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ECE 130C Homework 4 Solution
May 3, 2007
2
.
3
.
14
The dimension of
S
is
(a)
0
when
x
= 0 (b) 1
when
x
= (1
,
1
,
1
,
1)
(c) 3
when
x
= (1
,
1
,

1
,

1) because all arrangements of
this
x
are perpendicular to (1
,
1
,
1
,
1)
(d) 4
when the
x
’s are not equal and
don’t add to zero. No
x
gives dim
S
=2.
2
.
3
.
26
(a)
True. (b)
False because the basis vectors may not be in
S
.
2
.
3
.
34
If
v
1
, v
2
, v
3
is a basis for
V
, and
w
1
, w
2
, w
3
is a basis for
W
,
then these six vectors cannot be independent and some combination is zero:
∑
c
i
v
i
+
∑
d
i
w
i
= 0. This puts
∑
c
i
v
i
=

∑
d
i
w
i
in both subspaces.
2
.
3
.
36
n

r
= 17

11 = 6 is nullspace dimension; 64

11 = 53 is
dim
N
(
A
T
)
2
.
3
.
40
(a)
y
(
x
) =
e
2
x
(b)
y
=
x
(one basis vector in each case).
2
.
4
.
28
1
1
0
2
1
0
"
1
0
1
1
2
0
#
=
2
2
1
2
4
0
1
0
1
;
r
+ (
n

r
) =
n
= 3 but
2 + 2 is 4.
2
.
4
.
32
Row space =
yz
plane; column space =
xy
plane; nullspace =
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 Summer '10
 VolkanRodoplu

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