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Suppose a $6 \times 4$ matrix satisfies the following enter image description here

where $\alpha, \beta, \gamma, \theta, \sigma, \mu$ are non-zero. What are the conditions should be added so that any $4 \times 4$ submatrix has full rank?

I think the conditions are $\alpha_i \neq \beta_j \neq \gamma_k$ and $\theta_i \neq \sigma_j \neq \mu_k$ for $i,j,k=1,2,3$, in other words, all entries of both rank $3$ matrices are distinct. But I don't know whether the conditions are sufficient to conclude the statement. Can anyone help me?

EDIT: Okay, so the conditions stated above are not sufficient. Can I use an $6 \times 4$ Cauchy matrix instead? Because any square-submatrix of a cauchy matrix has full rank. But in this case, we have a few zeros in the matrix. So I don't know whether these zeros will affect the rank of submatrix or not. Also, what if I change to finite field?

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The necessary and sufficient conditions are precisely that the $\binom{6}{4} = 15$ minors of size $4$ are nonzero. However, $6$ of these minors are already guaranteed to be nonzero by the rank $3$ assumptions, so there are only $9$ conditions that need to be checked. Using the notation $I(x,y) = x_2y_3 - x_3y_2$, $J(x,y,z,w) = x_1(y_1 I(z,w) - w_1 I(z,y))$, these can be expressed in the rather pleasing form:

$$J(\alpha, \mu, \beta, \sigma) \ne J(\beta, \mu, \alpha, \sigma), \qquad J(\alpha, \mu, \beta, \theta) \ne J(\beta, \mu, \alpha, \theta), \qquad J(\alpha, \sigma, \beta, \theta) \ne J(\beta, \sigma, \alpha, \theta) \\ J(\alpha, \mu, \gamma, \sigma) \ne J(\gamma, \mu, \alpha, \sigma), \qquad J(\alpha, \mu, \gamma, \theta) \ne J(\gamma, \mu, \alpha, \theta), \qquad J(\alpha, \sigma, \gamma, \theta) \ne J(\gamma, \sigma, \alpha, \theta) \\ J(\beta, \mu, \gamma, \sigma) \ne J(\gamma, \mu, \beta, \sigma), \qquad J(\beta, \mu, \gamma, \theta) \ne J(\gamma, \mu, \beta, \theta), \qquad J(\beta, \sigma, \gamma, \theta) \ne J(\gamma, \sigma, \beta, \theta)$$

Note: since these conditions are complements of equalities, the $4 \times 4$ subdeterminants will be generically nonzero, i.e. for random choices of $\alpha_i, \ldots, \mu_i$, all size $4$ minors will be nonzero with probability $1$ (if $\text{char}(k) = 0$).

Update (Re your update): A Cauchy matrix has all nonzero entries, so your $6 \times 4$ matrix is never Cauchy. The conditions above are still necessary and sufficient over any field, but over a finite field the probability that all $9$ conditions hold is no longer $1$.

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  • $\begingroup$ What do you mean 'complements of equalities'? I understand that $I(x,y)$ is similar to a determinant of a $2 \times 2$ matrix. But I can't seem to figure out how you obtain the expression of $J(x,y,z,w)$. $\endgroup$ – Idonknow Sep 14 '14 at 16:39
  • $\begingroup$ @Idonknow: By "complement of equalities" I meant that the solutions to the $9$ conditions are the (intersection of) complements of solutions to (certain) equations. The expression for $J$ comes from a $3 \times 3$ minor - if you apply cofactor expansion to the condition that a given $4 \times 4$ subdeterminant is nonzero, you will see exactly where it comes from $\endgroup$ – zcn Sep 14 '14 at 22:47
  • $\begingroup$ I still don't get the complement of equalities part. Can you elaborate further? $\endgroup$ – Idonknow Nov 19 '14 at 18:30
  • $\begingroup$ Also, the probability that all $9$ conditions hold is no longer$1$, do you mean that there may be cases such that the rank $3$ and $4$ conditions are satisfied but there may exist some minor of size $4$ which is zero? $\endgroup$ – Idonknow Nov 19 '14 at 18:44
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The answer is NO. Consider for example

$$ A=\left(\begin{array}{ccc} \alpha_1 & \alpha_2 & \alpha_3 \\ \beta_1 & \beta_2 & \beta_3 \\ \gamma_1 & \gamma_2 & \gamma_3 \\ \end{array}\right)= \left(\begin{array}{ccc} 2 & 3 & 4 \\ 5 & 7 & 6 \\ 11 & 16 & 17 \\ \end{array}\right), \ \ B=\left(\begin{array}{ccc} \theta_1 & \theta_2 & \theta_3 \\ \sigma_1 & \sigma_2 & \sigma_3 \\ \mu_1 & \mu_2 & \mu_3 \\ \end{array}\right)= \left(\begin{array}{ccc} 1 & 8 & 9 \\ 10 & 71 & 18 \\ 12 & 85 & 19 \\ \end{array}\right) $$

Then $A$ and $B$ both have rank $3$, but

$$ C=\left(\begin{array}{cccc} \alpha_1 & 0 & \alpha_2 & \alpha_3 \\ \beta_1 & 0 & \beta_2 & \beta_3 \\ 0 & \theta_1 & \theta_2 & \theta_3 \\ 0 & \sigma_1 & \sigma_2 & \sigma_3 \end{array}\right)= \left(\begin{array}{ccc} 2 & 0 & 3 & 4 \\ 5 & 0 & 7 & 6 \\ 0 & 1 & 8 & 9 \\ 0 & 10 & 71 & 18 \\ \end{array}\right) $$

is not full rank (the vector $(10,55,-8,1)$ is in its kernel).

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  • $\begingroup$ so you are saying that the statement will not be true under any conditions? $\endgroup$ – Idonknow Sep 13 '14 at 15:35
  • $\begingroup$ @Idonknow well, it will obviously be true under the condition that all six $4\times4$ subdeterminants be nonzero :-) $\endgroup$ – Ewan Delanoy Sep 13 '14 at 15:37
  • $\begingroup$ May I know how you construct the counterexample above? From what I can see, you let $Cx=0$ and try to show that the $x$ is nontrivial. But I don't see how the coefficients are constructed. $\endgroup$ – Idonknow Sep 13 '14 at 15:41
  • $\begingroup$ You choose lots of coefficients at random, and then fix the others so that the counterexamples works. $\endgroup$ – Ewan Delanoy Sep 13 '14 at 15:52
  • $\begingroup$ In your first comment, you said that 'all six $4 \times 4$ subdeterminants be nonzero ', why is it $6$? $\endgroup$ – Idonknow Sep 13 '14 at 16:04

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