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Wize University Linear Algebra Textbook > Vector Spaces and Subspaces

Vector Spaces

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Vector Spaces

A vector space VVconsists of:
  1. A non-empty set of objects called vectors
  2. Two operations that must satisfy 10 axioms (assumptions):
  3. vector addition (+)\left(+\right)
  4. scalar multiplication ( ⋅ )(\ \cdot\ )
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Vector Space Axioms

The operations ++ and ⋅\cdot can be defined in any way as long as they satisfy the following:
  • Closure
  • u⃗+v⃗\vec u + \vec v must be in VV
  • k⋅v⃗k\cdot \vec v must be in VV
  • Associativity
  • (u⃗+v⃗)+w⃗ = u⃗+(v⃗+w⃗)(\vec u + \vec v) + \vec w \ =\ \vec u + (\vec v + \vec w)
  • k⋅(m⋅v⃗) = (km)⋅v⃗k\cdot(m\cdot\vec v) \ =\ (km)\cdot\vec v
  • Distributivity
  • k⋅(u⃗+v⃗) = k⋅u⃗ + k⋅v⃗k\cdot(\vec u + \vec v) \ =\ k\cdot\vec u \ +\ k\cdot \vec v
  • (k+m)⋅v⃗ = k⋅v⃗ + m⋅v⃗(k+m)\cdot\vec v \ =\ k\cdot \vec v \ +\ m\cdot\vec v
  • Commutativity
  • u⃗+v⃗= v⃗+u⃗\vec u + \vec v = \ \vec v + \vec u
And the space VV must contain:
  • Additive Identity
  • A zero vector 0⃗\vec 0 such that: v⃗+0⃗ = 0⃗+v⃗ = v⃗\vec v + \vec 0 \ =\ \vec 0 + \vec v \ =\ \vec v
  • Additive Inverse
  • A negative vector −v⃗-\vec v such that: v⃗+(−v⃗) = 0⃗\vec v + (-\vec v)\ =\ \vec 0
  • Multiplicative Identity
  • 1⋅v⃗ = v⃗1\cdot\vec v \ =\ \vec v

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Real Vector Spaces

Vector spaces Rn\mathbb{R}^n use component-wise addition and scalar multiplication.

Example

R2:{⟨x,y⟩  ∣  x,y∈R}\mathbb{R}^2:\{\lang x,y \rang\;|\;x,y\in \mathbb{R}\} with:
  • u⃗+v⃗ = ⟨u1,u2⟩+⟨v1,v2⟩ = ⟨u1+v1,u2+v2⟩\vec u + \vec v \ =\ \lang u_1,u_2\rang +\lang v_1,v_2\rang\ =\ \lang u_1+v_1,u_2+v_2\rang
  • k⋅v⃗ = k⋅⟨v1,v2⟩ = ⟨kv1,kv2⟩k\cdot \vec v\ =\ k\cdot\lang v_1,v_2\rang\ =\ \lang kv_1,kv_2\rang
  • 0⃗=⟨0,0⟩\vec{0} = \lang 0,0 \rang
  • −v⃗=⟨−v1,−v2⟩-\vec v=\lang -v_1,-v_2 \rang
Show that 0⃗\vec 0 is the additive identity:
v⃗+0⃗=⟨v1,v2⟩+⟨0,0⟩=⟨v1+0,v2+0⟩=⟨v1,v2⟩=v⃗\vec v + \vec 0 = \lang v_1,v_2\rang +\lang 0,0\rang = \lang v_1+0,v_2+0\rang =\lang v_1,v_2\rang =\vec v
Show that −v⃗- \vec v is the additive inverse:
v⃗+(−v⃗)=⟨v1,v2⟩+⟨−v1,−v2⟩=⟨v1+(−v1), v2+(−v2)⟩=⟨0,0⟩=0⃗\vec v + (-\vec v) = \lang v_1,v_2\rang +\lang -v_1,-v_2\rang =\lang v_1+( -v_1) ,\ v_2+(-v_2) \rang =\lang 0,0\rang =\vec 0

