Page 156 - A Course in Linear Algebra with Applications

140 Chapter Five: Basis and Dimension

Take the case of U + W first - it is the easier one. Let A
be the matrix whose columns are u i , . . . , u r : remember that
these are now n-column vectors. Also let B be the matrix
whose columns are w i , . . . , w s . Then U+W is just the column
space of the matrix M = [A \ B]. A basis for U + W can
therefore be found by putting M in reduced column echelon
form and deleting the zero columns.
Turning now to UDW, we look for scalars Ci and dj such
that
C1U1 + h c ru r — diwi H 1- d sw a :

for every element of U D W is of this form. Equivalently

C1U1 H h c ru r + -di)wi H h (-d 8)w 8 = 0.
(
Now this equation asserts that the vector

/ C l \

-di

\-d J
a
belongs to the null space of [A | B]. A method for finding a
basis for the null space of a matrix was described in 5.1. To
complete the process, read off the the first r entries of each
vector in the basis of the null space of [A \ B], and take these
entries to be c 1 ; ..., c r. The resulting vectors form a basis of
unw.
Example 5.3.5
Let
0 2 1 \
2 1 5 2
M =
2 -2 -1 - 1
\ 1 1 5 3 /