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Vectors and Vector Spaces

---:::note[Historical Context] The vector space was introduced by Giuseppe Peano (1888), building on Grassmann’s Ausdehnungslehre (1844). The modern axiomatic approach was established by Banach (1920). Every vector space has a basis (proved via Zorn’s Lemma, 1935). Vector spaces connect to representation theory, where groups act on vector spaces.

Section titled “---:::note[Historical Context] The vector space was introduced by Giuseppe Peano (1888), building on Grassmann’s Ausdehnungslehre (1844). The modern axiomatic approach was established by Banach (1920). Every vector space has a basis (proved via Zorn’s Lemma, 1935). Vector spaces connect to representation theory, where groups act on vector spaces.”

A vector space over a field FF ( R\mathbb{R} or C\mathbb{C}) is a set VV equipped With two operations:

  1. Vector addition: +:V×V→V+ : V \times V \to V
  2. Scalar multiplication: ⋅:F×V→V\cdot : F \times V \to V

Satisfying the following axioms for all u,v,w∈V\mathbf{u}, \mathbf{v}, \mathbf{w} \in V and all α,β∈F\alpha, \beta \in F:

  1. Commutativity: u+v=v+u\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}
  2. Associativity of addition: (u+v)+w=u+(v+w)(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})
  3. Additive identity: There exists 0∈V\mathbf{0} \in V such that v+0=v\mathbf{v} + \mathbf{0} = \mathbf{v}
  4. Additive inverse: For each v\mathbf{v}There exists −v-\mathbf{v} such that v+(−v)=0\mathbf{v} + (-\mathbf{v}) = \mathbf{0}
  5. Compatibility of scalar multiplication: α(βv)=(αβ)v\alpha(\beta \mathbf{v}) = (\alpha\beta)\mathbf{v}
  6. Identity element of scalar multiplication: 1⋅v=v1 \cdot \mathbf{v} = \mathbf{v}
  7. Distributivity over vector addition: α(u+v)=αu+αv\alpha(\mathbf{u} + \mathbf{v}) = \alpha\mathbf{u} + \alpha\mathbf{v}
  8. Distributivity over scalar addition: (α+β)v=αv+βv(\alpha + \beta)\mathbf{v} = \alpha\mathbf{v} + \beta\mathbf{v}

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.

Example 1. Rn\mathbb{R}^n with component-wise addition and scalar multiplication is a vector space Over R\mathbb{R}.

Example 2. The set Pn\mathcal{P}_n of all polynomials of degree at most nn with real coefficients, With the usual polynomial addition and scalar multiplication, is a vector space over R\mathbb{R}. Its dimension is n+1n + 1With standard basis {1,x,x2,…,xn}\{1, x, x^2, \ldots, x^n\}.

Example 3. The set C[a,b]C[a,b] of all continuous real-valued functions on [a,b][a,b]With point-wise Addition and scalar multiplication, is a vector space over R\mathbb{R}. This space is Infinite-dimensional.

Example 4. The set Mm×n(R)\mathcal{M}_{m \times n}(\mathbb{R}) of all m×nm \times n real matrices is a Vector space over R\mathbb{R}.

Example 5 (Function spaces). The set F(R,R)\mathcal{F}(\mathbb{R}, \mathbb{R}) of all functions f:R→Rf : \mathbb{R} \to \mathbb{R} is a vector space over R\mathbb{R} under point-wise addition (f+g)(x)=f(x)+g(x)(f + g)(x) = f(x) + g(x) and scalar multiplication (αf)(x)=α⋅f(x)(\alpha f)(x) = \alpha \cdot f(x). The spaces Ck(R)C^k(\mathbb{R}) of kk-times continuously differentiable functions and L2[a,b]L^2[a,b] of square-integrable functions are important subspaces of F(R,R)\mathcal{F}(\mathbb{R}, \mathbb{R}).

