Page 180 - Matrix Analysis & Applied Linear Algebra

4.2 Four Fundamental Subspaces 175

Spanning the Nullspace

To determine a spanning set for N (A), where rank (A m×n )= r, row
reduce A to a row echelon form U, and solve Ux = 0 for the basic
variables in terms of the free variables to produce the general solution
of Ax = 0 in the form

h n−r . (4.2.9)
x = x f 1 h 1 + x f 2 h 2 + ··· + x f n−r
By definition, the set H = {h 1 , h 2 ,..., h n−r } spans N (A). Moreover,
it can be proven that H is unique in the sense that H is independent
of the row echelon form U.

It was established in §2.4 that a homogeneous system Ax = 0 possesses a
unique solution (i.e., only the trivial solution x = 0 ) if and only if the rank of
the coefficient matrix equals the number of unknowns. This may now be restated
using vector space terminology.

Zero Nullspace
If A is an m × n matrix, then

• N (A)= {0} if and only if rank (A)= n; (4.2.10)

• N A T = {0} if and only if rank (A)= m. (4.2.11)

Proof. We already know that the trivial solution x = 0 is the only solution to
Ax = 0 if and only if the rank of A is the number of unknowns, and this is
T
what (4.2.10) says. Similarly, A y = 0 has only the trivial solution y = 0 if
T T
and only if rank A = m. Recall from (3.9.11) that rank A = rank (A)
in order to conclude that (4.2.11) holds.
T
Finally, let’s think about how to determine a spanning set for N A . Of
course, we can proceed in the same manner as described in Example 4.2.4 by
T
reducing A T to a row echelon form to extract the general solution for A x = 0.
However, the other three fundamental subspaces are derivable directly from E A
row
(or any other row echelon form U ∼ A ), so it’s rather awkward to have to
start from scratch and compute a new echelon form just to get a spanning set
T
for N A . It would be better if a single reduction to echelon form could
T
produce all four of the fundamental subspaces. Note that E A T = E , so E T
A A
T
won’t easily lead to N A . The following theorem helps resolve this issue.