Vector Spaces

Geometrically, a vector in the plane has a length and a direction. It is drawn as a directed line segment from the origin (its initial point) to a terminal point (x₁, x₂). The same ordered pair names the vector: x = (x₁, x₂), and x₁, x₂ are its components.

Two vectors u = (u₁, u₂) and v = (v₁, v₂) are equal if and only if u₁ = v₁ and u₂ = v₂.

Sum

u + v = (u₁ + v₁, u₂ + v₂). Geometrically the diagonal of the parallelogram with u and v as adjacent sides.

Scalar multiple

cv = (cv₁, cv₂). Multiplying by a real number c changes the "scale" of the vector, hence the word scalar.

Negative

−v = (−1)v = (−v₁, −v₂).

Difference

u − v = u + (−v) = (u₁ − v₁, u₂ − v₂): subtract v by adding its negative.

Zero vector

0 = (0, 0). Adding it changes nothing: (2, 1) + (0, 0) = (2, 1).

An ordered n-tuple (x₁, x₂, …, xₙ) is a vector in n-space. The set of all n-tuples is Rⁿ. An n-tuple can be read as a point with coordinates xᵢ or as a vector with components xᵢ. Two vectors in Rⁿ are equal exactly when every pair of corresponding components is equal.

• —Theorem 4.3: the additive identity is unique (v + u = v forces u = 0) and so is the additive inverse (v + u = 0 forces u = −v).
• —0v = 0 and c0 = 0. If cv = 0, then c = 0 or v = 0. And −(−v) = v.

Because the properties above match ordinary algebra, an equation in an unknown vector x is solved exactly like one in a number: expand, collect the x terms on one side, then divide by the scalar in front of x. Only at the end do you plug in the components.

A vector x is a linear combination of v₁, v₂, …, vₙ when there are scalars c₁, c₂, …, cₙ with

1. 1.Write c₁v₁ + ⋯ + cₙvₙ = x and compare components: each component gives one linear equation in the cᵢ.
2. 2.Solve the system (substitution, or Gauss-Jordan on the augmented matrix whose columns are v₁, …, vₙ and then x).
3. 3.A solution gives the coefficients. No solution means x is not a linear combination of the vᵢ.
4. 4.Check by multiplying out: the combination must give back every component of x.

Let V be a set with two operations, vector addition and scalar multiplication. If the ten axioms below hold for every u, v, w in V and every pair of scalars (real numbers) c and d, then V is a vector space and its elements are called vectors, whatever they look like.

• —Theorem 4.4, true in every vector space: 0v = 0, c0 = 0, if cv = 0 then c = 0 or v = 0, and (−1)v = −v.
• —Axioms 1–5 are about addition, 6–10 about scalar multiplication. The two closure axioms (1 and 6) are the ones that usually fail.

P₂ works because adding two polynomials of degree 2 or less, or scaling one, gives another of degree 2 or less. M₁,₁, the 1 × 1 matrices, is just R written with brackets, so it is a vector space too.

To show a set is not a vector space, find one axiom that fails and give one specific counterexample. One failure is enough.

From Exercise 2, these are vector spaces with the standard operations: M₁,₁, the singleton {0}, 2 × 2 matrices of the form [a b; c 0], {(x, x) | x ∈ R} and the 2 × 2 diagonal matrices. Third-degree polynomials and quadratic functions through the origin fail closure under addition the same way degree-2 polynomials do.

A nonempty subset W of a vector space V is a subspace of V when W is itself a vector space under the operations of V. You do not need all ten axioms: W inherits the rest from V.

• —Zero-vector shortcut: a subspace always contains 0 (take c = 0). If 0 is not in W, stop: W is not a subspace.
• —If 0 is in W you still have to check both closures; containing 0 is necessary, not sufficient.
• —The subspaces of R² are {0}, every line through the origin, and R². The subspaces of R³ are {0}, lines through the origin, planes through the origin, and R³.
• —A line or plane that misses the origin, like x + 2y = 1, is never a subspace.