Subspaces

Which subsets of a vector space are themselves vector spaces, using the same operations? This question sounds narrow. But its answer organizes almost everything that follows in linear algebra. A line through the origin in the plane, a plane through the origin in space, the solutions to a homogeneous linear system, the polynomials of degree at most three, the symmetric matrices — all of these are subspaces. Recognizing them as such instantly hands you a rich supply of theorems for free. The reason subspaces matter is efficiency. A vector space must satisfy ten axioms. These include closure under two operations, commutativity, associativity, identities, inverses, and distributive laws. Checking all ten from scratch for every subset would be exhausting and redundant. This is because a subset of a vector space automatically inherits associativity, commutativity, and the distributive laws from the ambient space. The subspace criterion isolates exactly the three conditions that can actually fail. It turns a ten-item checklist into a three-item one. The big picture is that subspaces are the natural "linear" subsets. They are the ones closed under the two operations that define a vector space, addition and scalar multiplication. Geometrically they are exactly the flat sets through the origin — lines, planes, and their higher-dimensional analogues. Algebraically they are the solution sets of homogeneous linear equations. They are also the kernels and images of linear maps, and the building blocks for the notion of span. This article gives the precise definition of a subspace. It proves the subspace criterion with full rigor. It works through standard examples, including solution spaces of homogeneous systems. It shows exactly which hypotheses cannot be dropped. It closes by connecting subspaces to the ideas of span and sums of subspaces. These ideas build directly on top of this material.

Key takeaways

A subspace is a subset of a vector space that is itself a vector space under the same operations.
The subspace criterion reduces ten vector space axioms to three checkable conditions.
The zero vector must lie in every subspace; a subset missing it is never a subspace.
Solution sets of homogeneous linear systems $Ax = 0$ are always subspaces; non-homogeneous ones never are.
Lines and planes through the origin are subspaces of $\mathbb{R}^2$ and $\mathbb{R}^3$ ; those not through the origin are not.
Key idea: to check a subset is a subspace, verify it contains $0$ and is closed under addition and scalar multiplication — everything else is inherited automatically.

Where this lives today

Subspaces are the organizing objects behind nearly every application of linear algebra. In differential equations, the solution set of a homogeneous linear ODE (or a homogeneous linear recurrence) forms a subspace of the space of functions (or sequences), which is exactly why superposition of solutions works: any linear combination of solutions is again a solution. In computer graphics and robotics, the set of velocity directions tangent to a constraint surface at a point forms a subspace (the tangent space), used to compute feasible motions. In coding theory and cryptography, linear codes are defined precisely as subspaces of $\mathbb{F}_q^n$ for a finite field $\mathbb{F}_q$ ; the fact that codewords form a subspace (rather than an arbitrary subset) is what makes efficient encoding and decoding algorithms possible, including syndrome decoding. In statistics, the column space of a design matrix in linear regression is a subspace of $\mathbb{R}^n$ , and least-squares fitting is literally the problem of projecting a data vector onto this subspace. In quantum mechanics, physical states with a fixed value of a conserved quantity (like total spin or particle number) span a subspace of the full Hilbert space, and observables act by mapping subspaces to subspaces. In numerical linear algebra, the null space and column space of a matrix — both subspaces — are the two fundamental subspaces used to analyze the solvability and rank of linear systems, directly generalizing Example 1 above. All of these depend on the same three-condition checklist proved in the Subspace Criterion.

Ask the tutor: Why is the null space of a matrix always a subspace but the solution set of an inhomogeneous system never is?

V

Intuition

Think of a vector space as a flat, infinitely extended geometric object anchored at the origin. A subspace is a "flat" piece of that object which itself passes through the origin and is closed under stretching and adding. In $\mathbb{R}^2$ , the subspaces are: the zero vector alone, every line through the origin, and the whole plane. Nothing else qualifies — not a line that misses the origin, not a single quadrant, not a circle. The word "closed" is doing real work. If you take two vectors in $W$ and add them, the result must land back in $W$ ; if you scale a vector in $W$ by any scalar (including $0$ and negative numbers), the result must land back in $W$ . Scaling by $0$ is exactly what forces $0 \in W$ — so condition 1 is almost a consequence of condition 3, except that condition 3 is stated as "for all $u \in W$ ," which is vacuous if $W$ is empty. That is why condition 1 is listed separately: it also guarantees $W$ is nonempty, which is needed before condition 3 says anything. Geometrically, closure under scalar multiplication means: if a vector is in $W$ , the entire line through the origin and that vector is in $W$ . Closure under addition means: if two vectors are in $W$ , every vector obtainable by the parallelogram rule from them is in $W$ . Together, these two closure properties are precisely what makes $W$ a union of lines through the origin that is also "flat" in the sense of containing all sums.

