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Linear Algebra
Geometry: solutions 4
1.
We consider
k
v
+
w
k
2
= (
v
+
w
)
·
(
v
+
w
)
=
v
·
v
+ 2
v
·
w
+
w
·
w
=
k
v
k
2
+
k
w
k
2
+ 2
v
·
w
Now we use Cauchy Schwarz
v
·
w
≤ 
v
·
w
 ≤ k
v
kk
w
k
and obtain
k
v
+
w
k
2
≤ k
v
k
2
+
k
w
k
2
+ 2
k
v
kk
w
k
= (
k
v
k
+
k
w
k
)
2
and taking the square root gives the triangle inequality.
2.
Let us choose
v
= (
a
1
,a
2
,
···
,a
N
)
∈
R
N
and
w
= (1
/N,
1
/N,
···
,
1
/N
)
∈
R
N
, then
v
·
w
=
a
1
+
a
2
+
···
+
a
N
N
,
and
k
v
k
2
=
a
2
1
+
a
2
2
+
···
+
a
2
N
,
k
w
k
2
= 1
/N
2
+ 1
/N
2
+
···
+ 1
/N
2
=
N/N
2
= 1
/N .
So using the Cauchy Schwarz inequality in the form (
v
·
w
)
2
≤ k
v
k
2
k
w
k
2
gives
±
a
1
+
a
2
+
···
+
a
N
N
²
2
≤
a
2
1
+
a
2
2
+
···
+
a
2
N
N
.
3.
V
is a subspace if
(i)
v
+
w
∈
V
for all
v
,
w
∈
V
(ii)
λ
v
∈
V
for all
λ
∈
R
and
v
∈
V
(a) The set of all vectors of the form
x
0
0
, with
x
∈
R
.
This is a subspace since: (i)
x
0
0
+
y
0
0
=
x
+
y
0
0
is of the same form and
(ii)
λ
x
0
0
=
λx
0
0
, too.
1
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View Full Document(b) The set of all vectors of the form
x
1
1
, with
x
∈
R
.
This is not a subspace since for instance 2
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 Spring '10
 Spyros
 Linear Algebra, Algebra, Geometry

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