Subspaces, Linear Combinations and the Linear Span

A vector space is usually too large to study all at once; the objects of real interest are its internal copies — subspaces — and the operations that build them out of a handful of vectors: linear combinations and span. This chapter develops the single most-used test in linear algebra (the subspace criterion) and proves the two structural facts every later chapter quietly relies on: span is always a subspace, and it is the smallest one containing the given vectors.

Have to hand: Chapter 1's vector space axioms and the proof style that goes with them (name every axiom used).

1. The Subspace Criterion

Definition (subspace). Let $V$ be a vector space over $\mathbb{F}$. A subset $U \subseteq V$ is a subspace of $V$ if $U$ is itself a vector space over $\mathbb{F}$ under the addition and scalar multiplication inherited from $V$.

Checking all eight vector space axioms directly would be wasteful — associativity, commutativity and the distributive laws are automatically inherited from $V$, since they are identities that already hold for every vector in $V$, hence in particular for vectors in $U$. Only three things can actually fail for a subset, which gives a three-condition test.

Proposition (subspace test). $U \subseteq V$ is a subspace if and only if:

1. $0 \in U$ (non-emptiness, anchored at the zero vector);
2. $u_1, u_2 \in U \implies u_1 + u_2 \in U$ (closure under addition);
3. $a \in \mathbb{F}, u \in U \implies a\cdot u \in U$ (closure under scalar multiplication).

Proof. ($\Rightarrow$) If $U$ is a subspace it is a vector space, hence has a zero vector; this zero vector must coincide with $0_V$ because the zero vector of $U$ satisfies $u + 0_U = u$ for $u \in U \subseteq V$, and $0_V$ is the unique element of $V$ with this property (Chapter 1), so $0_U = 0_V \in U$. Conditions 2 and 3 are immediate since $U$ is closed under its own operations by definition of being a vector space.

($\Leftarrow$) Suppose conditions 1–3 hold. Associativity, commutativity of $+$, and the four scalar-multiplication axioms (axioms 2–5 of Chapter 1) hold for all elements of $V$, hence restrict to hold for all elements of $U$ — these are universally quantified identities, inherited for free. Condition 1 gives the additive identity inside $U$. For additive inverses: given $u \in U$, condition 3 with $a = -1$ gives $(-1)\cdot u \in U$, and Chapter 1 proved $(-1)\cdot u = -u$; so $-u \in U$. Conditions 2 and 3 are exactly closure. Hence $U$ satisfies every vector space axiom and is a vector space in its own right — a subspace. $\blacksquare$

In practice, conditions 2 and 3 are usually combined and checked together as a single step: $a_1 u_1 + a_2 u_2 \in U$ for all $a_1, a_2 \in \mathbb{F}$, $u_1, u_2 \in U$. This is logically equivalent to conditions 2+3 (take $a_1=a_2=1$ for condition 2, $u_2=0$ for condition 3) and is faster to verify.

Worked example. Let $U = \{(x,y,z) \in \mathbb{R}^3 : x + 2y - z = 0\}$. Check: $0=(0,0,0)$ satisfies $0+0-0=0$, so $0\in U$. If $(x_1,y_1,z_1),(x_2,y_2,z_2)\in U$ and $a,b\in\mathbb{R}$, then the combination $a(x_1,y_1,z_1)+b(x_2,y_2,z_2)=(ax_1+bx_2, ay_1+by_2, az_1+bz_2)$ satisfies $(ax_1+bx_2)+2(ay_1+by_2)-(az_1+bz_2) = a(x_1+2y_1-z_1)+b(x_2+2y_2-z_2) = a\cdot0+b\cdot0=0$, using that both points satisfy the defining equation. So $U$ is a subspace — this is the general pattern: the solution set of a homogeneous linear equation (or system) is always a subspace, because the left-hand side is a linear expression in the coordinates.

2. Non-Subspace Examples and Why They Fail

• $W = \{(x,y)\in\mathbb{R}^2 : x^2=y\}$ fails: although $0\in W$, we have $(1,1)\in W$ but $2\cdot(1,1)=(2,2)\notin W$ since $2^2=4\ne2$. Closure under scalar multiplication fails because the defining equation is not linear.
• $W=\{(x,y) : x+y=1\}$ fails condition 1 immediately: $0+0=0\ne1$, so $0\notin W$. Any affine (shifted) set that does not pass through the origin is never a subspace.
• $W = \{(x,y)\in\mathbb{R}^2 : x\ge0\}$ fails condition 3: $(1,0)\in W$ but $(-1)\cdot(1,0)=(-1,0)\notin W$ (Chapter 1, Exercise).

