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Vectors and Vector Spaces

1.1 Definition of a Vector Space

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×VV+ : V \times V \to V
  2. Scalar multiplication: :F×VV\cdot : F \times V \to V

Satisfying the following axioms for all u,v,wV\mathbf{u}, \mathbf{v}, \mathbf{w} \in V and all α,βF\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 0V\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: 1v=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.

1.2 Examples

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:RRf : \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)=αf(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 2\ell^2 of all real sequences (a1,a2,a3,)(a_1, a_2, a_3, \ldots) With n=1an2<\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.

1.3 Subspaces

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

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

  1. u+vW\mathbf{u} + \mathbf{v} \in W (closed under addition)
  2. αuW\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 uW\mathbf{u} \in W. Then u=(1)uW-\mathbf{u} = (-1)\mathbf{u} \in W By closure under scalar multiplication, and u+(u)=0W\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,,vkW\mathbf{v}_1, \ldots, \mathbf{v}_k \in W and All α1,,αkF\alpha_1, \ldots, \alpha_k \in F

α1v1+α2v2++αkvkW\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, α1v1W\alpha_1 \mathbf{v}_1 \in W by closure under Scalar multiplication. Assume the result holds for k1k - 1 vectors. Then

α1v1++αkvk=(α1v1++αk1vk1)+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++αk1vk1W\alpha_1 \mathbf{v}_1 + \cdots + \alpha_{k-1} \mathbf{v}_{k-1} \in WAnd αkvkW\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

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

(a) W1={(x,y,z)R3:x+2yz=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+2y1z1=0x_1 + 2y_1 - z_1 = 0 and x2+2y2z2=0x_2 + 2y_2 - z_2 = 0. Then

(x1+x2)+2(y1+y2)(z1+z2)=(x1+2y1z1)+(x2+2y2z2)=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+vW1\mathbf{u} + \mathbf{v} \in W_1. For αR\alpha \in \mathbb{R}

(αx1)+2(αy1)(αz1)=α(x1+2y1z1)=α0=0(\alpha x_1) + 2(\alpha y_1) - (\alpha z_1) = \alpha(x_1 + 2y_1 - z_1) = \alpha \cdot 0 = 0

So αuW1\alpha \mathbf{u} \in W_1. Since W1W_1 is non-empty (e.g., 0W1\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=412^2 + 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

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 UWU \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.

UW={(x,y,z):z=0 and x=0}={(0,y,0):yR}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(UW)=2+21=3\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W) = 2 + 2 - 1 = 3. \checkmark \blacksquare

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}\{\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 b0\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.