Subspaces

The whole idea: a space living inside a space

We built vector spaces from ten axioms in the Vector Spaces node. Now a natural question: if $V$ is a vector space and $W$ is a subset of it ($W \subseteq V$), when is $W$ itself a vector space under the operations it inherits from $V$? When it is, we call $W$ a subspace of $V$.

A plane through the origin inside $\mathbb{R}^3$ feels like it should be a little 2D vector space living inside the big one. It is. A line through the origin too. The set $\{\mathbf{0}\}$ on its own — also a subspace. But a plane that misses the origin? We proved last node that anything missing $\mathbf{0}$ is dead on arrival. So the geometry is going to be unforgiving.

The three-condition test

Here's the shortcut. $W \subseteq V$ is a subspace if and only if:

1. $\mathbf{0} \in W$ (it contains the zero vector — equivalently, $W$ is nonempty).
2. Closed under addition: if $u, w \in W$ then $u + w \in W$.
3. Closed under scalar multiplication: if $w \in W$ and $c \in \mathbb{R}$ then $cw \in W$.

Three checks. That's it. Stop looking for a catch — there isn't one. The two closure conditions are the same as axioms 1–2 from last node; condition 1 just pins down that we haven't accidentally described the empty set.

Why three conditions buy you all ten

This is the part textbooks wave away with a footnote and hope you don't notice, and Dr. Möbius absolutely does not wave. Why is the test enough?

Because the other seven axioms are inherited for free. Commutativity ($u+w = w+u$), associativity, the scalar laws (7–10) — these are universal statements: they say "for all elements, such-and-such equation holds." If they hold for all elements of the big space $V$, they automatically hold for all elements of the smaller set $W \subseteq V$, since every element of $W$ is also an element of $V$. You can't break a "for all" law by looking at fewer things.

The only axioms that could fail on a subset are the existence axioms — does $\mathbf{0}$ live in $W$ (axiom 5)? Does each $w$ have its inverse $-w$ in $W$ (axiom 6)? — plus closure (1–2), which is the risk that the operations escape the subset. Condition 1 handles the zero. Conditions 2–3 handle closure. And the inverse? It comes free from condition 3: $-w = (-1)w$ (proved last node!), and condition 3 with $c = -1$ guarantees $(-1)w \in W$. So the three conditions cover every axiom that could possibly fail. Argued once, trusted forever.

The geometry of $\mathbb{R}^3$: a complete census

For $\mathbb{R}^3$ the subspaces are exactly four flavors, and you should be able to recite them cold:

• $\{\mathbf{0}\}$ — the zero subspace. Dimension $0$.
• Lines through the origin. Dimension $1$.
• Planes through the origin. Dimension $2$.
• All of $\mathbb{R}^3$. Dimension $3$.

That's the complete list. No others exist — not an infinite helix, not a sphere, not any curved shit. And the iron law uniting them: every subspace contains the origin. A line that doesn't pass through $\mathbf{0}$? Not a subspace — scale any of its points by $0$ and you'd need $\mathbf{0}$ inside, but it isn't. Through the origin or it's out. Tattoo it.

The null space is born

Now we cash this in on something from the Matrices stratum, and it's genuinely beautiful — the kind of thing that makes you feel guilty for not seeing it sooner. Take a matrix $A$ (say $m \times n$) and look at all solutions of the homogeneous system $A\mathbf{x} = \mathbf{0}$: $N(A) = \{\mathbf{x} \in \mathbb{R}^n : A\mathbf{x} = \mathbf{0}\}.$

Theorem. $N(A)$ is a subspace of $\mathbb{R}^n$.

Proof. Run the three-condition test.

1. $A\mathbf{0} = \mathbf{0}$, so $\mathbf{0} \in N(A)$. (Zero vector is always a solution of a homogeneous system.) ✓
2. Let $\mathbf{x}, \mathbf{y} \in N(A)$, so $A\mathbf{x} = \mathbf{0}$ and $A\mathbf{y} = \mathbf{0}$. By linearity of matrix multiplication (Matrices stratum), $A(\mathbf{x} + \mathbf{y}) = A\mathbf{x} + A\mathbf{y} = \mathbf{0} + \mathbf{0} = \mathbf{0}$, so $\mathbf{x} + \mathbf{y} \in N(A)$. ✓
3. Let $\mathbf{x} \in N(A)$, $c \in \mathbb{R}$. Then $A(c\mathbf{x}) = c(A\mathbf{x}) = c\mathbf{0} = \mathbf{0}$, so $c\mathbf{x} \in N(A)$. ✓

All three hold, so $N(A)$ is a subspace. $\blacksquare$ This set is so important it gets a name you'll use forever: the null space (or kernel) of $A$.

And the impostor: $A\mathbf{x} = \mathbf{b}$ with $\mathbf{b} \ne \mathbf{0}$

Watch the same idea fail the instant we make it inhomogeneous. The solution set of $A\mathbf{x} = \mathbf{b}$ with $\mathbf{b} \ne \mathbf{0}$ is not a subspace — not even close.

Why? Condition 1 already dies: $A\mathbf{0} = \mathbf{0} \ne \mathbf{b}$, so $\mathbf{0}$ is not a solution. No zero vector, not a subspace — first question asked, exam over. (You can also kill it with closure: if $A\mathbf{x} = \mathbf{b}$ then $A(2\mathbf{x}) = 2\mathbf{b} \ne \mathbf{b}$.) Geometrically it's a plane or line shifted off the origin — an "affine" set, the parallel copy of $N(A)$ pushed away from home. Same shape, wrong neighborhood.

Intersections yes, unions no

Two final structural facts. If $U$ and $W$ are subspaces of $V$, then:

• $U \cap W$ is always a subspace. (Anything in both is closed under the operations in both — three-condition test passes trivially. The intersection of the $xy$-plane and the $xz$-plane is the $x$-axis: a line through the origin. Still a subspace.)
• $U \cup W$ is usually NOT. Counterexample: in $\mathbb{R}^2$, let $U$ be the $x$-axis and $W$ the $y$-axis. Both are subspaces. But $(1,0) \in U$ and $(0,1) \in W$ are both in the union, while their sum $(1,1)$ is on neither axis — closure under addition fails. The union of two lines isn't a plane; it's a sad X that breaks the moment you add across the arms.

Remember the Set Theory stratum, where $\cap$ and $\cup$ behaved so symmetrically? Here the symmetry shatters — algebra doesn't give a damn about set-theory aesthetics. Subspaces care about algebraic structure, not just membership. Intersection preserves it; union smashes it into rubble.