Example 6 (Sequence spaces). The set ℓ2\ell^2 of all real sequences (a1,a2,a3,…)(a_1, a_2, a_3, \ldots) With ∑n=1∞an2<∞\sum_{n=1}^{\infty} a_n^2 \lt \infty is a vector space over R\mathbb{R}. This is the Infinite-dimensional analogue of Rn\mathbb{R}^n and is fundamental in functional analysis.

A subspace WW of a vector space VV is a subset W⊆VW \subseteq V that is itself a vector space Under the same operations.

Theorem 1.1 (Subspace Criterion). A non-empty subset W⊆VW \subseteq V is a subspace if and only If for all u,v∈W\mathbf{u}, \mathbf{v} \in W and all α∈F\alpha \in F:

  1. u+v∈W\mathbf{u} + \mathbf{v} \in W (closed under addition)
  2. αu∈W\alpha \mathbf{u} \in W (closed under scalar multiplication)

Proof. If WW is a subspace, closure is immediate from the definition. Conversely, if WW is Non-empty and closed under both operations, pick u∈W\mathbf{u} \in W. Then −u=(−1)u∈W-\mathbf{u} = (-1)\mathbf{u} \in W By closure under scalar multiplication, and u+(−u)=0∈W\mathbf{u} + (-\mathbf{u}) = \mathbf{0} \in W by closure Under addition. The remaining axioms are inherited from VV. ■\blacksquare

Proposition 1.2 (Closure under Linear Combinations). If WW is a subspace of VVThen WW is Closed under all finite linear combinations: for all v1,…,vk∈W\mathbf{v}_1, \ldots, \mathbf{v}_k \in W and All α1,…,αk∈F\alpha_1, \ldots, \alpha_k \in F

α1v1+α2v2+⋯+αkvk∈W\alpha_1 \mathbf{v}_1 + \alpha_2 \mathbf{v}_2 + \cdots + \alpha_k \mathbf{v}_k \in W

Proof. We proceed by induction on kk. For k=1k = 1, α1v1∈W\alpha_1 \mathbf{v}_1 \in W by closure under Scalar multiplication. Assume the result holds for k−1k - 1 vectors. Then

α1v1+⋯+αkvk=(α1v1+⋯+αk−1vk−1)+αkvk\alpha_1 \mathbf{v}_1 + \cdots + \alpha_k \mathbf{v}_k = (\alpha_1 \mathbf{v}_1 + \cdots + \alpha_{k-1} \mathbf{v}_{k-1}) + \alpha_k \mathbf{v}_k

By the inductive hypothesis, α1v1+⋯+αk−1vk−1∈W\alpha_1 \mathbf{v}_1 + \cdots + \alpha_{k-1} \mathbf{v}_{k-1} \in WAnd αkvk∈W\alpha_k \mathbf{v}_k \in W by closure under scalar multiplication. Their sum is in WW by Closure under addition. ■\blacksquare

Example 7. The set of all solutions to the homogeneous equation Ax=0A\mathbf{x} = \mathbf{0} forms a Subspace of Rn\mathbb{R}^nCalled the null space of AA.

1.4 Worked Example: Verifying Subspace Criteria

Section titled “1.4 Worked Example: Verifying Subspace Criteria”

Problem. Determine whether each of the following subsets of R3\mathbb{R}^3 is a subspace.

(a) W1={(x,y,z)∈R3:x+2y−z=0}W_1 = \{(x, y, z) \in \mathbb{R}^3 : x + 2y - z = 0\}

(b) W2={(x,y,z)∈R3:x2+y2=1}W_2 = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 = 1\}

(c) W3={(x,y,z)∈R3:x=0 and y=z}W_3 = \{(x, y, z) \in \mathbb{R}^3 : x = 0 \mathrm{~and~} y = z\}

Solution

(a) Let u=(x1,y1,z1)\mathbf{u} = (x_1, y_1, z_1) and v=(x2,y2,z2)\mathbf{v} = (x_2, y_2, z_2) be in W1W_1So x1+2y1−z1=0x_1 + 2y_1 - z_1 = 0 and x2+2y2−z2=0x_2 + 2y_2 - z_2 = 0. Then

