Vector spaces and subspaces

A vector space $V$ over a field $\mathbb{F}$ (take $\mathbb{F} = \mathbb{R}$ unless told otherwise) is a set with two operations, vector addition and scalar multiplication, satisfying six axioms for all $\mathbf u, \mathbf v, \mathbf w \in V$ and $\alpha, \beta \in \mathbb{F}$: closure of both operations, associativity and commutativity of addition, existence of an additive identity $\mathbf 0$ and additive inverses, compatibility of scalar multiplication ($\alpha(\beta \mathbf u) = (\alpha\beta)\mathbf u$, $1 \cdot \mathbf u = \mathbf u$), and the two distributive laws. Familiar examples reused throughout the unit: $\mathbb{R}^n$ with componentwise operations, the set $M_{m\times n}(\mathbb{R})$ of $m \times n$ matrices, and the set $P_n$ of polynomials of degree at most $n$. Checking all six axioms from scratch every time is wasteful once you already know $V$ is a vector space. A subspace $W$ of $V$ is a subset that is itself a vector space under $V$'s own operations, and since $W \subseteq V$, the inherited axioms (associativity, commutativity, distributivity) come for free. Only closure needs checking, packaged as the subspace test:

where: $W$ is the candidate subset of $V$, $\mathbf u, \mathbf v$ are arbitrary elements of $W$, and $\alpha, \beta$ are arbitrary scalars from $\mathbb{F}$ — the test says any linear combination of members of $W$ must land back inside $W$. In practice this splits into two separate checks that are easier to verify one at a time: closure under addition ($\mathbf u + \mathbf v \in W$) and closure under scalar multiplication ($\alpha\mathbf u \in W$), plus the quick necessary screen that $\mathbf 0 \in W$ — if the zero vector fails the defining condition of $W$, you can reject it as a subspace immediately, no further algebra needed. Every genuine subspace contains $\mathbf 0$; the trivial subspace $\{\mathbf 0\}$ and $V$ itself are always subspaces of any $V$. Geometry gives the fastest intuition in $\mathbb{R}^2$ and $\mathbb{R}^3$: lines and planes through the origin are subspaces, while lines and planes that miss the origin are not (they fail the zero-vector screen outright). This single picture — through the origin or not — resolves most exam-level subspace questions before you write a single equation, though you should still confirm it algebraically.

The most common exam task is applying the subspace test to a set defined by one equation, and the goal here is to establish once and for all, with full closure-under-addition and closure-under-scalar-multiplication algebra rather than a geometric guess, that the solution set of a single homogeneous linear equation in $\mathbb{R}^3$ is always a subspace, which is the fact behind "planes through the origin are subspaces."

Step 1. State the set and check the zero-vector screen. Let $W = \{(x,y,z) \in \mathbb{R}^3 : x + 2y - z = 0\}$. Testing $(0,0,0)$: $0 + 2(0) - 0 = 0$, so $\mathbf 0 \in W$ and the set survives the quick necessary check.

Step 2. Check closure under addition. Take $\mathbf u = (x_1,y_1,z_1)$ and $\mathbf v = (x_2,y_2,z_2)$, both satisfying the defining equation: $x_1 + 2y_1 - z_1 = 0$ and $x_2 + 2y_2 - z_2 = 0$. Their sum is $\mathbf u + \mathbf v = (x_1+x_2,\ y_1+y_2,\ z_1+z_2)$, and substituting into the defining equation gives $(x_1+x_2) + 2(y_1+y_2) - (z_1+z_2) = (x_1+2y_1-z_1) + (x_2+2y_2-z_2) = 0 + 0 = 0$, so $\mathbf u + \mathbf v \in W$.

Step 3. Check closure under scalar multiplication. For any $\alpha \in \mathbb{R}$, $\alpha\mathbf u = (\alpha x_1, \alpha y_1, \alpha z_1)$, and $\alpha x_1 + 2\alpha y_1 - \alpha z_1 = \alpha(x_1 + 2y_1 - z_1) = \alpha \cdot 0 = 0$, so $\alpha\mathbf u \in W$.

Step 4. Conclude. All three checks pass, so $W$ is a subspace of $\mathbb{R}^3$. The argument used nothing special about the coefficients $1, 2, -1$: for any homogeneous linear equation $ax + by + cz = 0$, the same three steps go through unchanged (the right-hand side stays $0$ under both addition and scaling), which is exactly why every homogeneous linear equation in $\mathbb{R}^n$ defines a subspace, and why a right-hand side other than zero would break Step 1 at the very first hurdle.

Problem. Is $W = \{(x,y) \in \mathbb{R}^2 : xy \ge 0\}$ a subspace of $\mathbb{R}^2$?

Step 1. Identify $W$ geometrically: it is every point where $x$ and $y$ have the same sign or one of them is zero — the closed first and third quadrants together.

Step 2. Search for a counterexample to closure under addition rather than trying to prove closure holds. Pick $\mathbf u = (1,1)$, with $xy = 1 \ge 0$ so $\mathbf u \in W$, and $\mathbf v = (-2,0)$, with $xy = 0 \ge 0$ so $\mathbf v \in W$.

Step 3. Form the sum and test it: $\mathbf u + \mathbf v = (1,1) + (-2,0) = (-1,1)$, and $(-1)(1) = -1 < 0$, so $\mathbf u + \mathbf v \notin W$.

Step 4. Conclude. Since $\mathbf u, \mathbf v \in W$ but their sum is not, closure under addition fails, so $W$ is not a subspace of $\mathbb{R}^2$ — it is the union of quadrants I and III, a set that is not even convex, let alone closed under vector addition.

Competitive stretch

Extend the same idea to $U = \{(x,y) \in \mathbb{R}^2 : x \ge 0,\ y \ge 0\}$ (the first quadrant alone). Closure under addition does hold here (adding two non-negative pairs gives a non-negative pair), so a hasty check might call it a subspace. But test scalar multiplication with a negative scalar: $\alpha = -1$ on $(1,1) \in U$ gives $(-1,-1) \notin U$. $U$ fails the scalar-multiplication half of the test and is not a subspace, which is exactly the trap the first common trap below warns about.

• Believing closure under addition alone is enough — scalar multiplication, especially by negative scalars, must be checked separately, and it is what breaks sets like the first quadrant.
• Skipping the quick $\mathbf 0 \in W$ screen — if zero fails the defining condition, you can reject the subset immediately, saving the addition and scalar-multiplication algebra entirely.
• Confusing "is a vector space on its own" with "is a subspace of a bigger space" — a subspace must reuse the ambient space's exact operations, not a new addition or scaling rule.
• Assuming any equation-defined subset is automatically a subspace — only a homogeneous linear equation (right-hand side zero, no products or powers of the variables) guarantees one; an inequality or a nonzero right-hand side can fail either half of the test.

Ask the tutor: unsure whether a particular set of polynomials, matrices, or a subset defined by an inequality is a vector space or a subspace? Ask the AI tutor to test it with you step by step.