Math 222 - Selected Homework Solutions for
Assignment 5
Instructor - Al Boggess
Fall 1998
Section 3.3
14 We are given that A is an m n matrix with linearly independent
columns a1 ; : : : a . We are to show the null space equals 0. This
means we must show that if Ax = 0 then x 0.
Now if x = (x1 : : : x ), then Ax = x1 a1 + : : : + x a . Since a1 ; : : : a
are linearly independent, we know that if x1 a1 + : : : + x a = 0, then
x1 = : : : x = 0 and thus x = 0 as desired.
15 We are given that x1 ; : : : ; Bx are linearly independent and that A is
a nonsingular matrix. We must show that if y = Ax for 1 i k
then y1 ; : : : ; y are linearly independent. This means we must show
that if c1 y1 + : : : + c y = 0 then c1 = : : : c = 0. We have
n
n
n
n
n
n
n
n
k
i
i
k
k
k
k
0 = c1 y1 + : : : + c y
= c1 Ax1 + : : : + c Ax
= A(c1 x1 + : : : + c x )
k
k
k
k
k
k
Thus,
0 = A(c1 x1 + : : : + c x )
Since A is nonsingular, it has an inverse; so
k
k
(c1 x1 + : : : + c x ) = A,1 0 = 0
k
k
Since x1 ; : : : ; x are linearly independent, we conclude that c1 = : : : =
c = 0 as desired.
k
k
1
16 We are given that v is spanned by v1 ; : : : ; v . Therefore
n
v = c1 v1 + : : : + c v for some choice of constants c1 ; : : : ; c . We are to show that v; v1 ; : : : ; v
n n
n
n
are NOT linearly independent. So we must produce some nontrivial
linear combination of these vectors which equals zero. Now we can
rearrange to read
v , c1 v1 , : : : , c v = 0
Since the coecient in front of v is 1 (which is not zero), the above
linear combination is nontrivial and so v; v1 ; : : : ; v are not linearly
n
n
n
independent.
17 We are given that v1 ; : : : ; v are linearly independent. We are to show
v2 ; : : : ; v cannot span. In particular, we will show that v1 cannot be
spanned by v2 ; : : : ; v . We will prove this by contradiction. Suppose
v1 can be spanned by v2 ; : : : ; v . Then
n
n
n
n
v1 = c2 v2 + : : : + c v
for some choice of constants c2 ; : : : c . We can rearrange this equation
n
n
n
to read
v1 , c2 v2 , : : : , c v = 0
Since the coecient in front of v1 is 1, the linear combination on the left
is nontrivial. Therefore we conclude that v1 ; v2 ; : : : ; v are not linearly
n
n
n
independent, which contradicts the fact they are linearly independent.
Section 3.4
17 We are given that U and V are two dimensional subspaces of R3 . The
dimension of U \ V must be at least one, as the following argument
demonstrates. Let u1 ; u2 and v1 ; v2 be bases for U and V , respectively. The collection of four vectors, u1 ; u2 ; v1 ; v2 , cannot be linearly
independent in R3 and so there are constants and , not all zero,
with
1 u1 + 2 u2 + 1 v1 + 2 v2 = 0 Since one of the constants is not zero, let us suppose 1 6= 0. The vector
u = 1 u1 + 2 u2 cannot be zero for otherwise, the above equation
i
2
i
would read 1 v1 + 2 v2 = 0 with 1 6= 0. However, since v1 and v2
are linearly independent, both 1 and 2 would have to be zero. Thus
u 6= 0. Equation can be rewritten as
u = ,1v1 , 2 v1 2 V
Clearly u belongs to U . The above equation implies u belongs to
V . Therefore, this nonzero vector, u, belongs to U \ V . Thus the
dimension of U \ V is at least one.
Geometrically, two two-dimensional planes which contain the origin
must intersect in a one-dimensional line, unless the planes are identical.
3