Let E,F be K-vector spaces. A map u:E→F is linear when
∀x,y∈E, ∀λ∈K,u(x+λy)=u(x)+λu(y).
Then u(0)=0 and u(∑λixi)=∑λiu(xi). The set L(E,F) of linear maps is itself a vector space; composition of linear maps is linear, and bilinear in the pair. An endomorphism is a linear E→E; an isomorphism is a bijective linear map (its inverse is then automatically linear); u−1 of an isomorphism, compositions of isomorphisms, are isomorphisms.
Examples
Example 20.8 (Building a map to specification)
Construct u∈L(R3) with keru=Vect(1,1,1) and imu={z=0}. Sanity first: rank–nullity demands 1+2=3 — consistent, so a solution may exist. Choose a basis adapted to the kernel, say ((1,1,1), e1, e2) (Example 19.7), and prescribe images (Proposition 20.2):
u(1,1,1)=0,u(e1)=e1,u(e2)=e2.
Then keru⊇Vect(1,1,1) and imu=Vect(e1,e2)={z=0}; rank–nullity forces dimkeru=1, so the kernel is exactly the prescribed line. Explicitly, decomposing (x,y,z)=z(1,1,1)+(x−z)e1+(y−z)e2:
u(x,y,z)=(x−z, y−z, 0).
The recipe generalizes: a linear map with prescribed kernel N and image I exists exactly when dimN+dimI=dimE — necessity is rank–nullity, sufficiency is this construction.
Example 20.10 (Interpolation, structurally)
Fix distinct x0,…,xn and let u:Rn[X]→Rn+1, P↦(P(x0),…,P(xn)): linear. Its kernel is {P:degP≤n, n+1 roots}={0} (Corollary 8.8). Equal dimensions n+1: u is an isomorphism — existence and uniqueness of the Lagrange interpolant (Theorem 8.23) in one line.
The same one-line pattern handles data mixing values and derivatives: v:R3[X]→R4, P↦(P(0),P′(0),P(1),P′(1)) is linear, and its kernel consists of degree-≤3 polynomials with double roots at 0 and 1, i.e. divisible by X2(X−1)2 of degree 4: only P=0. Equal dimensions again: every quadruple of data (P(0),P′(0),P(1),P′(1)) is realized by exactly one cubic — Hermite interpolation, granted by a kernel computation before any formula is written (the weekend problem of Chapter 22 meets its determinant).
Example 20.11 (Rank–nullity at work: the difference operator)
Let Δ:Rn[X]→Rn[X], P↦P(X+1)−P(X): linear. Kernel: if ΔP=0, then P(0)=P(1)=P(2)=…, so P−P(0) has infinitely many roots and vanishes (Corollary 8.8): kerΔ is the line of constants. Rank–nullity: rkΔ=(n+1)−1=n. Since degΔP<degP for nonconstant P (the top terms cancel), imΔ⊆Rn−1[X], which has dimension exactly n: the inclusion is an equality. Conclusion, with no computation of preimages: every polynomial Q of degree ≤n−1 is a difference Q=P(X+1)−P(X) — the discrete antiderivative exists. (Compare the weekend problem of Chapter 18, where Δ was inverted explicitly in the binomial basis.)