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Abstract Vector Spaces

Abstract vector spaces are vector spaces that satisfy the 10 axioms, but they do not behave like real vector spaces.
Example

V={⟨x,y⟩  ∣  x,y∈R+}V=\{\lang x,y \rang \;|\;x,y \in \mathbb{R}^+\} with operations:
  • u⃗+v⃗=⟨u1,u2⟩+⟨v1,v2⟩=⟨u1v1,u2v2⟩\vec u + \vec v = \lang u_1,u_2\rang + \lang v_1,v_2\rang=\lang u_1v_1,u_2v_2\rang
  • k⋅v⃗=k⋅⟨v1,v2⟩=⟨v1k,v2k⟩k\cdot \vec v = k\cdot \lang v_1,v_2 \rang = \lang v_1^k,v_2^k \rang
  • 0⃗=⟨1,1⟩\vec{0}= \lang 1,1 \rang
  • −v⃗=−⟨v1,v2⟩=⟨1v1,1v2⟩-\vec v =-\lang v_1, v_2 \rang=\left \lang \dfrac{1}{v_1},\dfrac{1}{v_2}\right \rang
Show that 0⃗\vec 0 is the additive identity:
v⃗+0⃗=⟨v1,v2⟩+⟨1,1⟩=⟨v1⋅1, v2⋅1⟩=⟨v1,v2⟩=v⃗\vec v + \vec 0 = \lang v_1,v_2\rang +\lang 1,1\rang = \lang v_1\cdot1,\ v_2 \cdot 1\rang =\lang v_1,v_2\rang =\vec v
Show that −v⃗- \vec v is the additive inverse:
v⃗+⟨−v⃗⟩=⟨v1,v2⟩+⟨1v1,1v2⟩=⟨v1⋅1v1, v2⋅1v2⟩=⟨1,1⟩=0⃗\vec v + \lang -\vec v\rang =\lang v_1,v_2\rang + \left\lang \dfrac{1}{v_1},\dfrac{1}{v_2} \right\rang = \left\lang v_1\cdot\dfrac{1}{v_1},\ v_2\cdot\dfrac{1}{v_2}\right\rang =\lang 1,1\rang =\vec 0

Show that 11 is the multiplicative identity:
1⋅v⃗=1⋅⟨v1,v2⟩=⟨v11,v21⟩=⟨v1,v2⟩=v⃗1\cdot \vec v = 1\cdot \lang v_1, v_2 \rang= \lang {v_1}^1, {v_2}^1 \rang = \lang v_1, v_2 \rang = \vec v
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Example: Additive Identity and Additive Inverse

Let V={(x,y)  ∣  x,y∈R}V=\{(x,y)\;|\;x,y\in\mathbb{R}\} be a vector space with the following operations:
  • (u1,u2)+(v1,v2)=(u1+v1−1, u2+v2+3)(u_1,u_2)+(v_1,v_2)=(u_1+v_1-1,\ u_2+v_2+3)
  • k⋅(v1,v2)=(kv1−k+1, kv2+3k−3)k\cdot(v_1,v_2)=(kv_1-k+1,\ kv_2+3k-3)

Part A)

Find the zero vector in this vector space.
We are looking for a vector (a1,a2)(a_1,a_2) that satisfies the following equation:
(v1,v2)+(a1,a2)=(v1,v2)(v_1,v_2)+(a_1,a_2)=(v_1,v_2)
Using the defined operation of vector addition we write this as:
(v1+a1−1,v2+a2+3)=(v1,v2)(v_1+a_1-1,v_2+a_2+3)=(v_1,v_2)
These vectors are equal if and only if:
v1+a1−1=v1  ⟹  a1=1v_1+a_1-1=v_1 \implies a_1 = 1
v2+a2+3=v2  ⟹  a2=−3v_2+a_2+3=v_2 \implies a_2=-3
So in this vector space the zero vector is 0⃗=(1,−3)\boxed{\vec 0 = (1,-3)}.
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Part B)

