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Let V={(u_1,u_2);|;u_1>0} and define the operation: u + v = (u_1v_1,u_2+v_2) …
Related Topics
Wize University Linear Algebra Textbook > Vector Spaces and Subspaces
Vector Spaces
5 Activities
Let
V
=
{
(
u
1
,
u
2
)
∣
u
1
>
0
}
V=\{(u_1,u_2)\;|\;u_1>0\}
V
=
{(
u
1
,
u
2
)
∣
u
1
>
0
}
and define the operation:
u
⃗
+
v
⃗
=
(
u
1
v
1
,
u
2
+
v
2
)
\vec u + \vec v = (u_1v_1,u_2+v_2)
u
+
v
=
(
u
1
v
1
,
u
2
+
v
2
)
If this set satisfies the vector space axioms, what is the zero vector in this space?
(
1
,
0
)
(1,0)
(
1
,
0
)
(
0
,
1
)
(0,1)
(
0
,
1
)
(
1
,
1
)
(1,1)
(
1
,
1
)
(
0
,
0
)
(0,0)
(
0
,
0
)
(
−
1
,
−
1
)
(-1,-1)
(
−
1
,
−
1
)
I don't know
Check Submission
More Vector Spaces Questions:
$\tkcth{7.6.}\tkcf{6}$_Ch_7.6_$\tkco{eg\_6}$_133_$\key{F.}$Builder_Vector_Spaces_$\tkct{}$_20.1W
Consider the set
S
=
{
x
}
\mathbb{S} = \big\{x\big\}
S
=
{
x
}
consisting of only one element:
x
x
x
. Define addition
⊕
\oplus
⊕
and scalar multiplication
⊗
\otimes
⊗
on
S
\mathbb{S}
S
by:
x
⊕
x
=
x
and
k
⊗
x
=
x
,
x\oplus x = x \quad \textrm{and} \quad k \otimes x = x,
x
⊕
x
=
x
and
k
⊗
x
=
x
,
for all
k
∈
R
k \, \in \, \mathbb{R}
k
∈
R
. Determine whether or not
S
\mathbb{S}
S
, along with these two operations, is a vector space.
$\tkcth{7.6.}\tkcf{14}$_Ch_7.6_$\tkco{eg\_14}$_133_$\key{F.}$Builder_Vector_Spaces_$\tkct{}$_20.1W
Determine whether the set
M
=
{
[
a
0
0
b
]
:
a
,
b
∈
R
}
\mathbb{M} = \Bigg\{ \sm{a}{0}{0}{b} \, : \, a,\, b \, \in \, \mathbb{R}\Bigg\}
M
=
{
[
a
0
0
b
]
:
a
,
b
∈
R
}
, along with the usual matrix addition and scalar multiplication, constitutes a vector space over
R
\mathbb{R}
R
.
$\tkcth{7.6.}\tkcf{13}$_Ch_7.6_$\tkco{eg\_13}$_133_$\key{F.}$Builder_Vector_Spaces_$\tkct{}$_20.1W
Let
S
\mathbb{S}
S
be a vector space. Prove that
(
−
k
)
v
⃗
=
−
(
k
v
⃗
)
=
k
(
−
v
⃗
)
=
(
−
1
)
k
v
⃗
(-k)\vv = -(k\vv) = k(-\vv) = (-1)k\vec{v}
(
−
k
)
v
=
−
(
k
v
)
=
k
(
−
v
)
=
(
−
1
)
k
v
, for all
k
∈
R
k \, \in \, \mathbb{R}
k
∈
R
and all
v
⃗
∈
S
\vv \, \in \, \mathbb{S}
v
∈
S
.
$\tkcth{7.6.}\tkcf{15}$_Ch_7.6_$\tkco{eg\_15}$_133_$\key{F.}$Builder_Vector_Spaces_$\tkct{}$_20.1W
Let
V
=
{
(
a
,
b
)
:
a
,
b
∈
R
,
b
>
0
}
\mathbb{V} = \big\{ (a,b) \, : \, a,\,b \, \in \, \mathbb{R}, \, b > 0 \big\}
V
=
{
(
a
,
b
)
:
a
,
b
∈
R
,
b
>
0
}
, and define vector addition by:
(
a
,
b
)
⊕
(
c
,
d
)
=
(
a
d
+
b
c
,
b
d
)
(a,b) \oplus (c,d) = (ad+bc,\, bd)
(
a
,
b
)
⊕
(
c
,
d
)
=
(
a
d
+
b
c
,
b
d
)
,
and define scalar multiplication by
Consider the set of all linear maps that consist of rotations of
R
2
\mathbb{R}^2
R
2
(recall that a rotation by some angle is a linear map). Is this a vector space, and if so, what is the dimension of this vector space?
