If rank of $(m+1)\times n$ matrix is $m+1$, then some $(m+1)\times (m+1)$ submatrix has non-zero determinant.

Question

I can't understand this :

If I have a $(m+1)\times n$ matrix and if its rank is $m+1$, then some $(m+1)\times (m+1)$ submatrix has non-zero determinant.

How is it so?... kindly help.

Answer 1

Hint: If the rank is $m+1$, then the matrix must have $m+1$ linearly independent columns.

Answer 2

Let $k$ be maximal such there is a non zero $k$ sub-determinant. Just assume that the subdeterminant is in the upper left corner. Then we claim that the rank is $k$. It cannot be less than $k$ since the first $k$ rows are independent. If we take any other row and consider the first $k$ elements of this row, they will be a unique linear combination of the upper left hand $k\times k$ rows. The coefficients of this linear combination give in fact a linear dependence between the entire rows, not just the first $k$ terms, since the combination is unique for the first $k$ terms. This shows that all $k+1$ rows are dependent and so the rank is at most $k$.