Figure

Subspaces vs. non-subspaces of

\mathbb{R}^2

: lines through the origin qualify, a shifted line and the axes-only set do not

Theorem: The Subspace Criterion

V

V

The proof technique here is direct verification of axioms, split into a forward direction that extracts necessary conditions from the definition of "vector space," and a backward direction that shows those same three conditions are sufficient because every other axiom is a universally quantified statement about $V$ that automatically restricts to any subset closed under the operations.

W

W \ne \emptyset

This corollary is the version most often used in practice: check $W \ne \emptyset$ (usually by exhibiting $0$ or noting some obvious element lies in $W$ ), then check closure under addition and scalar multiplication.

Worked examples

Example 1: The solution set of a homogeneous linear system. Let $A$ be a fixed $m \times n$ matrix with real entries, and let

W = \{\, x \in \mathbb{R}^n : Ax = 0 \,\}.

We claim $W$ is a subspace of $\mathbb{R}^n$ . Take, concretely, $n = 4$ and

A = \begin{pmatrix} 1 & 2 & 0 & -1 \\ 0 & 1 & 3 & 1 \end{pmatrix},

so $W = \{ (x_1,x_2,x_3,x_4) \in \mathbb{R}^4 : x_1 + 2x_2 - x_4 = 0,\ x_2 + 3x_3 + x_4 = 0\}$ . Check (i): The zero vector $0 = (0,0,0,0)$ satisfies $A0 = 0$ , since matrix multiplication sends $0$ to $0$ (each entry of $A0$ is a sum of terms each multiplied by $0$ ). So $0 \in W$ . Check (ii): Suppose $u, v \in W$ , so $Au = 0$ and $Av = 0$ . By linearity of matrix multiplication (distributivity, an established property of matrix-vector products),

A(u+v) = Au + Av = 0 + 0 = 0,

so $u + v \in W$ . Check (iii): Suppose $c \in \mathbb{R}$ and $u \in W$ , so $Au = 0$ . By linearity,

A(cu) = c(Au) = c \cdot 0 = 0,

so $cu \in W$ . By the Subspace Criterion, $W$ is a subspace of $\mathbb{R}^4$ . Concretely, let's verify with an actual solution. Set $x_3 = 1, x_4 = 0$ : the second equation gives $x_2 = -3x_3 - x_4 = -3$ , and the first gives $x_1 = x_4 - 2x_2 = 0 - 2(-3) = 6$ . So $u = (6,-3,1,0) \in W$ ; check: $6 + 2(-3) - 0 = 6-6=0$ ✓, and $-3+3(1)+0=0$ ✓. Scaling, $3u = (18,-9,3,0)$ also solves the system: $18+2(-9)-0=0$ ✓ and $-9+3(3)+0=0$ ✓, confirming closure under scalar multiplication in this instance. This example is the prototype for why homogeneity matters: the right-hand side must be $0$ , precisely so that $0$ is a solution and so that the linearity computations $A(u+v)=Au+Av$ and $A(cu) = cAu$ preserve the defining equation $=0$ . Compare with the counterexample below.

Example 2: Symmetric $2\times 2$ matrices form a subspace, but matrices of determinant $1$ do not. Let $V = M_{2\times 2}(\mathbb{R})$ , the vector space of all $2\times 2$ real matrices (a standard example of a vector space, with entrywise addition and scalar multiplication). Let

W = \{ A \in M_{2\times2}(\mathbb{R}) : A^T = A \}

be the set of symmetric matrices. We check $W$ is a subspace. Check (i): The zero matrix $0 = \begin{pmatrix} 0&0\\0&0\end{pmatrix}$ satisfies $0^T = 0$ , so $0 \in W$ . Check (ii): Let $A, B \in W$ , so $A^T = A$ and $B^T = B$ . Using the property $(A+B)^T = A^T + B^T$ (a standard fact about the transpose, itself provable entrywise: the $(i,j)$ entry of $(A+B)^T$ is the $(j,i)$ entry of $A+B$ , namely $a_{ji}+b_{ji}$ , which equals the $(i,j)$ entry of $A^T + B^T$ ),

(A+B)^T = A^T + B^T = A + B,

so $A + B \in W$ . Check (iii): Let $c \in \mathbb{R}$ , $A \in W$ . Using $(cA)^T = cA^T$ (again checkable entrywise),

(cA)^T = cA^T = cA,

so $cA \in W$ . By the Subspace Criterion, $W$ is a subspace of $M_{2\times 2}(\mathbb{R})$ . Let's verify with concrete matrices: take

A = \begin{pmatrix} 1 & 5 \\ 5 & -2 \end{pmatrix}, \qquad B = \begin{pmatrix} 3 & -4 \\ -4 & 7\end{pmatrix},

both symmetric. Then

A + B = \begin{pmatrix} 4 & 1 \\ 1 & 5 \end{pmatrix},

which is symmetric (the $(1,2)$ entry $1$ equals the $(2,1)$ entry $1$ ), confirming closure under addition in this instance. And $7A = \begin{pmatrix} 7 & 35\\ 35 & -14\end{pmatrix}$ is symmetric as well. Now contrast this with

U = \{ A \in M_{2\times2}(\mathbb{R}) : \det(A) = 1 \}.

