Definition 3.3.22 A submatrix of a matrix A is the rectangular array of numbers obtained by deleting some rows and columns of A. Let A be an m × n matrix. The determinant rank of the matrix equals r where r is the largest number such that some r × r submatrix of A has a non zero determinant. The row rank is defined to be the dimension of the span of the rows. The column rank is defined to be the dimension of the span of the columns.
Theorem 3.3.23 If A, an m × n matrix has determinant rank r, then there exist r rows of the matrix such that every other row is a linear combination of these r rows.
Proof: Suppose the determinant rank of A =

and the r rows whose indices are

I want to show that every row is a linear combination of these rows. Consider the l^{th} row and let p be an index between 1 and n. Form the following

Of course you can assume l
Expand the determinant of the above matrix along the last column. Let C_{k} denote the cofactor associated with the entry a_{ikp}. This is not dependent on the choice of p. Remember, you delete the column and the row the entry is in and take the determinant of what is left and multiply by −1 raised to an appropriate power. Let C denote the cofactor associated with a_{lp}. This is given to be nonzero, it being the determinant of the matrix r × r matrix in the upper left corner. Thus

which implies

Since this is true for every p and since m_{k} does not depend on p, this has shown the l^{th} row is a linear combination of the i_{1},i_{2},
Proof: From Theorem 3.3.23, every row is in the span of r rows where r is the determinant rank. Therefore, the row rank (dimension of the span of the rows) is no larger than the determinant rank. Could the row rank be smaller than the determinant rank? If so, it follows from Theorem 3.3.23 that there exist p rows for p < r ≡ determinant rank, such that the span of these p rows equals the row space. But then you could consider the r ×r sub matrix which determines the determinant rank and it would follow that each of these rows would be in the span of the restrictions of the p rows just mentioned. By Theorem 2.6.4, the exchange theorem, the rows of this sub matrix would not be linearly independent and so some row is a linear combination of the others. By Corollary 3.3.11 the determinant would be 0, a contradiction. ■
Corollary 3.3.25 If A has determinant rank r, then there exist r columns of the matrix such that every other column is a linear combination of these r columns. Also the column rank equals the determinant rank.
Proof: This follows from the above by considering A^{T}. The rows of A^{T} are the columns of A and the determinant rank of A^{T} and A are the same. Therefore, from Corollary 3.3.24, column rank of A = row rank of A^{T} = determinant rank of A^{T} = determinant rank of A. ■
The following theorem is of fundamental importance and ties together many of the ideas presented above.
Theorem 3.3.26 Let A be an n × n matrix. Then the following are equivalent.
Proof: Suppose det

Now consider the column vector, x ≡

Since also A0 = 0, it follows A is not one to one. Similarly, A^{T} is not one to one by the same argument applied to A^{T}. This verifies that 1.) implies 2.).
Now suppose 2.). Then since A^{T} is not one to one, it follows there exists x≠0 such that

Taking the transpose of both sides yields

where the 0^{T} is a 1 × n matrix or row vector. Now if Ay = x, then

contrary to x≠0. Consequently there can be no y such that Ay = x and so A is not onto. This shows that 2.) implies 3.).
Finally, suppose 3.). If 1.) does not hold, then det
Corollary 3.3.27 Let A be an n × n matrix. Then the following are equivalent.
Proof: This follows immediately from the above theorem.