[Solution] $\tkcth{7.6.}\tkcf{6}$_Ch_7.6_$\tkco{eg\_…

Consider the set S={x}\mathbb{S} = \big\{x\big\}S={x} consisting of only one element: xxx. Define addition ⊕\oplus⊕ and scalar multiplication ⊗\otimes⊗ on S\mathbb{S}S by:
x⊕x=xandk⊗x=x,
x\oplus x = x \quad \textrm{and} \quad k \otimes x = x,
x⊕x=xandk⊗x=x,
for all k ∈ Rk \, \in \, \mathbb{R}k∈R. Determine whether or not S\mathbb{S}S, along with these two operations, is a vector space.  

(u⃗⊕v⃗)∈S\bcb{(\bco{\vu} \oplus \bcth{\vv}) \in \bcb{
\mathbb{
S
}
}}(u⊕v)∈S, for all u⃗\bco{\vu}u,v⃗ ∈ S\bcth{\vv} \bcb{\, \in \,} \bcb{
%
\mathbb{
S
}
%
}v∈S. 

u⃗⊕v⃗=v⃗⊕u⃗\bcb{\bco{\vu} \oplus \bcth{\vv} = \bcth{\vv} \oplus \bco{\vu}}u⊕v=v⊕u, ∀ u⃗, v⃗∈S\bcb{\forall \, \bco{\vu},\, \bcth{\vv} \in \bcb{
\mathbb{
S
}
}}∀u,v∈S

(u⃗⊕v⃗)⊕w⃗=u⃗⊕(v⃗⊕w⃗), ∀ u⃗, v⃗, w⃗ ∈ S\bcb{(\bco{\vu} \oplus \bcth{\vv}) \oplus \bco{\color{Purple}\vw} = \bco{\vu} \oplus (\bcth{\vv} \oplus \bco{\color{Purple}\vw}), \ \forall \ \bco{\vu},\ \bcth{\vv},\ \bco{\color{Purple} \vw} \, \in \, \bcb{
%
\mathbb{
S
}
%
}
}(u⊕v)⊕w=u⊕(v⊕w), ∀ u, v, w∈S

∃ 0⃗∈S : 0⃗⊕v⃗=v⃗,∀  v⃗∈S\bcb{\exists \, \bcf{\color{orange}\vz} \in \bcb{
%
\mathbb{
S
}%
}
 \, : \, \bcf{\color{orange}\vz} \oplus \bcth{\vv} = \bcth{\vv}, \quad \forall \; \bcth{\vv} \in 
%
\mathbb{
S
}
%
}∃0∈S:0⊕v=v,∀v∈S

∃ −v⃗∈S : v⃗⊕(−v⃗)=0⃗∀  v⃗∈S\bcb{\exists \, \bct{-\vv} \in \bcb{
%
\mathbb{
S
}%
} \, : \, \bcth{\vv} \oplus (\bct{-\vv}) = \bcf{\color{orange}\vz} \quad \forall \; \bcth{\vv} \in \bcb{
%
\mathbb{
S
}%
}
}∃−v∈S:v⊕(−v)=0∀v∈S

(k⊙v⃗)∈S\bcb{(\bco{\color{Magenta}k} \odot \bcth{\vv}) \in \bcb{
%
\mathbb{
S
}%
}
}(k⊙v)∈S, for all k∈R\bcb{\bco{\color{Magenta} k} \in \mathbb{R} }k∈R, and all v⃗∈S\bcb{ \bcth{\vv} \in \bcb{
%
\mathbb{
S
}%
}}v∈S

k⊙(u⃗⊕v⃗)=(k⊙u⃗)⊕(k⊙v⃗)\bcb{\bco{\color{Magenta}k}\odot(\bco{\vu} \oplus \bcth{\vv}) = (\bco{\color{Magenta}k}\odot \bco{\vu}) \oplus (\bco{\color{Magenta}k}\odot \bcth{\vv})}k⊙(u⊕v)=(k⊙u)⊕(k⊙v), for all k∈R\bcb{\bco{\color{Magenta} k} \in \mathbb{R} }k∈R, and all u⃗, v⃗∈S\bcb{ \bco{\vu},\ \bcth{\vv} \in \bcb{
%
\mathbb{
S
}%
}}u, v∈S

