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Robert Israel
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I tried the $n=2$ case, using Maple to find a "plex" Groebner basis for the equations. The result is rather complicated, and too large to show here. The first member of the basis, for example, is $y_1$ times an irreducible polynomial of $143$ terms and total degree $8$ in the entries of $A$ and $B$, which must be $0$ in order to have a solution with $y_1 \ne 0$. So I don't think you'll get a "simple" criterion.

EDIT: You want the $2n \times n$ matrix $$ \pmatrix{X A - I\cr A - X B - 2 I\cr}$$ to have rank $< n$. That is equivalent to all its ${2n} \choose n$ $n \times n$ submatrices (obtained by selecting $n$ of the rows) having determinant $0$. Each of those determinants is a polynomial ...

I tried the $n=2$ case, using Maple to find a "plex" Groebner basis for the equations. The result is rather complicated, and too large to show here. The first member of the basis, for example, is $y_1$ times an irreducible polynomial of $143$ terms and total degree $8$ in the entries of $A$ and $B$, which must be $0$ in order to have a solution with $y_1 \ne 0$. So I don't think you'll get a "simple" criterion.

I tried the $n=2$ case, using Maple to find a "plex" Groebner basis for the equations. The result is rather complicated, and too large to show here. The first member of the basis, for example, is $y_1$ times an irreducible polynomial of $143$ terms and total degree $8$ in the entries of $A$ and $B$, which must be $0$ in order to have a solution with $y_1 \ne 0$. So I don't think you'll get a "simple" criterion.

EDIT: You want the $2n \times n$ matrix $$ \pmatrix{X A - I\cr A - X B - 2 I\cr}$$ to have rank $< n$. That is equivalent to all its ${2n} \choose n$ $n \times n$ submatrices (obtained by selecting $n$ of the rows) having determinant $0$. Each of those determinants is a polynomial ...

Source Link
Robert Israel
  • 54.2k
  • 1
  • 76
  • 152

I tried the $n=2$ case, using Maple to find a "plex" Groebner basis for the equations. The result is rather complicated, and too large to show here. The first member of the basis, for example, is $y_1$ times an irreducible polynomial of $143$ terms and total degree $8$ in the entries of $A$ and $B$, which must be $0$ in order to have a solution with $y_1 \ne 0$. So I don't think you'll get a "simple" criterion.