Math 222 - Selected Homework Solutions from
Chapter 5.2
Instructor - Al Boggess
Fall 1998
Section 5.2
1d The given matrix is
2
3
4 ,2
6 1
7
6
A := 64 2 31 775
3 4
The two column vectors are clearly linearly independent (one is not a
scalar multiple of the other). So a basis of the column space is the two
column vectors. The nullspace is f0g. The transpose of A is
"
A := ,42 13 21 34
#
t
Its row echelon form is
2
5 5 3
1
0
6
14 14 77
6
4
5
0 1 47 11
7
Therefore the rst two columns are linearly independent and a basis of
the column space is (4; ,2) and (1; 3) . The third and fourth column
variables are free; letting x3 = 1 and x4 = 0 gives a basis vector of
(,5=14; ,4=7; 1; 0). Letting x3 = 0 and x4 = 1 gives a basis vector
of (,5=14; ,11=7; 0; 1). Note that these two vectors are orthogonal to
the columns of A.
t
t
1
7. If a is a nonzero column vector of A, then a cannot be in N (A )
because N (A ) consists of those vectors which are orthogonal to the
rows of A which are the columns of A. In particular a would have to
be orthogonal to itself, which cannot happen unless a = 0.
8. If y 2 S , then y v = 0 for all v of the form
j
t
j
t
t
j
j
t
v = 1 x1 + : : : + x v k
k
Letting = 1 and all other = 0, we get
i
j
0 =yv =yx
i
as desired. Conversely, suppose y x = 0 for each 1 i k. We must
show that y v = 0 where v is given by . We have
i
yv = y
= k
X
x
i
i=1
k
X
(y x )
i
i=1
i
i
= 0
as desired.
15. Suppose W = U V , then each w can be expressed uniquely as w =
u + v where u 2 U and v 2 V . Suppose w 2 U \ V , then w belongs to
both U and V and we have
w = w+0 w 2U 02V
w = 0+w 02U w 2V
Uniqueness now states that w = 0 (since there is only one way to write
w is a sum of an element in U with an element in V ).
2