(k+c)⊙v⃗=(k⊙v⃗)⊕(c⊙v⃗)\bcb{(\bco{\color{Magenta}k} + \bco{\color{Brown}c})\odot \bcth{\vv} = (\bco{\color{Magenta}k}\odot \bcth{\vv}) \oplus (\bco{\color{Brown}c}\odot \bcth{\vv})}(k+c)⊙v=(k⊙v)⊕(c⊙v), for all k, c∈R\bcb{\bco{\color{Magenta} k},\ {\color{Brown}c} \in \mathbb{R} }k, c∈R, and all v⃗∈S\bcb{ \bcth{\vv} \in \bcb{
%
\mathbb{
S
}%
}}v∈S

(k⋅c)⊙v⃗=k⊙(c⊙v⃗)\bcb{(\bco{\color{Magenta}k}\cdot \bco{\color{Brown}c})\odot \bcth{\vv} = \bco{\color{Magenta}k}\odot( \bco{\color{Brown}c} \odot \bcth{\vv})}(k⋅c)⊙v=k⊙(c⊙v), for all k, c∈R\bcb{\bco{\color{Magenta} k},\ {\color{Brown}c} \in \mathbb{R} }k, c∈R, and all v⃗∈S\bcb{ \bcth{\vv} \in \bcb{
%
\mathbb{
S
}%
}}v∈S

1⊙v⃗=v⃗\bcb{1\odot \bcth{\vv} = \bcth{\vv}}1⊙v=v, for all v⃗∈S\bcb{ \bcth{\vv} \in \bcb{
%
\mathbb{
S
}%
}}v∈S, and 1∈R\bcb{1 \in \mathbb{R}}1∈R.

S\bcb{
%
\mathbb{
S
}
%
}S, with the given operations, is a vector space.

S\bcb{
%
\mathbb{
S
}
%
}S, with the given operations is not a vector space.

I don't know

$\tkcth{7.6.}\tkcf{14}$_Ch_7.6_$\tkco{eg\_14}$_133_$\key{F.}$Builder_Vector_Spaces_$\tkct{}$_20.1W

Determine whether the set M={[a00b] : a, b ∈ R}\mathbb{M} = \Bigg\{ \sm{a}{0}{0}{b}  \, : \, a,\, b \, \in \, \mathbb{R}\Bigg\}M={[a0​0b​]: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)v⃗=−(kv⃗)=k(−v⃗)=(−1)kv⃗(-k)\vv = -(k\vv) = k(-\vv) = (-1)k\vec{v}(−k)v=−(kv)=k(−v)=(−1)kv, for all k ∈ Rk \, \in \, \mathbb{R}k∈R and all v⃗ ∈ 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)=(ad+bc, bd)(a,b) \oplus (c,d) = (ad+bc,\, bd)(a,b)⊕(c,d)=(ad+bc,bd), 
and define scalar multiplication by

Consider the set of all linear maps that consist of rotations of R2\mathbb{R}^2R2 (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 ∈ Rk \, \in \, \mathbb{R}k∈R, k0⃗=0⃗k\vz = \vzk0=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)v⃗=−v⃗(-1) \vv = -\vv(−1)v=−v for all v⃗∈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:
(u1,u2)+(v1,v2)=(u1+v1,u2+v2)(u_1,u_2)+(v_1,v_2)=(u_1+v_1,u_2+v_2)(u1​,u2​)+(v1​,v2​)=(u1​+v1​,u2​+v2​)
k⋅(v1,v2)=(k2v1,kv2)k\cdot(v_1,v_2)=(k^2v_1,kv_2)k⋅(v1​,v2​)=(k2v1​,kv2​)

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:
⟨u1,u2,u3⟩+⟨v1,v2,v3⟩=⟨u1v1, u2v2, u3+v3⟩\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⟨u1​,u2​,u3​⟩+⟨v1​,v2​,v3​⟩=⟨u1​v1​, u2​v2​, u3​+v3​⟩

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:
(u1,u2,u3)+(v1,v2,v3)=(u1v1,u2v2,u3+v3)(u_1,u_2,u_3)+(v_1,v_2,v_3)=(u_1v_1,u_2v_2,u_3+v_3)(u1​,u2​,u3​)+(v1​,v2​,v3​)=(u1​v1​,u2​v2​,u3​+v3​)
k⋅(v1,v2,v3)=(v1k,v2k,kv3)k\cdot(v_1,v_2,v_3)=(v_1^k,v_2^k,kv_3)k⋅(v1​,v2​,v3​)=(v1k​,v2k​,kv3​)

