A Vector Space is a rigorous mathematical "playground" defined not by the nature of its objects, but by how those objects behave. Whether you are dealing with arrows in $\mathbf{R}^n$, matrices in $\mathbf{M}$, or continuous functions, the same rules apply.
The Eight Axioms of Space
Any collection of objects is a vector space if it obeys these fundamental rules:
- 1. Commutativity: $x + y = y + x$
- 2. Associativity: $x + (y + z) = (x + y) + z$
- 3. Zero Vector: There is a unique $0$ such that $x + 0 = x$
- 4. Inverses: For each $x$, there is a unique $-x$ such that $x + (-x) = 0$
- 5. Identity: $1x = x$
- 6. Scalar Associativity: $(c_1c_2)x = c_1(c_2x)$
- 7. Distributivity (I): $c(x + y) = cx + cy$
- 8. Distributivity (II): $(c_1 + c_2)x = c_1x + c_2x$
Defining Subspaces
A subspace $S$ of $V$ is a subset that is "closed" under the operations of the larger space. You can never escape the subset by adding its members or scaling them.
The Closure Theorem
A subset $S$ is a subspace if and only if for every $v, w \in S$ and every scalar $c, d$:
$$cv + dw \in S$$
This implies $S$ must contain the zero vector ($0 \in S$) because $0v = 0$.The Span and the Sum
The span of a set $S$ is the smallest subspace containing all vectors in $S$:
$$SS = \text{all } c_1v_1 + \dots + c_nv_n$$
Furthermore, given two subspaces $S$ and $T$, their sum $S + T$ (containing all vectors $s+t$) forms a new subspace. Note that the union $S \cup T$ is almost never a subspace!
🎯 The "Zero" Litmus Test
The fastest way to disqualify a subset from being a subspace is to check for the zero vector. If $x=0$ is not included, it cannot be a subspace. Common pitfalls include planes shifted away from the origin or quadrants that exclude negative values.