rank of a submatrix

Question

Suppose the $8 \times 4$ matrix $A$ has rank $4$. Is it always true that any $4 \times 4$ submatrix of $A$ has rank $4$? I am doing research on coding theory and I am wondering whether this is true.

My guess is that it is always true. Since $A$ has rank $4$, any $4$ rows are linearly independent.

Remark: I am considering matrix of the form

$$ \left(\begin{array}{cccc} \alpha_{1} & 0 & \alpha_{2} & \alpha_{3} \\ \beta_{1} & 0 & \beta_{2} & \beta_{3} \\ \gamma_{1} & 0 & \gamma_{2} & \gamma_{3} \\ 0 & \theta_{1} & \theta_{2} & \theta_{3} \\ 0 & \sigma_{1} & \sigma_{2} & \sigma_{3} \\ 0 & \mu_{1} & \mu_{2} & \mu_{3} \end{array}\right) $$

where $\alpha, \beta, \gamma$ are non-zero. In this case, my question is: any $3\times 3$ submatrix of the matrix above has rank $3$. Is the statement true? Note that the matrix above is assumed to have rank $4$.

Answer 1

Nope. Counterexample: $$ \pmatrix{ 1 & & &\\ &1&&\\ &&1&\\ &&&1\\ 0 &&\cdots&0\\ \vdots & && \vdots\\ 0 & \cdots && 0 } $$

For your matrix of consideration: note we can still have a matrix of rank 4 with $\theta_i = \sigma_i = \mu_i = 1$ for all $i$. However, the matrix $$ \pmatrix{ \gamma_1 & 0 & \gamma_2 & \gamma_3\\ 0&1&1&1\\ 0&1&1&1\\ 0&1&1&1\\ } $$ Will never have full rank.