Page 177 - Matrix Analysis & Applied Linear Algebra

172 Chapter 4 Vector Spaces

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Conversely, if R A = R B , then
span {A 1∗ , A 2∗ ,..., A m∗ } = span {B 1∗ , B 2∗ ,..., B m∗ } ,

so each row of B is a combination of the rows of A, and vice versa. On the
basis of this fact, it can be argued that it is possible to reduce A to B by using
row
only row operations (the tedious details are omitted), and thus A ∼ B. The
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proof of (4.2.6) follows by replacing A and B with A T and B .
Example 4.2.2
Testing Spanning Sets. Two sets {a 1 , a 2 ,..., a r } and {b 1 , b 2 ,..., b s } in
n
span the same subspace if and only if the nonzero rows of E A agree with
the nonzero rows of E B , where A and B are the matrices containing the a i ’s
and b i ’s as rows. This is a corollary of (4.2.5) because zero rows are irrelevant
in considering the row space of a matrix, and we already know from (3.9.9) that
row
A ∼ B if and only if E A = E B .
Problem: Determine whether or not the following sets span the same subspace:
1 2 3 0 1
         
   
   
A =  2   4   6  , B =  0   2  .
2 1 1 1 3
  ,   ,     ,  
   
   
3 3 4 1 4
Solution: Place the vectors as rows in matrices A and B, and compute
   
1223 1201
A =  2413  →  0011  = E A
3614 0000
and

0011 1201
B = → = E B .
1234 0011
Hence span {A} = span {B} because the nonzero rows in E A and E B agree.
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We already know that the rows of A span R A , and the columns of A
span R (A), but it’s often possible to span these spaces with fewer vectors than
the full set of rows and columns.

Spanning the Row Space and Range
Let A be an m × n matrix, and let U be any row echelon form derived
from A. Spanning sets for the row and column spaces are as follows:
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• The nonzero rows of U span R A . (4.2.7)
• The basic columns in A span R (A). (4.2.8)