1.1 Definition of a Vector Space
A vector space over a field F ( R or C) is a set V equipped With two operations:
- Vector addition: +:V×V→V
- Scalar multiplication: ⋅:F×V→V
Satisfying the following axioms for all u,v,w∈V and all α,β∈F:
- Commutativity: u+v=v+u
- Associativity of addition: (u+v)+w=u+(v+w)
- Additive identity: There exists 0∈V such that v+0=v
- Additive inverse: For each vThere exists −v such that v+(−v)=0
- Compatibility of scalar multiplication: α(βv)=(αβ)v
- Identity element of scalar multiplication: 1⋅v=v
- Distributivity over vector addition: α(u+v)=αu+αv
- Distributivity over scalar addition: (α+β)v=αv+βv
Intuition. The abstract definition captures the algebraic structure shared by diverse objects: Geometric arrows, polynomials, functions, matrices. The axioms encode exactly what we need for Linear combinations to behave reasonably.
1.2 Examples
Example 1. Rn with component-wise addition and scalar multiplication is a vector space Over R.
Example 2. The set Pn of all polynomials of degree at most n with real coefficients, With the usual polynomial addition and scalar multiplication, is a vector space over R. Its dimension is n+1With standard basis {1,x,x2,…,xn}.
Example 3. The set C[a,b] of all continuous real-valued functions on [a,b]With point-wise Addition and scalar multiplication, is a vector space over R. This space is Infinite-dimensional.
Example 4. The set Mm×n(R) of all m×n real matrices is a Vector space over R.
Example 5 (Function spaces). The set F(R,R) of all functions f:R→R is a vector space over R under point-wise addition (f+g)(x)=f(x)+g(x) and scalar multiplication (αf)(x)=α⋅f(x). The spaces Ck(R) of k-times continuously differentiable functions and L2[a,b] of square-integrable functions are important subspaces of F(R,R).
Example 6 (Sequence spaces). The set ℓ2 of all real sequences (a1,a2,a3,…) With ∑n=1∞an2<∞ is a vector space over R. This is the Infinite-dimensional analogue of Rn and is fundamental in functional analysis.
1.3 Subspaces
A subspace W of a vector space V is a subset W⊆V that is itself a vector space Under the same operations.
Theorem 1.1 (Subspace Criterion). A non-empty subset W⊆V is a subspace if and only If for all u,v∈W and all α∈F:
- u+v∈W (closed under addition)
- αu∈W (closed under scalar multiplication)
Proof. If W is a subspace, closure is immediate from the definition. Conversely, if W is Non-empty and closed under both operations, pick u∈W. Then −u=(−1)u∈W By closure under scalar multiplication, and u+(−u)=0∈W by closure Under addition. The remaining axioms are inherited from V. ■
Proposition 1.2 (Closure under Linear Combinations). If W is a subspace of VThen W is Closed under all finite linear combinations: for all v1,…,vk∈W and All α1,…,αk∈F
α1v1+α2v2+⋯+αkvk∈W
Proof. We proceed by induction on k. For k=1, α1v1∈W by closure under Scalar multiplication. Assume the result holds for k−1 vectors. Then
α1v1+⋯+αkvk=(α1v1+⋯+αk−1vk−1)+αkvk
By the inductive hypothesis, α1v1+⋯+αk−1vk−1∈WAnd αkvk∈W by closure under scalar multiplication. Their sum is in W by Closure under addition. ■
Example 7. The set of all solutions to the homogeneous equation Ax=0 forms a Subspace of RnCalled the null space of A.
1.4 Worked Example: Verifying Subspace Criteria
Problem. Determine whether each of the following subsets of R3 is a subspace.
(a) W1={(x,y,z)∈R3:x+2y−z=0}
(b) W2={(x,y,z)∈R3:x2+y2=1}
(c) W3={(x,y,z)∈R3:x=0 and y=z}
Solution
(a) Let u=(x1,y1,z1) and v=(x2,y2,z2) be in W1So x1+2y1−z1=0 and x2+2y2−z2=0. Then
(x1+x2)+2(y1+y2)−(z1+z2)=(x1+2y1−z1)+(x2+2y2−z2)=0+0=0
So u+v∈W1. For α∈R
(αx1)+2(αy1)−(αz1)=α(x1+2y1−z1)=α⋅0=0
So αu∈W1. Since W1 is non-empty (e.g., 0∈W1), it is a subspace.
(b) W2 is not a subspace. For instance, (1,0,0)∈W2 since 12+02=1But 2⋅(1,0,0)=(2,0,0)∈/W2 since 22+02=4=1. So W2 is not closed Under scalar multiplication.
(c) Let u=(0,a,a) and v=(0,b,b) be in W3. Then u+v=(0,a+b,a+b)∈W3 and αu=(0,αa,αa)∈W3. Since (0,0,0)∈W3It is a non-empty subspace.
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1.5 Worked Example: Sum and Intersection of Subspaces
Problem. Let U={(x,y,z)∈R3:z=0} (the xy-plane) and W={(x,y,z)∈R3:x=0} (the yz-plane). Find U+W and U∩W And verify the dimension formula.
Solution
U has basis {(1,0,0),(0,1,0)} and dim(U)=2. W has basis {(0,1,0),(0,0,1)} and dim(W)=2.
U∩W={(x,y,z):z=0 and x=0}={(0,y,0):y∈R} Which has basis {(0,1,0)} and dimension 1.
U+W=span{(1,0,0),(0,1,0),(0,1,0),(0,0,1)}=span{(1,0,0),(0,1,0),(0,0,1)}=R3 So dim(U+W)=3.
Verify: dim(U+W)=dim(U)+dim(W)−dim(U∩W)=2+2−1=3. ✓ ■
1.6 Common Pitfalls
- The empty set is not a vector space. The subspace criterion requires the subset to be non-empty. The trivial subspace {0} is the smallest subspace of any vector space.
- Non-homogeneous conditions do not define subspaces. The set of solutions to Ax=b with b=0 is not a subspace (it is an affine subspace, or coset of the null space).
- Closure must hold for all scalars. A set that is closed under addition and multiplication by positive scalars is not necessarily a subspace; it must also be closed under multiplication by −1.