2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.