$\tkcth{7.6.}\tkcf{10}$_Ch_7.6_$\tkco{eg\_10}$_133_$\key{F.}$Builder_Vector_Spaces_$\tkct{}$_20.1W
Let
S
\mathbb{S}
S
be a vector space. Prove that for all
k
∈
R
k \, \in \, \mathbb{R}
k
∈
R
,
k
0
⃗
=
0
⃗
k\vz = \vz
k
0
=
0
.
$\tkcth{7.6.}\tkcf{12}$_Ch_7.6_$\tkco{eg\_12}$_133_$\key{F.}$Builder_Vector_Spaces_$\tkct{}$_20.1W
Let
S
\mathbb{S}
S
be a vector space. Prove that
(
−
1
)
v
⃗
=
−
v
⃗
(-1) \vv = -\vv
(
−
1
)
v
=
−
v
for all
v
⃗
∈
S
\vv \in \mathbb{S}
v
∈
S
.
Let
V
=
{
(
x
,
y
)
∣
x
,
y
∈
R
}
V=\{(x,y)\;|\;x,y\in\mathbb{R}\}
V
=
{(
x
,
y
)
∣
x
,
y
∈
R
}
and define the following operations:
(
u
1
,
u
2
)
+
(
v
1
,
v
2
)
=
(
u
1
+
v
1
,
u
2
+
v
2
)
(u_1,u_2)+(v_1,v_2)=(u_1+v_1,u_2+v_2)
(
u
1
,
u
2
)
+
(
v
1
,
v
2
)
=
(
u
1
+
v
1
,
u
2
+
v
2
)
k
⋅
(
v
1
,
v
2
)
=
(
k
2
v
1
,
k
v
2
)
k\cdot(v_1,v_2)=(k^2v_1,kv_2)
k
⋅
(
v
1
,
v
2
)
=
(
k
2
v
1
,
k
v
2
)
Vector Spaces
Practice: Finding the Additive Identity and Additive Inverse
Let
W
=
{
⟨
x
,
y
,
z
⟩
∣
x
,
y
>
0
,
z
∈
R
}
W=\{\lang x,y,z \rang\;|\;x,y>0,z\in\mathbb{R}\}
W
=
{⟨
x
,
y
,
z
⟩
∣
x
,
y
>
0
,
z
∈
R
}
with the following operations:
⟨
u
1
,
u
2
,
u
3
⟩
+
⟨
v
1
,
v
2
,
v
3
⟩
=
⟨
u
1
v
1
,
u
2
v
2
,
u
3
+
v
3
⟩
\lang u_1,u_2,u_3 \rang+\lang v_1,v_2,v_3 \rang=\lang u_1v_1,\ u_2v_2,\ u_3+v_3\rang
⟨
u
1
,
u
2
,
u
3
⟩
+
⟨
v
1
,
v
2
,
v
3
⟩
=
⟨
u
1
v
1
,
u
2
v
2
,
u
3
+
v
3
⟩
Let
W
=
{
(
x
,
y
,
z
)
∣
x
,
y
>
0
,
z
∈
R
}
W=\{(x,y,z)\;|\;x,y>0,z\in\mathbb{R}\}
W
=
{(
x
,
y
,
z
)
∣
x
,
y
>
0
,
z
∈
R
}
with the following operations:
(
u
1
,
u
2
,
u
3
)
+
(
v
1
,
v
2
,
v
3
)
=
(
u
1
v
1
,
u
2
v
2
,
u
3
+
v
3
)
(u_1,u_2,u_3)+(v_1,v_2,v_3)=(u_1v_1,u_2v_2,u_3+v_3)
(
u
1
,
u
2
,
u
3
)
+
(
v
1
,
v
2
,
v
3
)
=
(
u
1
v
1
,
u
2
v
2
,
u
3
+
v
3
)
k
⋅
(
v
1
,
v
2
,
v
3
)
=
(
v
1
k
,
v
2
k
,
k
v
3
)
k\cdot(v_1,v_2,v_3)=(v_1^k,v_2^k,kv_3)
k
⋅
(
v
1
,
v
2
,
v
3
)
=
(
v
1
k
,
v
2
k
,
k
v
3
)
Let
Z
=
{
x
⃗
=
(
x
)
∣
x
>
3
}
Z=\{\vec x=(x)\;|\;x>3\}
Z
=
{
x
=
(
x
)
∣
x
>
3
}
with the following operations:
u
⃗
+
v
⃗
=
u
v
−
3
(
u
+
v
)
+
12
\vec u+\vec v=uv-3(u+v)+12
u
+
v
=
uv
−
3
(
u
+
v
)
+
12
k
