z 2011-06-21T12:33:17Z <p>Let $G$ be the Gram matrix, with $ij$-entry $\langle Q_i,Q_j\rangle$. Then the rank of $G$ is the dimension of the space spanned by the operators $Q_i$. So the answer to your first question is no. (The point is that this all works in any inner product space, changing from "vectors" to "operators" makes no difference.)</p> <p>For the second question, vectors can be encoded as diagonal operators. So this question would only make sense if you were working with some quite restricted class of operators.</p> http://mathoverflow.net/questions/68354/can-finitely-many-hermitian-positive-semidefinite-operators-always-be-embedded/71889#71889 Answer by Martin Schwarz for Can finitely many hermitian (positive-semidefinite) operators always be embedded into a small dimensional space preserving inner product? Martin Schwarz 2011-08-02T15:26:22Z 2011-08-02T15:26:22Z <p>Treating the linear operators as vectors, wouldn't the <a href="http://en.wikipedia.org/wiki/Johnson%E2%80%93Lindenstrauss_lemma" rel="nofollow">Johnson-Lindenstrauss</a> lemma give a low-distortion embedding?</p>