Page 221 - Matrix Analysis & Applied Linear Algebra

216 Chapter 4 Vector Spaces

Rank isn’t changed by row or column operations, so r = rank (A)= rank (M),
and thus M is nonsingular. Now suppose that W is any other nonsingu-
lar submatrix of A, and let P and Q be permutation matrices such that
W X
PAQ = . If
Y Z
−1
I 0 I −W X −1
E = −1 , F = , and S = Z − YW X,
−YW I 0 I
then

W 0 W 0
EPAQF = =⇒ A ∼ , (4.5.8)
0 S 0 S
and hence r = rank (A)= rank (W)+ rank (S) ≥ rank (W) (recall Example
3.9.3). This guarantees that no nonsingular submatrix of A can have order
greater than r = rank (A).
Example 4.5.2
1 2 1

Problem: Determine the rank of A = 2 4 1 .
3 6 1
Solution: rank (A) = 2 because there is at least one 2 × 2 nonsingular sub-
matrix (e.g., there is one lying on the intersection of rows 1 and 2 with columns
2 and 3), and there is no larger nonsingular submatrix (the entire matrix is sin-
gular). Notice that not all 2 × 2 matrices are nonsingular (e.g., consider the one
lying on the intersection of rows 1 and 2 with columns 1 and 2).

Earlier in this section we saw that it is impossible to increase the rank by
means of matrix multiplication—i.e., (4.5.2) says rank (AE) ≤ rank (A). In
a certain sense there is a dual statement for matrix addition that says that it
is impossible to decrease the rank by means of a “small” matrix addition—i.e.,
rank (A + E) ≥ rank (A) whenever E has entries of small magnitude.

Small Perturbations Can’t Reduce Rank
If A and E are m × n matrices such that E has entries of sufficiently
small magnitude, then

rank (A + E) ≥ rank (A). (4.5.9)

The term “sufficiently small” is further clarified in Exercise 5.12.4.