Key distinction. A subset can satisfy closure under addition without satisfying closure under scalar multiplication, or vice versa — the three conditions are logically independent and all three must be checked. A common shortcut, checking only condition 1, catches none of the genuinely interesting failures.

3. Linear Combinations and Span

Definition. A linear combination of vectors $v_1,\dots,v_n \in V$ is any vector of the form $a_1v_1 + \cdots + a_nv_n$ with $a_1,\dots,a_n \in \mathbb{F}$.

Definition (span). The span of a subset $S \subseteq V$, written $\operatorname{span}(S)$, is the set of all linear combinations of finitely many vectors of $S$. By convention $\operatorname{span}(\varnothing) = \{0\}$.

Theorem. For any $S \subseteq V$, $\operatorname{span}(S)$ is a subspace of $V$, and it is the smallest subspace of $V$ containing $S$ (i.e. if $U$ is any subspace with $S \subseteq U$, then $\operatorname{span}(S) \subseteq U$).

Proof. Subspace. $0 \in \operatorname{span}(S)$ (empty combination, or $0 = 0\cdot s$ for any $s \in S$ if $S \ne \varnothing$). If $u = \sum_{i=1}^m a_i v_i$ and $w = \sum_{j=1}^k b_j w_j$ are two linear combinations of vectors of $S$, then for $c,d \in \mathbb{F}$, $cu + dw = \sum_i (ca_i)v_i + \sum_j (db_j)w_j$ is again a (finite) linear combination of vectors of $S$ — this uses axioms 3, 4, 5 of Chapter 1 to redistribute the scalars — so $\operatorname{span}(S)$ is closed under the combined operation of condition 2+3 above, hence a subspace.

Smallest. Let $U$ be any subspace with $S \subseteq U$. Every vector $s \in S$ lies in $U$. By the subspace criterion, $U$ is closed under linear combinations, so for $v_1,\dots,v_n \in S \subseteq U$, $a_1v_1+\cdots+a_nv_n \in U$. Hence every element of $\operatorname{span}(S)$ lies in $U$, i.e. $\operatorname{span}(S) \subseteq U$. $\blacksquare$

This theorem is the precise sense in which "span is the subspace generated by $S$": it is simultaneously a subspace containing $S$, and smaller than (contained in) every other such subspace.

Worked example. Let $S = \{(1,0,0),(0,1,0)\}$ in $\mathbb{R}^3$. Then $\operatorname{span}(S) = \{a(1,0,0)+b(0,1,0) : a,b\in\mathbb{R}\} = \{(a,b,0) : a,b\in\mathbb{R}\}$, the $xy$-plane. Checking the smallest-subspace property directly: any subspace $U$ containing $(1,0,0)$ and $(0,1,0)$ must, by closure, contain $a(1,0,0)+b(0,1,0) = (a,b,0)$ for every $a,b$, so $U \supseteq \{(a,b,0)\}$ necessarily — the $xy$-plane cannot be avoided once those two generators are required to be in $U$.

4. Sums of Subspaces

Definition. If $U_1, \dots, U_k$ are subspaces of $V$, their sum is $U_1 + \cdots + U_k := \{u_1+\cdots+u_k : u_i \in U_i\}$.

Proposition. $U_1+\cdots+U_k$ is a subspace of $V$, and in fact $U_1+\cdots+U_k = \operatorname{span}(U_1\cup\cdots\cup U_k)$.

Proof sketch. Closure under the combined linear-combination operation follows entrywise exactly as for span. For the equality: every element of $U_1+\cdots+U_k$ is manifestly a linear combination of elements of $U_1\cup\cdots\cup U_k$ (namely $1\cdot u_1+\cdots+1\cdot u_k$), so $U_1+\cdots+U_k \subseteq \operatorname{span}(U_1\cup\cdots\cup U_k)$. Conversely, any linear combination of elements drawn from the $U_i$ can be regrouped, using closure of each $U_i$ under scalar multiplication and addition, into a sum of one element from each $U_i$, giving the reverse inclusion. $\blacksquare$

5. Direct Sums

Definition. $V$ is the direct sum of subspaces $U_1,\dots,U_k$, written $V = U_1 \oplus \cdots \oplus U_k$, if $V = U_1+\cdots+U_k$ and every $v \in V$ can be written as $u_1+\cdots+u_k$ ($u_i \in U_i$) in exactly one way.