This is not a subspace. First, $0 \notin U$ since $\det(0) = 0 \ne 1$ , already violating condition (i) — so by the Subspace Criterion (contrapositive of the forward direction, or just directly: a subspace must contain $0$ ), $U$ cannot be a subspace. Concretely, take

A = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \in U \quad (\det A = 1),

and note $A + A = \begin{pmatrix} 2&0\\0&2\end{pmatrix}$ has determinant $4 \ne 1$ , so $U$ also fails closure under addition. Both failures independently show $U$ is not a subspace.

A = \begin{pmatrix} 1 & 2 & 0 & -1 \\ 0 & 1 & 3 & 1 \end{pmatrix},

Scope, edge cases & common traps

Situation	Is it a subspace?	Why
$\{0\}$ alone (the trivial subspace)	Yes	Vacuously/trivially satisfies all three conditions
$V$ itself	Yes	Every vector space is a subspace of itself
A line through the origin in $\mathbb{R}^2$	Yes	Closed under addition (collinear vectors add to collinear vectors on the same line) and scaling
A line not through the origin	No	Fails to contain $0$
Solution set of $Ax=0$	Yes	Example 1 above; this is the kernel or null space of $A$
Solution set of $Ax=b$ , $b\ne 0$	No	Counterexample above; this is an affine subspace, not linear
Union of two distinct lines through origin in $\mathbb{R}^2$	No	Fails closure under addition (sum of one vector from each line is generally on neither)
Intersection of two subspaces	Yes	Both closure conditions and containment of $0$ pass through intersection (proved similarly, foreshadowing Sums of Subspaces)
First-quadrant vectors in $\mathbb{R}^2$	No	Closed under addition and contains $0$ , but fails scalar closure: $(-1)\cdot(1,1) = (-1,-1)$ leaves the set
Vectors with integer entries in $\mathbb{R}^2$	No	Closed under addition and contains $0$ , but fails scalar closure over $\mathbb{R}$ : $\tfrac12 (1,0) \notin \mathbb{Z}^2$

Common trapThe classic error: checking closure but forgetting the zero vector, or checking over the wrong field

Two errors recur constantly. First, students verify closure under addition and scalar multiplication but skip checking $0 \in W$ , missing sets like "vectors with positive first coordinate" that are closed under addition (sum of two positives is positive) yet fail because $0$ is excluded and negative scalars break scalar closure anyway — but subtler examples exist where addition and some scalars work and only the omission of $0$ signals trouble immediately. Second, and more subtle: closure under scalar multiplication must hold for every scalar in the field $\mathbb{F}$ , not just integers or positive reals. The set $\mathbb{Z}^2 \subset \mathbb{R}^2$ is closed under addition and contains $0$ , but is not an $\mathbb{R}$ -subspace because scalar multiplication by $\tfrac12$ (a perfectly legal real scalar) leaves the set. Always ask: closed under addition, closed under all scalars from the ambient field, and does it contain the zero vector?

The mathematician behind it

Origins · 1 minGiuseppe Peano and how this was discovered

Giuseppe Peano

1858–1932 · Italy, foundational analysis and axiomatic mathematics

Giuseppe Peano was an Italian mathematician best known for axiomatizing the natural numbers, but he was also the first to give a fully explicit, modern axiomatic definition of a vector space, in his 1888 book Calcolo Geometrico. Building on Hermann Grassmann's earlier, more geometric Ausdehnungslehre (1844), Peano isolated the abstract algebraic properties — closure, associativity, distributivity — that define a "linear system," anticipating by decades the abstract treatment of vector spaces and their subspaces that became standard in the twentieth century. His work made precise exactly the kind of axiom-checking that the subspace criterion exploits: once a structure satisfies a short list of axioms, an entire calculus of linear combinations, spans, and subspaces follows mechanically.

History

- 1844: Hermann Grassmann publishes Die lineale Ausdehnungslehre, introducing linear combinations and extended magnitudes, a geometric precursor to vector spaces.

- 1888: Giuseppe Peano gives the first modern axiomatic definition of a real vector space in Calcolo Geometrico, including explicit closure axioms.

