Assume I have a $n\times n$ positive semidefinite matrix $G$ of rank $p$ satisfying a set of $np - p(p-1)/2$ equations $v^T_jGv_j = 1$, $j = 1 \ldots np - p(p-1)/2$ for some given vectors $v_j$. It is assumed these equations are linearly independent. Note here that the number of equations is exactly equal to the degrees of freedom in $G$. Is it then true that there are only finitely many matrices $G$ satisfying these equations? Thanks in advance.
6
-
1$\begingroup$ You are asking whether a non degenerate system of $n$ linear equations in $n$ unknowns has a finite number of solutions? The answer is "yes", and one can sharpen the result to "a unique solution". $\endgroup$– Igor RivinCommented Nov 2, 2011 at 13:40
-
$\begingroup$ @Igor. This is not a linear system, unless $p=0$. There is the nonlinear constraint that $G$ has rank $p$. $\endgroup$– Denis SerreCommented Nov 2, 2011 at 14:21
-
$\begingroup$ @Igor. Just try the case $n=2$ and $p=1$. You'll find two solutions. $\endgroup$– Denis SerreCommented Nov 2, 2011 at 14:26
-
$\begingroup$ Also, isn't it the case that there are potentially infinitely many solutions in the case that the entries come from a skew field? The poster has not made it clear from where the matrix entries come. Gerhard "Ask Me About System Design" Paseman, 2011.11.02 $\endgroup$– Gerhard PasemanCommented Nov 2, 2011 at 16:29
-
$\begingroup$ AH, I have missed the rank constraint. $\endgroup$– Igor RivinCommented Nov 2, 2011 at 17:35
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
2
In the case $n=3$,$p=2$, your 5 constraints for $v_1 = (1,0,0)^T$, $v_2 = (0,1,0)^T$, $v_3 = (0,0,1)^T$, $v_4 = (1,-2,0)^T$ and $v_5 = (1,-1,1)^T$ have solution $G = \pmatrix{1 & 1 & t\cr 1 & 1 & t\cr t & t & 1\cr}$, which has rank 2 and is positive semidefinite if $-1 < t < 1$.
answered Nov 2, 2011 at 23:32
-
$\begingroup$ How did you arrive at this solution? I guess I really mean: where did v4 and v5 come from? $\endgroup$ Commented Nov 2, 2011 at 23:37
-
$\begingroup$ Using Maple, I looked for $v_4 = (a,b,0)^T$, $v_5 = (c,d,e)^T$ for which the determinant of a symmetric solution $G$ is identically 0. $\endgroup$ Commented Nov 4, 2011 at 13:29