⋅
v
⃗
=
(
v
−
3
)
a
+
3
k\cdot \vec v = (v-3)^a+3
k
⋅
v
=
(
v
−
3
)
a
+
3
$\tkcth{7.6.}\tkcf{3}$_Ch_7.6_$\tkco{eg\_3}$_133_$\key{F.}$Builder_Vector_Spaces_$\tkct{}$_20.1W
Consider the set
F
[
0
,
1
]
=
{
f
:
[
0
,
1
]
→
R
:
f
is a function
}
\mathcal{F}[0,1] = \big\{ f \, : \, [ 0,\, 1] \, \to \, \mathbb{R} \, :\, f \textrm{ is a function } \big\}
F
[
0
,
1
]
=
{
f
:
[
0
,
1
]
→
R
:
f
is a function
}
, with addition and scalar multiplication defined by:
f
⊕
g
=
(
f
+
g
)
(
x
)
=
f
(
x
)
+
g
(
x
)
and
k
⊙
f
(
x
)
=
(
k
f
)
(
x
)
=
k
(
f
(
x
)
)
f\oplus g = (f+g)(x) = f(x) + g(x) \quad \textrm{and} \quad k\odot f(x) = (kf)(x) = k\big(f(x)\big)
f
⊕
g
=
(
f
+
g
)
(
x
)
=
f
(
x
)
+
g
(
x
)
and
k
⊙
f
(
x
)
=
(
k
f
)
(
x
)
=
k
(
f
(
x
)
)
Determine whether or not
F
[
0
,
1
]
\mathcal{F}[0,\,1]
F
[
0
,
1
]
is a vector space.
$\tkcth{7.6.}\tkcf{8}$_Ch_7.6_$\tkco{eg\_8}$_133_$\key{F.}$Builder_Vector_Spaces_$\tkct{}$_20.1W
Let
S
\mathbb{S}
S
be a vector space. Prove that the zero vector is unique. That is, suppose there are two vectors
x
⃗
\vx
x
and
y
⃗
\vec{y}
y
in
S
\mathbb{S}
S
with the property
x
⃗
+
v
⃗
=
v
⃗
\vx + \vv = \vv
x
+
v
=
v
and
y
⃗
+
v
⃗
=
v
⃗
\vec{y}+\vv = \vv
y
+
v
=
v
for all
v
⃗
∈
S
\vv \, \in \, \mathbb{S}
v
∈
S
, then show that this implies
x
⃗
=
y
⃗
\vx = \vec{y}
x
=
y
.
$\tkcth{7.6.}\tkcf{9}$_Ch_7.6_$\tkco{eg\_9}$_133_$\key{F.}$Builder_Vector_Spaces_$\tkct{}$_20.1W
Let
S
\mathbb{S}
S
be a vector space. Prove that every vector has a
unique
negative.
That is, for
v
⃗
∈
S
\vv \, \in \mathbb{S}
v
∈
S
, suppose there are two vectors
x
⃗
\vx
x
and
y
⃗
\vec{y}
y
in
S
\mathbb{S}
S
with the property
v
⃗
+
x
⃗
=
0
⃗
\vv + \vx = \vz
v
+
x
=
0
and
v
⃗
+
y
⃗
=
0
⃗
\vv + \vec{y} = \vz
v
+
y
=
0
; show that this implies
x
⃗
=
y
⃗
\vx = \vec{y}
x
=
y
.
Let
V
=
R
2
V=\mathbb{R}^2
V
=
R
2
but with the defined the operations:
u
⃗
+
v
⃗
=
(
u
1
+
v
2
,
u
2
+
v
1
)
\vec u + \vec v = (u_1+v_2,u_2+v_1)
u
+
v
=
(
u
1
+
v
2
,
u
2
+
v
1
)
In this space the zero vector is
0
⃗
=
(
0
,
0
)
\vec 0 = (0,0)\qquad
0
=
(
0
,
0
)
(can you prove that?)
