# $\operatorname{rank}(A) =$max number of rows of submatrix $B$; Proof

I don't understand how to proof the following:

The rank of a matrix $A \in M$ ($m \times n$, Field) equals the maximum number of rows of a square submatrix $B$ of $A$ with $\det (B) \neq 0$.

The definitions I use:

• The column rank of a matrix $A$ is the maximum number of linearly independent column vectors of $A$.
• The row rank of a matrix $A$ is the maximum number of linearly independent row vectors of $A$.

I know that column rank equals row rank.

• You should probably add your definition on what "rank" is, so people can help you starting from that definition. – Eff Jan 27 '15 at 18:56
• The column rank of a matrix A is the maximum number of linearly independent column vectors of A. The row rank of a matrix A is the maximum number of linearly independent row vectors of A. Column rank = row rank. – Arthur D Jan 27 '15 at 19:00

let $r$ be the common value of row and column rank. then there is a $r \times r$ sub matrix made of linearly independent rows/columns. this submatrix has a nonzero determinant.
we will show that square submatrix of asize bigger than $r$ has zero determinant. first if the submatrix does not contain all $r$ linearly independent rows, then it has at least two linearly dependent rows, therefore the determinant is zero. so now we only need to worry about having $r$ linearly independent rows, a basis of the row space and one extra row. using row exchanges we can assume that the first $r$ rows are linearly independent. since the last row is a linearly dependent on the first $r$ rows, we can zero out the last row by adding the correct multiple of the first $r$ rows. therefore the determinant of this matrix is zero.