Recipe

Test whether a set is a subspace

Use when: a problem hands you a subset W of a vector space, defined by an equation, inequality, or other condition on its coordinates, and asks whether W is a subspace

Goal: a proof that W passes all three axioms for an arbitrary element and scalar, or one concrete counterexample showing exactly where it fails

Taught in Testing whether a set is a subspace

1. 1

   Plug the zero vector into W's defining condition. This check is run first only because it's usually the fastest of the three to evaluate — the instant it fails, W is disqualified and the other two checks are pointless.

   Example

   For W = \{(x,y,z) \in \mathbb{R}^3 : x + 2y - z = 0\}, plug in (0,0,0): 0 + 2(0) - 0 = 0 \ \checkmark so \mathbf 0 \in W, and the set survives to Step 2.

   • \mathbf 0 \notin W — stop; W fails the zero axiom and cannot be a subspace, no matter what the other two checks would show.
   • \mathbf 0 \in W — continue to Step 2.

   ⚡ Shortcut

   If W's defining condition is a homogeneous linear equation (or several) — every term is a constant times a single coordinate, every right-hand side is 0, and nothing is squared, multiplied by another variable, or otherwise nonlinear — you can skip straight past all three checks. The solution set of a homogeneous linear system is always a subspace, for exactly the algebraic reason Steps 2 and 3 spell out in general below. The equation above, x+2y-z=0, is exactly this shape, so it's automatically a subspace with no further work. A condition with a nonzero constant (like x+2y-z=5) or a nonlinear term (like x^2+y^2=9) does NOT qualify for this shortcut — those need the full three-step check, and usually fail it.

   Why: every subspace, by definition, must contain \mathbf 0 — it's one of the three axioms, and the cheapest one to rule out.

2. 2

   Pick two arbitrary elements of W — written with subscripted variables standing for unknown numbers, not two numbers you chose — and check whether their sum still satisfies W's defining condition.

   Example

   Let \mathbf u = (x_1,y_1,z_1) and \mathbf v = (x_2,y_2,z_2) be arbitrary elements of W, so x_1+2y_1-z_1=0 and x_2+2y_2-z_2=0. Add them componentwise, \mathbf u+\mathbf v = (x_1{+}x_2,\ y_1{+}y_2,\ z_1{+}z_2), then substitute into the defining condition: (x_1{+}x_2) + 2(y_1{+}y_2) - (z_1{+}z_2) = \underbrace{(x_1+2y_1-z_1)}_{=\,0} + \underbrace{(x_2+2y_2-z_2)}_{=\,0} = 0 so \mathbf u + \mathbf v \in W too.

   • the sum fails the condition for some specific pair you can point to — stop; W is not a subspace, and that pair is your counterexample.
   • the algebra shows the sum satisfies the condition for every choice of x_1,y_1,z_1,x_2,y_2,z_2 — continue to Step 3.

   Why: a handful of numeric examples that happen to work can never rule out some other pair that doesn't — only an argument using symbols for an arbitrary pair covers every possible element of W at once.

3. 3

   Pick one arbitrary element of W and an arbitrary scalar c, and check whether the scaled vector still satisfies W's defining condition.

   Example

   Let \mathbf u = (x_1,y_1,z_1) \in W, so x_1+2y_1-z_1=0, and let c be an arbitrary scalar. Scale componentwise, c\mathbf u = (cx_1,\ cy_1,\ cz_1), then substitute: cx_1 + 2(cy_1) - cz_1 = c(x_1+2y_1-z_1) = c \cdot 0 = 0 so c\mathbf u \in W for every scalar c, including negative and fractional ones.

   Why: an arbitrary scalar c covers every possible scaling at once, including shrinking, flipping sign, and enlarging — a set that only stays put under, say, positive whole-number scaling hasn't passed this check.