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Economics 204
Problem Set 4 Solutions
Exercise 1
a) First note that
S
is a subset of
R
3
;
which is a vector space (over
R
;
with the operations assumed). Hence, all we have to show is that
0
vector is
contained in
S
and that
8
±
2
R
, and
x;y
2
S
, we have
+
±y
2
S
. But
this is pretty obvious:
i
)
take
c
= 0
to show that
0
2
S
;
ii
)
if
x
=
c
1
v
and
y
=
c
2
v
, then
+
±y
= (
1
+
±c
2
)
v
and if we let
c
=
1
+
±c
2
then it follows
that
+
±y
2
S:
The space is one dimensional, and
f
v
g
is a basis for
S
.
b) Same argument applies here:
i
) 0
vector is obviously in
S
.
ii
)
Now take
±
2
R
and
x;y
2
S
; let
z
:=
+
±y
, then
z
1
+
z
2
+
z
3
= (
1
+
±y
1
) +
(
2
+
±y
2
) + (
3
+
±y
3
) =
(
x
1
+
x
2
+
x
3
) +
±
(
y
1
+
y
2
+
y
3
) =
0 +
±
0 = 0;
and
z
1
+ 2
z
2
= (
1
+
±y
1
) + 2(
2
+
±y
2
) = (
1
+ 2
2
) + (
1
+ 2
±y
2
) =
(
x
1
+ 2
x
2
) +
±
(
y
1
+ 2
y
2
) = 0
. The space is again one dimensional (Note that
x
2
, then
x
1
and
x
3
are determined).
f
(1
;
1
;
0)
g
is a basis for
S
.
c)
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This note was uploaded on 10/04/2010 for the course ECON 204 taught by Professor Anderson during the Summer '08 term at University of California, Berkeley.
 Summer '08
 ANDERSON
 Economics

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