Page 181 - Matrix Analysis & Applied Linear Algebra

176 Chapter 4 Vector Spaces

Left-Hand Nullspace
If rank (A m×n )= r, and if PA = U, where P is nonsingular and
U is in row echelon form, then the last m − r rows in P span the

left-hand nullspace of A. In other words, if P = P 1 , where P 2 is
P 2
(m − r) × m, then
T T
N A = R P 2 . (4.2.12)

C
Proof. If U = , where C r×n , then PA = U implies P 2 A = 0, and
0
T T
this says R P ⊆ N A . To show equality, demonstrate containment in
2
T
the opposite direction by arguing that every vector in N A must also be in
T T T −1
R P . Suppose y ∈ N A , and let P Q 2 ) to conclude that
2 =( Q 1
T
T
T
T
0 = y A = y P −1 U = y Q 1 C =⇒ 0 = y Q 1
T −1 −1
because N C = {0} by (4.2.11). Now observe that PP = I = P P
insures P 1 Q 1 = I r and Q 1 P 1 = I m − Q 2 P 2 , so
T T T
0 = y Q 1 =⇒ 0 = y Q 1 P 1 = y (I − Q 2 P 2 )
T T T
=⇒ y = y Q 2 P 2 = y Q 2 P 2
T T T
=⇒ y ∈ R P 2 =⇒ y ∈ R P 2 .
Example 4.2.5
1 2 2 3

T
Problem: Determine a spanning set for N A , where A = 2 4 1 3 .
3 6 1 4
Solution: To find a nonsingular matrix P such that PA = U is in row echelon
form, proceed as described in Exercise 3.9.1 and row reduce the augmented

matrix A | I to U | P . It must be the case that PA = U because P
is the product of the elementary matrices corresponding to the elementary row
operations used. Since any row echelon form will suffice, we may use Gauss–
Jordan reduction to reduce A to E A as shown below:
   
1223 100 1201 −1/3 2/30
2413 010 −→ 0011 2/3 −1/30
   
3614 001 0000 1/3 −5/31

  
−1/3 2/30  1/3 
T
P =  2/3 −1/30  , so (4.2.12) implies N A = span  −5/3  .
1/3 −5/31  1 