rank of a submatrix 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 $4$.

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.