Let Z={x⃗=(x)  ∣  x>3} Z=\{\vec x=(x)\;|\;x>3\}Z={x=(x)∣x>3} with the following operations:
u⃗+v⃗=uv−3(u+v)+12\vec u+\vec v=uv-3(u+v)+12u+v=uv−3(u+v)+12
k⋅v⃗=(v−3)a+3k\cdot \vec v = (v-3)^a+3k⋅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)andk⊙f(x)=(kf)(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)andk⊙f(x)=(kf)(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 x⃗\vxx and y⃗\vec{y}y​ in S\mathbb{S}S with the property x⃗+v⃗=v⃗\vx + \vv = \vvx+v=v and y⃗+v⃗=v⃗\vec{y}+\vv = \vvy​+v=v for all v⃗ ∈ S\vv \, \in \, \mathbb{S}v∈S, then show that this implies x⃗=y⃗\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 v⃗ ∈S\vv \, \in \mathbb{S}v∈S, suppose there are two vectors x⃗\vxx and y⃗\vec{y}y​ in S\mathbb{S}S with the property v⃗+x⃗=0⃗\vv + \vx  = \vzv+x=0 and v⃗+y⃗=0⃗\vv + \vec{y} = \vzv+y​=0;  show that this implies x⃗=y⃗\vx = \vec{y}x=y​.

Let V=R2V=\mathbb{R}^2V=R2 but with the defined the operations:
u⃗+v⃗=(u1+v2,u2+v1)\vec u + \vec v = (u_1+v_2,u_2+v_1)u+v=(u1​+v2​,u2​+v1​)
In this space the zero vector is 0⃗=(0,0)\vec 0 = (0,0)\qquad0=(0,0) (can you prove that?)

Let V={(u1,u2)  ∣  u1>0}V=\{(u_1,u_2)\;|\;u_1>0\}V={(u1​,u2​)∣u1​>0} and define the operation:
u⃗+v⃗=(u1v1,u2+v2)\vec u + \vec v = (u_1v_1,u_2+v_2)u+v=(u1​v1​,u2​+v2​)
If this set satisfies the vector space axioms, what is the zero vector in this space?

Abstract vector spaces

Let V={x⃗=(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 VVV 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:
x⃗+y⃗=xy−3(x+y)+12\vec x + \vec y = xy-3(x+y)+12x+y​=xy−3(x+y)+12

Practice: Vector Properties
Prove the following vector property:
u⃗+v⃗=v⃗+u⃗\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:
(u1,u2,u3)+(v1,v2,v3)=(u1v1,u2v2,u3+v3)(u_1,u_2,u_3)+(v_1,v_2,v_3)=(u_1v_1,u_2v_2,u_3+v_3)(u1​,u2​,u3​)+(v1​,v2​,v3​)=(u1​v1​,u2​v2​,u3​+v3​)
k⋅(v1,v2,v3)=(v1k,v2k,kv3)k\cdot(v_1,v_2,v_3)=(v_1^k,v_2^k,kv_3)k⋅(v1​,v2​,v3​)=(v1k​,v2k​,kv3​)

Let Z={x⃗=(x)  ∣  x>3}Z=\{\vec x = (x)\;|\;x>3\}Z={x=(x)∣x>3} with the following operations:
u⃗+v⃗=uv−3(u+v)+12\vec u+\vec v=uv-3(u+v)+12u+v=uv−3(u+v)+12
k⋅v⃗=(v−3)a+3k\cdot \vec v = (v-3)^a+3k⋅v=(v−3)a+3

Consider the set R2\mathbb{R}^2R2 where the addition operator ⊕\oplus⊕ is defined to be
(v1,w1)⊕(v2,w2)=(w1+w2,v1+v2)(v_1, w_1) \oplus (v_2, w_2) = (w_1 + w_2, v_1 + v_2)(v1​,w1​)⊕(v2​,w2​)=(w1​+w2​,v1​+v2​)
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:
⟨u1,u2,u3⟩+⟨v1,v2,v3⟩=⟨u1v1, u2v2, u3+v3⟩\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⟨u1​,u2​,u3​⟩+⟨v1​,v2​,v3​⟩=⟨u1​v1​, u2​v2​, u3​+v3​⟩

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:
(u1,u2)+(v1,v2)=(u1+v1+1, u2+v2+1)(u_1,u_2)+(v_1,v_2)=(u_1+v_1+1,\ u_2+v_2+1)(u1​,u2​)+(v1​,v2​)=(u1​+v1​+1, u2​+v2​+1)

$\tkcth{7.6.}\tkcf{6}$_Ch_7.6_$\tkco{eg\_6}$_133_$\key{F.}$Builder_Vector_Space…

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