Find the additive inverse, −v⃗-\vec v, of v⃗=(v1,v2)\vec v=(v_1,v_2).
We are looking for a vector (a1,a2)(a_1,a_2) that satisfies the following equation (using the zero vector found in Part A):
v⃗+(−v⃗)=0⃗  ⟹  (v1,v2)+(a1,a2)=(1,−3)\vec v + (-\vec v) = \vec 0\implies (v_1,v_2)+(a_1,a_2)=(1,-3)
Using the defined operation of vector addition, we write this as:
(v1+a1−1, v2+a2+3)=(1,−3)(v_1+a_1-1,\ v_2+a_2+3)=(1,-3)
These vectors are equal if and only if:
v1+a1−1=1  ⟹  a1=2−v1v_1+a_1-1=1 \implies a_1 = 2-v_1
v2+a2+3=−3  ⟹  a2=−6−v2v_2+a_2+3=-3 \implies a_2=-6-v_2
So in this vector space the additive inverse of v⃗\vec v is:
−(v1,v2)=(2−v1,−6−v2)\boxed{-(v_1,v_2)=(2-v_1,-6-v_2)}
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Example: Verifying Vector Space Axioms

Let V={(x,y)  ∣  x,y∈R}V=\{(x,y)\;|\;x,y\in\mathbb{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)
  • k⋅(v1,v2)=(2kv1,2kv2)k\cdot(v_1,v_2)=(2kv_1,2kv_2)
Determine which of the following vector space axioms are satisfied.
A) k⋅(u⃗+v⃗)=k⋅u⃗+k⋅v⃗k\cdot(\vec u + \vec v) = k\cdot\vec u + k\cdot\vec v
Let u⃗=(u1,u2)\vec u = (u_1,u_2) and v⃗=(v1,v2)\vec v = (v_1,v_2), then:
k⋅(u⃗+v⃗)=k⋅((u1,u2)+(v1,v2))=k⋅(u1+v2,u2+v2)=(2k(u1+v1),2k(u2+v2))k\cdot(\vec u + \vec v) = k\cdot((u_1,u_2)+(v_1,v_2)) = k\cdot(u_1+v_2,u_2+v_2) = \colorOne{(2k(u_1+v_1),2k(u_2+v_2))}
Meanwhile:
k⋅u⃗+k⋅v⃗=k⋅(u1,u2)+k⋅(v1,v2)=(2ku1,2ku2)+(2kv1,2kv2)=(2ku1+2kv1,2ku2+2kv2)k\cdot\vec u + k\cdot \vec v = k\cdot(u_1,u_2)+k\cdot(v_1,v_2)=(2ku_1,2ku_2)+(2kv_1,2kv_2)=\colorThree{(2ku_1+2kv_1,2ku_2+2kv_2)}
Since (2k(u1+v1),2k(u2+v2))=(2ku1+2kv1,2ku2+2kv2)\colorOne{(2k(u_1+v_1),2k(u_2+v_2))} = \colorThree{(2ku_1+2kv_1,2ku_2+2kv_2)}, this axiom is satisfied.
B) (k+m)⋅v⃗=k⋅v⃗+m⋅v⃗(k+m)\cdot\vec v = k\cdot\vec v + m\cdot\vec v
(k+m)⋅v⃗=(k+m)⋅(v1,v2)=(2(k+m)v1,2(k+m)v2)(k+m)\cdot\vec v = (k+m)\cdot(v_1,v_2) = \colorOne{(2(k+m)v_1,2(k+m)v_2)}
Meanwhile:
k⋅v⃗+m⋅v⃗=(2kv1,2kv2)+(2mv1,2mv2)=(2kv1+2mv1,2kv2+2mv2)k\cdot\vec v + m\cdot\vec v = (2kv_1,2kv_2)+(2mv_1,2mv_2)=\colorThree{(2kv_1+2mv_1,2kv_2+2mv_2)}
Since (2(k+m)v1,2(k+m)v2)=(2kv1+2mv1,2kv2+2mv2)\colorOne{(2(k+m)v_1,2(k+m)v_2)}=\colorThree{(2kv_1+2mv_1,2kv_2+2mv_2)}, this axiom is satisfied.
C) k⋅(m⋅v⃗)=(km)⋅v⃗k\cdot(m\cdot\vec v) = (km)\cdot\vec v
k⋅(m⋅v⃗)=k⋅(m⋅(v1,v2))=k⋅(2mv1,2mv2)=(2k(2mv1),2k(2mv2))=(4kmv1,4kmv2)k\cdot(m\cdot\vec v) = k\cdot(m\cdot(v_1,v_2)) =k\cdot(2mv_1,2mv_2) = (2k(2mv_1),2k(2mv_2)) = \colorOne{(4kmv_1,4kmv_2)}
Meanwhile:
(km)⋅v⃗=(km)⋅(v1,v2)=(2kmv1,2kmv2)(km)\cdot\vec v = (km)\cdot(v_1,v_2)=\colorThree{(2kmv_1,2kmv_2)}
Since(4kmv1,4kmv2)≠(2kmv1,2kmv2)\colorOne{(4kmv_1,4kmv_2)} \textcolor{red}{\neq}\colorThree{(2kmv_1,2kmv_2)}, this axiom is not satisfied.

