Subspaces

You already know a vector space $V$: a set of objects (vectors) that you can add together and scale by numbers (scalars), where all the usual rules hold — addition is commutative and associative, there is a zero vector $\mathbf{0}$, every vector has a negative, and scalar multiplication distributes nicely. The standard example is $\mathbb{R}^n$, the set of $n$-tuples of real numbers.

Often we care about a smaller collection of vectors living inside $V$ — for example, all the points on a particular line in $\mathbb{R}^2$. The natural question is: is that smaller collection a vector space in its own right, using the exact same addition and scalar multiplication it inherits from $V$?

When the answer is yes, we call the subset a subspace. The key word is same operations: we do not get to invent a new way to add vectors — we must use $V$'s addition and scaling and check that we never "fall out" of the subset.

Intuitively, a subspace is a subset that is self-contained: start with vectors inside it, add them or scale them, and you always land back inside it. Geometrically in $\mathbb{R}^2$ and $\mathbb{R}^3$, this self-containment forces subspaces to be perfectly flat objects that pass through the origin.

Here is the beautiful shortcut. A subset $W$ of a vector space $V$ is a subspace if and only if it passes three checks:

1. Contains the zero vector: $\mathbf{0} \in W$. (In particular $W$ is nonempty.)
2. Closed under addition: if $\mathbf{u} \in W$ and $\mathbf{v} \in W$, then $\mathbf{u} + \mathbf{v} \in W$.
3. Closed under scalar multiplication: if $\mathbf{u} \in W$ and $c$ is any scalar, then $c\,\mathbf{u} \in W$.

Why only three checks, when a vector space has many axioms? Because the other axioms (commutativity, associativity, distributivity, and so on) are identities that already hold for every vector in $V$ — so they automatically hold for the vectors in $W \subseteq V$. The only thing that could go wrong is falling out of the set, and conditions 2 and 3 forbid exactly that. Condition 1 guarantees $W$ is nonempty and pins the zero vector inside; it also rules out the empty set.

A practical tip: check for $\mathbf{0}$ first. If $\mathbf{0} \notin W$, you are done immediately — $W$ is not a subspace, no further work needed. Conditions 2 and 3 are sometimes bundled into a single line: $W$ is a subspace iff it is nonempty and $c\,\mathbf{u} + d\,\mathbf{v} \in W$ for all $\mathbf{u}, \mathbf{v} \in W$ and all scalars $c, d$ (closure under linear combinations).

In low dimensions the subspaces are easy to picture, and the picture is instructive.

Subspaces of $\mathbb{R}^2$:

• the zero subspace $\{\mathbf{0}\}$ (a single point, the origin);
• every line through the origin, e.g. $\{(x,y): y = mx\}$;
• all of $\mathbb{R}^2$.

Subspaces of $\mathbb{R}^3$:

• the zero subspace $\{\mathbf{0}\}$;
• every line through the origin;
• every plane through the origin;
• all of $\mathbb{R}^3$.

Notice the recurring phrase: through the origin. This is forced by condition 1 of the test — a subspace must contain $\mathbf{0}$. A line or plane that misses the origin cannot be a subspace.

There is also a clean algebraic source of these objects. The solution set of a homogeneous linear equation such as $a x + b y + c z = 0$ (note the right-hand side is $0$) is always a subspace: the origin solves it, the sum of two solutions is a solution, and any scalar multiple of a solution is a solution. More generally, the solution set of any homogeneous linear system $A\mathbf{x} = \mathbf{0}$ is a subspace — this is the null space of $A$, which you will meet soon.

A single failed condition is enough to disqualify a set. The most common culprits:

• Shifted lines/planes (missing the origin). The line $y = 2x + 1$ in $\mathbb{R}^2$ does not pass through $(0,0)$, so it fails condition 1. Likewise any $a x + b y = k$ with $k \ne 0$.

• The first quadrant $\{(x,y): x \ge 0,\ y \ge 0\}$. This one is sneaky: it does contain $\mathbf{0}$, and it is closed under addition. But it fails closure under scalar multiplication: take $(1,1)$ and multiply by $-1$ to get $(-1,-1)$, which is outside. One failure is fatal.

• Curves. The unit circle $\{(x,y): x^2 + y^2 = 1\}$ does not even contain the origin, and is closed under neither operation.

• Unions of subspaces. The union of the $x$-axis and the $y$-axis contains $\mathbf{0}$ and is closed under scalar multiplication, yet fails closure under addition: $(1,0) + (0,1) = (1,1)$ lies on neither axis.

• Sets defined by a nonlinear condition, like $\{(x,y,z): x^2 = y^2\}$. It contains $\mathbf{0}$ and is closed under scaling, but $(1,1,0) + (1,-1,0) = (2,0,0)$ has $x^2 = 4 \ne 0 = y^2$, breaking closure under addition.

The moral: closure can fail on just one of the two operations, so you must check both — unless condition 1 already fails, which ends the discussion.