(x1+x2)+2(y1+y2)−(z1+z2)=(x1+2y1−z1)+(x2+2y2−z2)=0+0=0(x_1 + x_2) + 2(y_1 + y_2) - (z_1 + z_2) = (x_1 + 2y_1 - z_1) + (x_2 + 2y_2 - z_2) = 0 + 0 = 0

So u+v∈W1\mathbf{u} + \mathbf{v} \in W_1. For α∈R\alpha \in \mathbb{R}

(αx1)+2(αy1)−(αz1)=α(x1+2y1−z1)=α⋅0=0(\alpha x_1) + 2(\alpha y_1) - (\alpha z_1) = \alpha(x_1 + 2y_1 - z_1) = \alpha \cdot 0 = 0

So αu∈W1\alpha \mathbf{u} \in W_1. Since W1W_1 is non-empty (e.g., 0∈W1\mathbf{0} \in W_1), it is a subspace.

(b) W2W_2 is not a subspace. For instance, (1,0,0)∈W2(1, 0, 0) \in W_2 since 12+02=11^2 + 0^2 = 1But 2⋅(1,0,0)=(2,0,0)∉W22 \cdot (1, 0, 0) = (2, 0, 0) \notin W_2 since 22+02=4≠12^2 + 0^2 = 4 \neq 1. So W2W_2 is not closed Under scalar multiplication.

(c) Let u=(0,a,a)\mathbf{u} = (0, a, a) and v=(0,b,b)\mathbf{v} = (0, b, b) be in W3W_3. Then u+v=(0,a+b,a+b)∈W3\mathbf{u} + \mathbf{v} = (0, a + b, a + b) \in W_3 and αu=(0,αa,αa)∈W3\alpha \mathbf{u} = (0, \alpha a, \alpha a) \in W_3. Since (0,0,0)∈W3(0, 0, 0) \in W_3It is a non-empty subspace.

■\blacksquare

1.5 Worked Example: Sum and Intersection of Subspaces

Section titled “1.5 Worked Example: Sum and Intersection of Subspaces”

Problem. Let U={(x,y,z)∈R3:z=0}U = \{(x, y, z) \in \mathbb{R}^3 : z = 0\} (the xyxy-plane) and W={(x,y,z)∈R3:x=0}W = \{(x, y, z) \in \mathbb{R}^3 : x = 0\} (the yzyz-plane). Find U+WU + W and U∩WU \cap W And verify the dimension formula.

Solution

UU has basis {(1,0,0),(0,1,0)}\{(1, 0, 0), (0, 1, 0)\} and dim⁡(U)=2\dim(U) = 2. WW has basis {(0,1,0),(0,0,1)}\{(0, 1, 0), (0, 0, 1)\} and dim⁡(W)=2\dim(W) = 2.

U∩W={(x,y,z):z=0 and x=0}={(0,y,0):y∈R}U \cap W = \{(x, y, z) : z = 0 \mathrm{~and~} x = 0\} = \{(0, y, 0) : y \in \mathbb{R}\} Which has basis {(0,1,0)}\{(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)}=R3U + W = \mathrm{span}\{(1,0,0), (0,1,0), (0,1,0), (0,0,1)\} = \mathrm{span}\{(1,0,0), (0,1,0), (0,0,1)\} = \mathbb{R}^3 So dim⁡(U+W)=3\dim(U + W) = 3.

Verify: dim⁡(U+W)=dim⁡(U)+dim⁡(W)−dim⁡(U∩W)=2+2−1=3\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W) = 2 + 2 - 1 = 3. ✓\checkmark ■\blacksquare

  • The empty set is not a vector space. The subspace criterion requires the subset to be non-empty. The trivial subspace {0}\{\mathbf{0}\} is the smallest subspace of any vector space.
  • Non-homogeneous conditions do not define subspaces. The set of solutions to Ax=bA\mathbf{x} = \mathbf{b} with b≠0\mathbf{b} \neq \mathbf{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 −1-1.