Can finitely many hermitian (positive-semidefinite) operators always be embedded into a small dimensional space preserving inner product? - MathOverflow

Can finitely many hermitian (positive-semidefinite) operators always be embedded into a small dimensional space preserving inner product?

2011-06-21T09:13:12Z

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 $poly(n)$ ?

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?

For vectors, the questions above are trivial. I wonder if there are any known results for operators? Thanks.

Answer by Chris Godsil for Can finitely many hermitian (positive-semidefinite) operators always be embedded into a small dimensional space preserving inner product?

2011-06-21T12:33:17Z

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.)

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.

Answer by Martin Schwarz for Can finitely many hermitian (positive-semidefinite) operators always be embedded into a small dimensional space preserving inner product?

2011-08-02T15:26:22Z

Treating the linear operators as vectors, wouldn't the Johnson-Lindenstrauss lemma give a low-distortion embedding?