176 CHAPTER 8. RANK OF A MATRIX

Corollary 8.6.8 The determinant rank equals the row rank.

Proof: From Theorem 8.6.7, the row rank is no larger than the determinant rank. Couldthe row rank be smaller than the determinant rank? If so, there exist p rows for p < r suchthat the span of these p rows equals the row space. But this implies that the r×r sub-matrixwhose determinant is nonzero also has row rank no larger than p which is impossible if itsdeterminant is to be nonzero because at least one row is a linear combination of the others.■

Corollary 8.6.9 If A has determinant rank r, then there exist r columns of the matrix suchthat every other column is a linear combination of these r columns. Also the column rankequals the determinant rank.

Proof: This follows from the above by considering AT . The rows of AT are the columnsof A and the determinant rank of AT and A are the same. Therefore, from Corollary 8.6.8,column rank of A = row rank of AT = determinant rank of AT = determinant rank of A. ■

The following theorem is of fundamental importance and ties together many of theideas presented above.

Theorem 8.6.10 Let A be an n×n matrix. Then the following are equivalent.

1. det(A) = 0.

2. A,AT are not one to one.

3. A is not onto.

Proof: Suppose det(A) = 0. Then the determinant rank of A = r < n. Therefore, thereexist r columns such that every other column is a linear combination of these columnsby Theorem 8.6.7. In particular, it follows that for some m, the mth column is a linearcombination of all the others. Thus letting A =

(a1 · · · am · · · an

)where the

columns are denoted by ai, there exists scalars, α i such that

am = ∑k ̸=m

αkak.

Now consider the column vector x≡(

α1 · · · −1 · · · αn

)T. Then

Ax =−am + ∑k ̸=m

αkak = 0.

Since also A0 = 0, it follows A is not one to one. Similarly, AT is not one to one by thesame argument applied to AT . This verifies that 1.) implies 2.).

Now suppose 2.). Then since AT is not one to one, it follows there exists x ̸= 0 suchthat AT x = 0. Taking the transpose of both sides yields xT A = 0. where the 0 is a 1× nmatrix or row vector. Now if Ay = x, then

|x|2= xT (Ay) =(xT A

)y = 0y = 0

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(A) ̸= 0 but then from Theorem7.1.14 A−1 exists and so for every y ∈ Fn there exists a unique x ∈ Fn such that Ax = y. Infact x = A−1y. Thus A would be onto contrary to 3.). This shows 3.) implies 1.) ■