Practice: Verifying Vector Space Axioms

Let V={(x,y)  ∣  x,y∈R}V=\{(x,y)\;|\;x,y\in\mathbb{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)
  • k⋅(v1,v2)=(kv1,kv2)k\cdot(v_1,v_2)=(kv_1,kv_2)
Determine which of the following vector space axioms, if any, are satisfied. [Select all that apply]

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}\} 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
  • k⋅⟨v1,v2,v3⟩=⟨v1k,v2k,kv3⟩k\cdot\lang v_1,v_2,v_3\rang=\lang v_1^k,v_2^k,kv_3 \rang
Find the zero vector, 0⃗∈W\vec 0 \in W.
Extra Practice
$\tkcth{7.6.}\tkcf{6}$_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\} consisting of only one element: xx. Define addition ⊕\oplus and scalar multiplication ⊗\otimes on S\mathbb{S} by:
x⊕x=xandk⊗x=x, x\oplus x = x \quad \textrm{and} \quad k \otimes x = x,
for all k ∈ Rk \, \in \, \mathbb{R}. Determine whether or not S\mathbb{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={[a00b] : a, b ∈ R}\mathbb{M} = \Bigg\{ \sm{a}{0}{0}{b} \, : \, a,\, b \, \in \, \mathbb{R}\Bigg\}, along with the usual matrix addition and scalar multiplication, constitutes a vector space over R\mathbb{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} be a vector space. Prove that (−k)v⃗=−(kv⃗)=k(−v⃗)=(−1)kv⃗(-k)\vv = -(k\vv) = k(-\vv) = (-1)k\vec{v}, for all k ∈ Rk \, \in \, \mathbb{R} and all v⃗ ∈ S\vv \, \in \, \mathbb{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\}, and define vector addition by:
(a,b)⊕(c,d)=(ad+bc, bd)(a,b) \oplus (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}^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} be a vector space. Prove that for all k ∈ Rk \, \in \, \mathbb{R}, k0⃗=0⃗k\vz = \vz.
$\tkcth{7.6.}\tkcf{12}$_Ch_7.6_$\tkco{eg\_12}$_133_$\key{F.}$Builder_Vector_Spaces_$\tkct{}$_20.1W
Let S\mathbb{S} be a vector space. Prove that (−1)v⃗=−v⃗(-1) \vv = -\vv for all v⃗∈S\vv \in \mathbb{S}.
Let V={(x,y)  ∣  x,y∈R}V=\{(x,y)\;|\;x,y\in\mathbb{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)
k⋅(v1,v2)=(k2v1,kv2)k\cdot(v_1,v_2)=(k^2v_1,kv_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}\} 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
Let W={(x,y,z)  ∣  x,y>0,z∈R}W=\{(x,y,z)\;|\;x,y>0,z\in\mathbb{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)
k⋅(v1,v2,v3)=(v1k,v2k,kv3)k\cdot(v_1,v_2,v_3)=(v_1^k,v_2^k,kv_3)
Let Z={x⃗=(x)  ∣  x>3} Z=\{\vec x=(x)\;|\;x>3\} with the following operations:
