For $r_1\neq r_2\neq r_3\neq r_4$, and for nonzero $b_1, b_2, b_3, b_4$, look at the matrix

$$\begin{pmatrix} 1&r_1&r_1^3&r_1^2+b_1 r_1&b_1&b_1 r_1^2&b_1 r_1^3\\ 1&r_2& r_2^3& r_2^2+b_2 r_2& b_2& b_2 r_2^2& b_2 r_2^3\\1&r_3& r_3^3& r_3^2+b_3 r_3& b_3& b_3 r_3^2& b_3 r_3^3\\1&r_4& r_4^3& r_4^2+b_4 r_4& b_4& b_4 r_4^2& b_4 r_4^3 \end{pmatrix}.$$
Does this matrix have rank 4?

  • $\begingroup$ Put b1=r1=-r2=-b2 , b3=r3=-b4=-r4. What is the rank? $\endgroup$ – Martín Vacas Vignolo Dec 13 '15 at 6:46
  • $\begingroup$ I tried several special cases. It is still 4. $\endgroup$ – user298283 Dec 13 '15 at 6:53

Setting $$ b_1 = \frac{r_2 r_3 + r_3 r_4 + r_4 r_2}{r_2 + r_3 + r_4} $$ and similarly for $b_2$, $b_3$, $b_4$ (permuting the indices) results in a matrix of rank $3$. Indeed, with this choice of $b_1,b_2,b_3,b_4$, the vector $$ \begin{pmatrix} -(r_2 - r_3) (r_3 - r_4) (r_2 - r_4)\\ (r_1 - r_3) (r_3 - r_4) (r_1 - r_4)\\ -(r_1 - r_2) (r_2 - r_4) (r_1 - r_4)\\ (r_1 - r_2) (r_2 - r_3) (r_1 - r_3) \end{pmatrix} $$ is orthogonal to each column (and is not the zero vector thanks to the assumptions on $r_1,r_2,r_3,r_4$) and is therefore not in the column space. Moreover, $b_1,b_2,b_3,b_4$ are generically nonzero.

I found this example by computing (by machine) several determinants of $4\times 4$ minors and algebraically solving for values of $b_1,b_2,b_3,b_4$ that made them all vanish.

  • $\begingroup$ I was not expecting counter examples. Thanks so much! I am about to repeat what you did. It seems this is the only solution of b_i ? $\endgroup$ – user298283 Dec 13 '15 at 13:20

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