For two subspaces, uniqueness of decomposition reduces to a single clean test.

Theorem. $V = U_1 \oplus U_2$ if and only if $V = U_1+U_2$ and $U_1 \cap U_2 = \{0\}$.

Proof. ($\Rightarrow$) Suppose $v \in U_1\cap U_2$ with $v \ne 0$. Then $v = v + 0$ (with $v\in U_1,0\in U_2$) and $v = 0+v$ (with $0\in U_1, v\in U_2$) are two different decompositions of $v$ as $u_1+u_2$ (different because $v \ne 0$ means the pairs $(v,0)$ and $(0,v)$ are distinct), contradicting uniqueness. So $U_1\cap U_2=\{0\}$.

($\Leftarrow$) Suppose $U_1\cap U_2=\{0\}$ and $v = u_1+u_2 = u_1'+u_2'$ are two decompositions. Then $u_1-u_1' = u_2'-u_2$. The left side lies in $U_1$ (subspace, closed under subtraction: $u_1+(-1)u_1'$), the right side in $U_2$; since they're equal, this common vector lies in $U_1\cap U_2=\{0\}$, forcing $u_1=u_1'$ and $u_2=u_2'$. $\blacksquare$

A frequent confusion: for two subspaces, $U_1\cap U_2=\{0\}$ must genuinely be checked as an intersection of subspaces — it is not enough that the subspaces merely look different. For three or more subspaces, pairwise trivial intersection is not sufficient for a direct sum (a standard counterexample: three distinct lines through the origin in $\mathbb{R}^2$ pairwise intersect only at $0$, but their sum is all of $\mathbb{R}^2$ and decompositions are not unique).

Common Exam Pitfalls

• Checking only $0 \in U$ and declaring $U$ a subspace. All three conditions are independent; a set can contain $0$ and still fail closure (e.g. $\{(x,y):xy=0\}$, the union of the two axes, contains $0$ but $(1,0)+(0,1)=(1,1)$ is not in the set).
• Testing a nonlinear defining equation and expecting a subspace. Only homogeneous linear equations (and intersections of such) define subspaces; anything with squares, products of variables, or a nonzero constant term fails.
• Confusing span with union. $\operatorname{span}(\{u,w\})$ contains far more than $\{u,w\}$ itself — all linear combinations, generically an entire plane through the origin, not just the two original vectors.
• Treating "sum" and "direct sum" as synonyms. $U_1+U_2$ always makes sense and is always a subspace; it is a direct sum only when $U_1\cap U_2=\{0\}$, which must be separately verified.
• Assuming pairwise trivial intersection suffices for $k\ge3$ direct summands. The correct general criterion (developed further via dimension in Chapter 3) requires more than pairwise disjointness of intersections.
• Forgetting the "smallest subspace" half of the span theorem. Many students can show $\operatorname{span}(S)$ is a subspace but forget that proving it is the smallest one containing $S$ is an equally essential, separate half of the theorem.

Summary

• Subspace criterion: $U\subseteq V$ is a subspace iff $0\in U$, $U$ is closed under addition, and $U$ is closed under scalar multiplication; the remaining axioms (associativity and commutativity of $+$, and axioms 2–5 of Chapter 1) are automatically inherited from $V$.
• Linear (not affine, not polynomial) homogeneous equations define subspaces; a nonzero constant term or a nonlinear term breaks this.
• Span: $\operatorname{span}(S)$ is the set of all finite linear combinations of elements of $S$; it is always a subspace, and it is the smallest subspace containing $S$ — both halves require proof.
• Sum of subspaces: $U_1+\cdots+U_k=\{u_1+\cdots+u_k\}=\operatorname{span}(U_1\cup\cdots\cup U_k)$, always a subspace.
• Direct sum: $V=U_1\oplus U_2$ iff $V=U_1+U_2$ and $U_1\cap U_2=\{0\}$; for $k\ge3$ summands, pairwise trivial intersection is not by itself sufficient.