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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

![enter image description here][1]

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 $4$.

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    $\begingroup$ $R(A)=4\implies$ at least one $4\times 4$ determinant is $\neq 0$. $\endgroup$ – thanasissdr Sep 10 '14 at 4:24
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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.

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  • $\begingroup$ What about $4 \times 4$ submatrix? I think it always have full rank right? $\endgroup$ – Idonknow Sep 10 '14 at 5:04
  • $\begingroup$ See my latest edit $\endgroup$ – Omnomnomnom Sep 10 '14 at 5:07

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