- 1918: Hermann Weyl, in Raum, Zeit, Materie, uses an axiomatic vector space formulation to underpin affine and Euclidean geometry, popularizing the abstract approach in physics.

- 1920s–1930s: Stefan Banach and others generalize vector spaces to infinite dimensions with norms, making the subspace concept central to functional analysis.

- 1940s: The subspace criterion becomes a standard textbook tool as linear algebra is systematized for university curricula following the work of Halmos and others.

Does every vector space have at least one subspace?

Yes — every vector space $V$ has at least two subspaces automatically: the trivial subspace $\{0\}$ , consisting of only the zero vector, and $V$ itself. Both satisfy the subspace criterion vacuously or trivially: $\{0\}$ contains zero, and $0+0=0 \in \{0\}$ , $c\cdot 0 = 0 \in \{0\}$ for any scalar $c$ ; and $V$ trivially satisfies all three conditions since it is closed under its own operations by definition. These are sometimes called the trivial subspaces. A vector space may have many more subspaces (infinitely many, if $\dim V \ge 1$ over an infinite field), or, if $V = \{0\}$ itself, these two trivial subspaces coincide.

Is a subspace required to be a proper subset, or can it equal the whole space?

A subspace can equal the whole space $V$ ; there is no requirement of properness in the general definition. When authors want to exclude this case, they use the term proper subspace, meaning $W \subsetneq V$ (strictly contained). Similarly, some authors reserve "subspace" to exclude the trivial subspace $\{0\}$ using the phrase "nonzero subspace," but by default, both $\{0\}$ and $V$ count as subspaces of $V$ . Always check context or explicit qualifiers like "proper" or "nontrivial" when reading a claim about subspaces.

Why do we need to check scalar multiplication for all scalars, including negative ones and non-integers?

Because the field $\mathbb{F}$ over which $V$ is a vector space includes all its elements as legitimate scalars, and the subspace must be closed under multiplication by every one of them, not just a convenient subset. Over $\mathbb{R}$ , this includes negative numbers, irrational numbers, and zero. A common trap is a set like the first quadrant $\{(x,y): x,y\ge 0\}$ , which is closed under addition and contains $0$ , yet fails because multiplying by $c=-1$ produces a vector outside the set. The scalar field matters too: $\mathbb{Q}^2$ is a subspace of $\mathbb{Q}^2$ over $\mathbb{Q}$ but not of $\mathbb{R}^2$ over $\mathbb{R}$ , since irrational scalars break closure.

How does the subspace criterion relate to the concept of span, which comes next?

The subspace criterion tells you how to recognize a subspace once you have a candidate subset. The concept of span tells you how to construct a subspace from a given set of vectors: the span of a set $S$ is defined as the set of all linear combinations of vectors in $S$ , and one proves (using exactly the subspace criterion) that this span is always a subspace — indeed the smallest subspace containing $S$ . So the criterion proved here is the tool used immediately afterward to justify that span produces a genuine subspace, rather than merely some subset.

Is the empty set ever a subspace?

No, never. Every subspace must contain the zero vector by condition (i) of the subspace criterion, so the empty set — which contains nothing — automatically fails to be a subspace of any vector space with at least one element (and every vector space has at least the zero vector, so every vector space is nonempty). This is a frequent point of confusion because the empty set is vacuously closed under addition and scalar multiplication (there are no elements to violate those conditions), but vacuous closure is not enough; the explicit requirement $0 \in W$ rules the empty set out immediately.

Are the row space, column space, and null space of a matrix all subspaces?

Yes, all three are subspaces, though of different ambient spaces. For an $m \times n$ matrix $A$ : the column space (span of the columns) is a subspace of $\mathbb{R}^m$ ; the row space (span of the rows) is a subspace of $\mathbb{R}^n$ ; and the null space $\{x \in \mathbb{R}^n : Ax = 0\}$ is a subspace of $\mathbb{R}^n$ , exactly as verified in Example 1 of this article. All three are proved subspaces either directly via the subspace criterion (as done here for the null space) or via the fact that any span of a set of vectors is automatically a subspace, a fact that follows immediately from the same criterion.

W_1

and

W_2

are both subspaces of

V

, is

W_1 \cup W_2

always a subspace?

No, generally not — only $W_1 \cap W_2$ is guaranteed to be a subspace. The union $W_1 \cup W_2$ is a subspace if and only if $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$ ; otherwise, closure under addition typically fails, as in the $x$ -axis/ $y$ -axis example above (each axis is a one-dimensional subspace, but their union is not closed under addition). This is exactly why the notion of sum of subspaces, $W_1 + W_2 = \{w_1+w_2 : w_1\in W_1, w_2\in W_2\}$ , is introduced: it is the smallest subspace that does contain both $W_1$ and $W_2$ , correcting for the failure of the union to be closed under addition.