Page 182 - Matrix Analysis & Applied Linear Algebra

4.2 Four Fundamental Subspaces 177
Example 4.2.6

Problem: Suppose rank (A m×n )= r, and let P = P 1 be a nonsingular
P 2

matrix such that PA = U = C r×n , where U is in row echelon form. Prove
0
R (A)= N (P 2 ). (4.2.13)

Solution: The strategy is to first prove R (A) ⊆ N (P 2 ) and then show the
reverse inclusion N (P 2 ) ⊆ R (A). The equation PA = U implies P 2 A = 0, so
all columns of A are in N (P 2 ), and thus R (A) ⊆ N (P 2 ) . To show inclusion
in the opposite direction, suppose b ∈ N (P 2 ), so that

P 1 b
P 1 d r×1
Pb = b = = .
P 2 P 2 b 0

C d
Consequently, P A | b = PA | Pb = , and this implies
0 0
rank[A|b]= r = rank (A).
Recall from (2.3.4) that this means the system Ax = b is consistent, and thus
b ∈ R (A) by (4.2.3). Therefore, N (P 2 ) ⊆ R (A), and we may conclude that
N (P 2 )= R (A).

It’s often important to know when two matrices have the same nullspace (or
left-hand nullspace). Below is one test for determining this.

Equal Nullspaces
For two matrices A and B of the same shape:
row
• N (A)= N (B) if and only if A ∼ B. (4.2.14)
col
T T
• N A = N B if and only if A ∼ B. (4.2.15)

T T
Proof. We will prove (4.2.15). If N A = N B , then (4.2.12) guarantees
T T
R P 2 = N B , and hence P 2 B = 0. But this means the columns of B
are in N (P 2 ). That is, R (B) ⊆ N (P 2 )= R (A) by using (4.2.13). If A is
replaced by B in the preceding argument—and in (4.2.13)— the result is that
R (A) ⊆ R (B), and consequently we may conclude that R (A)= R (B) . The
desired conclusion (4.2.15) follows from (4.2.6). Statement (4.2.14) now follows
by replacing A and B by A T and B T in (4.2.15).