# 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

![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 $4$.

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

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.
• What about $4 \times 4$ submatrix? I think it always have full rank right? – Idonknow Sep 10 '14 at 5:04