Determining the rank of a matrix based on its minors I was wondering if the following lemma was easy to prove.  I got a little tripped up when I saw the explicit condition that the matrix could have elements in any field not necessarily finite, so I didn't know if a standard proof from linear algebra would still apply.

Let $M$ be a matrix with entries  in a (possibly infinite) field $F$.
Suppose that there exists a minor $m_n$ of order $n$ for $M$ that is nonzero and such that all minors of order $n+1$ which contain $m_n$ are zero.
How do you show the rank of the matrix is $n$?

 A: That there exists a minor of order $n$ which is non-zero suggests that there exists at least $n$ linearly independent rows/columns of the matrix (namely the rows/columns of the submatrix corresponding to the minor). The rank is thus at least $n$. If the rank is larger than $n$ then there exists at least $n+1$ linearly independent rows, meaning the $n+1$ order submatrix taken using those rows will be invertible and have non-zero determinant contrary to the fact that the all $n+1$ order minors are zero. This shows that the rank must be $n$.
Edit: As Jyrki pointed out, only the minors which contain $m_n$ are zero. So let us proceed another way. The whole matrix cannot have full rank since it must contain $m_n$. That means there is at least one linearly dependent row and column which is not in $m_n$ since all rows and columns in $m_n$ are linearly independent. Removing the row and the column produces a smaller matrix which again contains $m_n$. We can repeat this procedure until all that remains is $m_n$, proving the rank is $n$.
