Linear Algebra, Theory And Applications, 2012a

3.3. THE MATHEMATICAL THEORY OF DETERMINANTS 95
Of course you can assume l /∈ {i 1 , ··· ,i r } because there is nothing to prove if the l th
row is one of the chosen ones. The above matrix has determinant 0. This is because if
p/∈{j 1 , ··· ,j r } then the above would be a submatrix of A which is too large to have non
zero determinant. On the other hand, if p ∈{j 1 , ··· ,j r } then the above matrix has two
columns which are equal so its determinant is still 0.
Expand the determinant of the above matrix along the last column. Let C k denote the
cofactor associated with the entry a ik p. 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
⎛
Thus
which implies
a lp =
⎜
⎝
⎞
a i1j 1
··· a i1j r
.
⎟
. ⎠
a irj 1
··· a irj r
0=a lp C +
r∑
k=1
r∑
C k a ik p
k=1
−C k
C
a i k p ≡
r∑
m k a ik p
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 , ··· ,i r rows.
k=1
Corollary 3.3.24 The determinant rank equals the row rank.
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