Abstract vector spaces
Let
V
=
{
x
⃗
=
(
x
)
∣
x
∈
R
,
x
>
3
}
V=\{\vec x=(x)\;\big|\;x\in\mathbb{R},x>3\}
V
=
{
x
=
(
x
)
x
∈
R
,
x
>
3
}
, that is
V
V
V
is the set of all
vectors
that are real numbers in the
interval
(
3
,
∞
)
(3,\infty)
(
3
,
∞
)
Define the following operations of vector addition and scalar multiplication:
x
⃗
+
y
⃗
=
x
y
−
3
(
x
+
y
)
+
12
\vec x + \vec y = xy-3(x+y)+12
x
+
y
=
x
y
−
3
(
x
+
y
)
+
12
Practice: Vector Properties
Prove the following vector property:
u
⃗
+
v
⃗
=
v
⃗
+
u
⃗
\vec{u}+\vec{v}=\vec{v}+\vec{u}
u
+
v
=
v
+
u
Let
W
=
{
(
x
,
y
,
z
)
∣
x
,
y
>
0
,
z
∈
R
}
W=\{(x,y,z)\;|\;x,y>0,z\in\mathbb{R}\}
W
=
{(
x
,
y
,
z
)
∣
x
,
y
>
0
,
z
∈
R
}
with the following operations:
(
u
1
,
u
2
,
u
3
)
+
(
v
1
,
v
2
,
v
3
)
=
(
u
1
v
1
,
u
2
v
2
,
u
3
+
v
3
)
(u_1,u_2,u_3)+(v_1,v_2,v_3)=(u_1v_1,u_2v_2,u_3+v_3)
(
u
1
,
u
2
,
u
3
)
+
(
v
1
,
v
2
,
v
3
)
=
(
u
1
v
1
,
u
2
v
2
,
u
3
+
v
3
)
k
⋅
(
v
1
,
v
2
,
v
3
)
=
(
v
1
k
,
v
2
k
,
k
v
3
)
k\cdot(v_1,v_2,v_3)=(v_1^k,v_2^k,kv_3)
k
⋅
(
v
1
,
v
2
,
v
3
)
=
(
v
1
k
,
v
2
k
,
k
v
3
)
Let
Z
=
{
x
⃗
=
(
x
)
∣
x
>
3
}
Z=\{\vec x = (x)\;|\;x>3\}
Z
=
{
x
=
(
x
)
∣
x
>
3
}
with the following operations:
u
⃗
+
v
⃗
=
u
v
−
3
(
u
+
v
)
+
12
\vec u+\vec v=uv-3(u+v)+12
u
+
v
=
uv
−
3
(
u
+
v
)
+
12
k
⋅
v
⃗
=
(
v
−
3
)
a
+
3
k\cdot \vec v = (v-3)^a+3
k
⋅
v
=
(
v
−
3
)
a
+
3
Consider the set
R
2
\mathbb{R}^2
R
2
where the addition operator
⊕
\oplus
⊕
is defined to be
(
v
1
,
w
1
)
⊕
(
v
2
,
w
2
)
=
(
w
1
+
w
2
,
v
1
+
v
2
)
(v_1, w_1) \oplus (v_2, w_2) = (w_1 + w_2, v_1 + v_2)
(
v
1
,
w
1
)
⊕
(
v
2
,
w
2
)
=
(
w
1
+
w
2
,
v
1
+
v
2
)
and scalar multiplication is
Vector Space and Subspace
Consider the vector space,
C
(
[
0
,
1
]
)
\mathcal{C}([0, 1])
C
([
0
,
1
])
, the space of continuous functions on the interval
[
0
,
1
]
[0, 1]
[
0
,
1
]
. Is this vector space finite-dimensional, or infinite-dimensional? Why?
Vector Spaces
Practice: Finding the Additive Identity and Additive Inverse
Let
W
=
{
⟨
x
,
y
,
z
⟩
∣
x
,
y
>
0
,
z
∈
R
}
W=\{\lang x,y,z \rang\;|\;x,y>0,z\in\mathbb{R}\}
W
=
{⟨
x
,
y
,
z
⟩
∣
x
,
y
>
0
,
z
∈
R
}
with the following operations:
⟨
u
1
,
u
2
,
u
3
⟩
+
⟨
v
1
,
v
2
,
v
3
⟩
=
⟨
u
1
v
1
,
u
2
v
2
,
u
3
+
v
3
⟩
\lang u_1,u_2,u_3 \rang+\lang v_1,v_2,v_3 \rang=\lang u_1v_1,\ u_2v_2,\ u_3+v_3\rang
⟨
u
1
,
u
2
,
u
3
⟩
+
⟨
v
1
,
v
2
,
v
3
⟩
=
⟨
u
1
v
1
,
u
2
v
2
,
u
3
+
v
3
⟩
Vector Spaces
Practice: Verifying Vector Space Axioms
Let
V
=
{
(
x
,
y
)
∣
x
,
y
∈
R
}
V=\{(x,y)\;|\;x,y\in\mathbb{R}\}
V
=
{(
x
,
y
)
∣
x
,
y
∈
R
}
and define the following operations:
(
u
1
,
u
2
)
+
(
v
1
,
v
2
)
=
(
u
1
+
v
1
+
1
,
u
2
+
v
2
+
1
)
(u_1,u_2)+(v_1,v_2)=(u_1+v_1+1,\ u_2+v_2+1)
(
u
1
,
u
2
)
+
(
v
1
,
v
2
)
=
(
u
1
+
v
1
+
1
,
u
2
+
v
2
+
1
)