What are the conditions should be added so that submatrix has full rank

Question

Suppose a $6 \times 4$ matrix satisfies the following

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?

Answer 1

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 $1$.

Answer 2

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).