u⃗+v⃗=uv−3(u+v)+12\vec u+\vec v=uv-3(u+v)+12
k⋅v⃗=(v−3)a+3k\cdot \vec 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\}, 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)
Determine whether or not F[0, 1]\mathcal{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} be a vector space. Prove that the zero vector is unique. That is, suppose there are two vectors x⃗\vx and y⃗\vec{y} in S\mathbb{S} with the property x⃗+v⃗=v⃗\vx + \vv = \vv and y⃗+v⃗=v⃗\vec{y}+\vv = \vv for all v⃗ ∈ S\vv \, \in \, \mathbb{S}, then show that this implies x⃗=y⃗\vx = \vec{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} be a vector space. Prove that every vector has a unique negative.
That is, for v⃗ ∈S\vv \, \in \mathbb{S}, suppose there are two vectors x⃗\vx and y⃗\vec{y} in S\mathbb{S} with the property v⃗+x⃗=0⃗\vv + \vx = \vz and v⃗+y⃗=0⃗\vv + \vec{y} = \vz; show that this implies x⃗=y⃗\vx = \vec{y}.
Let V=R2V=\mathbb{R}^2 but with the defined the operations:
u⃗+v⃗=(u1+v2,u2+v1)\vec u + \vec v = (u_1+v_2,u_2+v_1)
In this space the zero vector is 0⃗=(0,0)\vec 0 = (0,0)\qquad (can you prove that?)
Let V={(u1,u2)  ∣  u1>0}V=\{(u_1,u_2)\;|\;u_1>0\} and define the operation:
u⃗+v⃗=(u1v1,u2+v2)\vec u + \vec v = (u_1v_1,u_2+v_2)
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\}, that is VV is the set of all vectors that are real numbers in the interval (3,∞)(3,\infty)
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)+12
Practice: Vector Properties
Prove the following vector property:
u⃗+v⃗=v⃗+u⃗\vec{u}+\vec{v}=\vec{v}+\vec{u}
Let W={(x,y,z)  ∣  x,y>0,z∈R}W=\{(x,y,z)\;|\;x,y>0,z\in\mathbb{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)
k⋅(v1,v2,v3)=(v1k,v2k,kv3)k\cdot(v_1,v_2,v_3)=(v_1^k,v_2^k,kv_3)
Let Z={x⃗=(x)  ∣  x>3}Z=\{\vec x = (x)\;|\;x>3\} with the following operations:
u⃗+v⃗=uv−3(u+v)+12\vec u+\vec v=uv-3(u+v)+12
k⋅v⃗=(v−3)a+3k\cdot \vec v = (v-3)^a+3
Consider the set R2\mathbb{R}^2 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)
and scalar multiplication is
Vector Space and Subspace

Consider the vector space, C([0,1])\mathcal{C}([0, 1]) , the space of continuous functions on the interval [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}\} 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
Vector Spaces
Practice: Verifying Vector Space Axioms
Let V={(x,y)  ∣  x,y∈R}V=\{(x,y)\;|\;x,y\in\mathbb{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)