- #1
Petawa
- 7
- 0
I've been struggling with this problem for about two weeks. My supervisor is also stumped - though the problem is easy to state, I don't know the proper approach to tackle it. I'd appreciate any hints or advice.
Let V be an random k-dimensional vector subspace of ℝn, chosen uniformly over all possible k-dimensional subspaces. Let X[itex]\in[/itex]ℝn[itex]\times[/itex]n be a symmetric matrix whose column space is contained in V. Now I add constraints to X: given some pairs (i,j) such that [itex]1\leq i < j\leq n[/itex], I need Xij=0. The pairs are fixed and independent of V. How many of these zero constraints can I satisfy before (with high probability) the only solution is X=0?
I've found a sufficient condition for a non-zero solution to exist: the number of constraints q must satisfy q< k(k+1)/2. From simulations, I think it is also a necessary condition (with probability one), but I can't seem to show it. I'd appreciate any ideas on how I might proceed.
Let V be an random k-dimensional vector subspace of ℝn, chosen uniformly over all possible k-dimensional subspaces. Let X[itex]\in[/itex]ℝn[itex]\times[/itex]n be a symmetric matrix whose column space is contained in V. Now I add constraints to X: given some pairs (i,j) such that [itex]1\leq i < j\leq n[/itex], I need Xij=0. The pairs are fixed and independent of V. How many of these zero constraints can I satisfy before (with high probability) the only solution is X=0?
I've found a sufficient condition for a non-zero solution to exist: the number of constraints q must satisfy q< k(k+1)/2. From simulations, I think it is also a necessary condition (with probability one), but I can't seem to show it. I'd appreciate any ideas on how I might proceed.
Proof of sufficient condition
Let MV be the space of symmetric n by n matrices whose column space is contained in V. An orthogonal basis for MV is {Q(ei ejT+ej eiT)QT : [itex]1\leq i \leq j \leq n[/itex]} where the columns of Q[itex]\in[/itex]ℝn[itex]\times[/itex]k form an orthonormal basis for V and ei are the standard basis vectors for ℝk, so dim(M_V)=k(k+1)/2.
Let MX be the space of symmetric n[itex]\times[/itex]n matrices that satisfy the q zero constraints. Now suppose no non-zero X satisfying the constraints exist: this implies MV[itex]\cap[/itex] MX={0}. Hence dim(MV+MX) = dim(MV)+dim(MX) = k(k+1)/2+n(n+1)/2-q. Since MV+MX is contained within the space of symmetric matrices, its dimension is bounded by n(n+1)/2. Thus "no non-zero X" implies q[itex]\geq[/itex]k(k+1)/2. The contrapositive implies there is a non-zero solution when q<k(k+1)/2.