Vector Spaces and Isomorphism

A vector space is an abelian group over a field with scalar multiplication — Rn\mathbb{R}^n, sequence spaces, and spaces of continuous functions are the same structure in different packaging. Move (a,b)(a, b) and the line y=a+bxy = a+bx and the point (a,b)(a, b) move together.

(V,+) abelian over F,c(v+w)=cv+cw(V,+) \text{ abelian over } F,\qquad c(v+w) = cv + cw
Φ:P2→R3,a+bx+cx2↦(a,b,c)\Phi: P_2 \to \mathbb{R}^3,\quad a+bx+cx^2 \mapsto (a,b,c)

Examples & Exercises

Show that the set SS of all real sequences, with termwise addition (x+y)n=xn+yn(x+y)_n = x_n + y_n and scalar multiplication (cx)n=cxn(cx)_n = cx_n, is a vector space.

(x+y)n=xn+yn,(cx)n=cxn(x+y)_n = x_n + y_n,\qquad (cx)_n = cx_n
Sliders a=1a = 1, b=0.5b = 0.5 — the line y=1+0.5xy = 1 + 0.5x and the point (1,0.5)(1, 0.5).
Show proof
  1. Closure: if x,y∈Sx, y \in S then every term of x+y=(xn+yn)x + y = (x_n + y_n) is real, so x+y∈Sx + y \in S. Likewise every term of cx=(cxn)cx = (cx_n) is real, so cx∈Scx \in S.
  2. (S,+)(S, +) is an abelian group: commutativity and associativity are inherited termwise from R\mathbb{R}, the identity is the zero sequence (0,0,… )(0, 0, \dots), and the inverse of xx is (−xn)(-x_n).
  3. The remaining scalar axioms hold termwise too: (c(dx))n=c(dxn)=(cd)xn=((cd)x)n(c(dx))_n = c(dx_n) = (cd)x_n = ((cd)x)_n, and (1x)n=1⋅xn=xn(1x)_n = 1 \cdot x_n = x_n.
  4. The distributive laws check out at the sequence level: (c(x+y))n=c(xn+yn)=cxn+cyn=(cx+cy)n(c(x+y))_n = c(x_n + y_n) = cx_n + cy_n = (cx + cy)_n, and ((c+d)x)n=(c+d)xn=cxn+dxn=(cx+dx)n((c+d)x)_n = (c+d)x_n = cx_n + dx_n = (cx + dx)_n. Hence SS is a vector space over R\mathbb{R}.

Prove that Φ(a,b)=a+bi\Phi(a, b) = a + bi is an isomorphism R2→C\mathbb{R}^2 \to \mathbb{C} of real vector spaces.

Φ(a,b)=a+bi,Ψ(a+bi)=(a,b)\Phi(a, b) = a + bi,\qquad \Psi(a + bi) = (a, b)
Φ(0,1)=i\Phi(0, 1) = i — the imaginary unit ii sits at the point (0,1)(0, 1).
Show proof
  1. Additivity: Φ((a,b)+(c,d))=Φ(a+c,b+d)=(a+c)+(b+d)i=(a+bi)+(c+di)=Φ(a,b)+Φ(c,d)\Phi((a,b) + (c,d)) = \Phi(a+c, b+d) = (a+c) + (b+d)i = (a+bi) + (c+di) = \Phi(a,b) + \Phi(c,d).
  2. Homogeneity for real scalars: for real tt, Φ(t(a,b))=ta+tbi=t(a+bi)=tΦ(a,b)\Phi(t(a,b)) = ta + tbi = t(a+bi) = t\Phi(a,b). The scalars being real is the point — this is an isomorphism of real vector spaces.
  3. Injectivity: a+bi=c+dia + bi = c + di forces a=ca = c and b=db = d by uniqueness of real and imaginary parts, so Φ\Phi is injective.
  4. Surjectivity: with Ψ(a+bi)=(a,b)\Psi(a+bi) = (a,b) we have Ψ(Φ(a,b))=(a,b)\Psi(\Phi(a,b)) = (a,b) and Φ(Ψ(a+bi))=a+bi\Phi(\Psi(a+bi)) = a+bi, so Ψ\Psi is the inverse of Φ\Phi. A bijective linear map is an isomorphism.

Prove that a linear isomorphism T:V→WT: V \to W sends linearly independent sets to linearly independent sets.

∑i=1nci T(vi)=0  ⟹  c1=⋯=cn=0\sum_{i=1}^{n} c_i\,T(v_i) = 0 \;\Longrightarrow\; c_1 = \cdots = c_n = 0
Sliders a=1a = 1, b=−1b = -1 — the line y=1−xy = 1 - x and the point (1,−1)(1, -1).
Show proof
  1. Assume {v1,…,vn}⊂V\{v_1, \dots, v_n\} \subset V is linearly independent and ∑ciT(vi)=0\sum c_i T(v_i) = 0. By linearity, ∑ciT(vi)=T(∑civi)\sum c_i T(v_i) = T(\sum c_i v_i), so T(∑civi)=0T(\sum c_i v_i) = 0.
  2. Linear maps send 00 to 00 (T(0)=T(0⋅v)=0⋅T(v)=0T(0) = T(0 \cdot v) = 0 \cdot T(v) = 0), so injectivity of TT gives ∑civi=0\sum c_i v_i = 0.
  3. Independence of {vi}\{v_i\} forces c1=⋯=cn=0c_1 = \cdots = c_n = 0. Hence {T(v1),…,T(vn)}\{T(v_1), \dots, T(v_n)\} is linearly independent.

1. Number Systems