Everything in this course is built on top of four ideas Linear Algebra I already made precise: a vector space is a set closed under addition and scalar multiplication, a subspace is a smaller such set living inside a bigger one, the span of a set of vectors is every linear combination you can build from them, and a set is linearly independent when no vector in it is redundant — none is a combination of the others. Nothing below changes any of that. What changes, starting in chapter 01, is the ambient set these definitions get tested against: not just \mathbb{R}^n, but polynomials, matrices, and vector spaces over fields that aren't \mathbb{R} at all.
Think of span as "everything reachable." Give someone two vectors and ask what they can build by scaling and adding them. That reachable set is the span. It might turn out to be a single point, a line, a plane, or all of space. Independence just asks whether you handed over any vector you didn't need. If one vector was already reachable from the others, it added nothing, and the set is dependent. A subspace is simply a reachable set that's self-contained. Combine anything inside it using only addition and scaling, and you never land outside it again.
A vector space is a set of objects, called vectors, that can be added to each other and scaled by numbers from a field, with both operations obeying the ordinary rules of algebra (associativity, a zero vector, scaling by 1 changes nothing). The span of a set of vectors is every vector reachable by adding and scaling them — every linear combination you can build. A subspace test only ever needs three checks — the zero vector belongs, the set is closed under addition, and it's closed under scalar multiplication — because every other closure property (subtraction, arbitrary linear combinations) is a consequence of those three, not an independent fact to verify. Independence is the sharper of the two ideas here: a spanning set can be wasteful (more vectors than the space actually needs), but an independent set is never wasteful, and the two properties meeting at once is exactly what a basis will turn out to be in the next section.
Theorem
The Subspace Test
Recalled from Linear Algebra — Vectors, Matrices, and the Geometry of Data · Testing whether a set is a subspace — first taught there.
Let V be a vector space and let W \subset V be a nonempty subset. Then W is a subspace of V if and only if both of the following hold:
\mathbf{u}, \mathbf{v} \in W \implies \mathbf{u} + \mathbf{v} \in Wc \in \mathbb{R},\ \mathbf{u} \in W \implies c\mathbf{u} \in W(\implies) If W is a subspace, it is itself a vector space under the addition and scalar multiplication inherited from V, so both closure properties hold automatically — that direction is immediate from the definition of "subspace" and needs no further argument.
(\impliedby) This is the direction that does real work: assume W is nonempty and closed under addition and scalar multiplication, and check that W inherits every vector space axiom from V.
First, \mathbf{0} \in W. Since W is nonempty, pick any \mathbf{u} \in W. Closure under scalar multiplication with c = 0 gives 0 \cdot \mathbf{u} = \mathbf{0} \in W. So W contains the zero vector without that having to be assumed separately — it falls out of nonemptiness plus scalar closure.
Next, every \mathbf{u} \in W has its negative in W: closure under scalar multiplication with c = -1 gives (-1)\mathbf{u} = -\mathbf{u} \in W.
With \mathbf{0} \in W and negatives available inside W, the two closure hypotheses supply exactly the operations W needs: addition of any two elements of W stays in W, and scaling any element of W by any real number stays in W. So W is closed under both vector space operations, and it contains an additive identity and additive inverses drawn from that same closure.
Every remaining axiom — commutativity and associativity of addition, associativity of scalar multiplication, the distributive laws, and the identity 1 \cdot \mathbf{u} = \mathbf{u} — is an equation that holds for all vectors in V, by hypothesis that V is a vector space. Since every element of W is in particular an element of V, each of these equations holds for the elements of W too; W is not required to satisfy some different, weaker set of axioms, only the same ones V already satisfies for every vector it contains. So none of these need separate verification — they are inherited, not re-derived.
Having the two closure properties, an identity element, inverses, and every other axiom inherited for free, W satisfies the full definition of a vector space under the operations of V, restricted to W. That makes W a subspace of V.
Why the test stops at two conditions. A first guess might list three requirements — closure under addition, closure under scalar multiplication, and \mathbf{0} \in W — and that three-condition version is also correct and commonly used. The proof above shows why the third condition is redundant whenever W is already known to be nonempty: scaling any single element of W by 0 produces \mathbf{0}, so nonemptiness plus scalar closure already forces \mathbf{0} \in W. The nonemptiness hypothesis cannot be dropped, though — \varnothing satisfies both closure conditions vacuously (there are no pairs or elements to violate them) but is not a vector space, since it has no zero vector.
Sanity check: take V = \mathbb{R}^2 and W = \{(x, y) : y = x\}, the line through the origin. Adding two points on that line, (a, a) + (b, b) = (a+b, a+b), lands back on the line, and scaling, c(a,a) = (ca, ca), does too — both closure conditions hold, so the test confirms W is a subspace, matching the geometric fact that a line through the origin is a subspace of the plane. Shift the line off the origin to W' = \{(x, y) : y = x + 1\} and closure under scalar multiplication already fails: (0,1) \in W' but 2(0,1) = (0,2) \notin W', correctly flagging that a line missing the origin is not a subspace.
Theorem
The Spanning Set Theorem
Recalled from Linear Algebra — Vectors, Matrices, and the Geometry of Data · Basis — first taught there.
Let S = \{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_p\} be a set of vectors in a vector space V, and let H = \operatorname{span}(S).
- If some \mathbf{v}_k \in S is a linear combination of the remaining vectors in S, then the set S' = S \setminus \{\mathbf{v}_k\} formed by removing \mathbf{v}_k still satisfies \operatorname{span}(S') = H.
- If H \neq \{\mathbf{0}\}, some subset of S is a basis for H.
First claim. Suppose \mathbf{v}_k is a linear combination of the other vectors in S, say \mathbf{v}_k = c_1\mathbf{v}_1 + \cdots + c_{k-1}\mathbf{v}_{k-1} + c_{k+1}\mathbf{v}_{k+1} + \cdots + c_p\mathbf{v}_p. Write S' = S \setminus \{\mathbf{v}_k\}. Since S' \subseteq S, every vector reachable as a combination of S' is automatically reachable as a combination of S, so \operatorname{span}(S') \subseteq H.
For the reverse inclusion, take any \mathbf{y} \in H. Then \mathbf{y} is some combination \mathbf{y} = a_1\mathbf{v}_1 + \cdots + a_k\mathbf{v}_k + \cdots + a_p\mathbf{v}_p. Substitute the expression for \mathbf{v}_k above in place of \mathbf{v}_k and collect terms: each \mathbf{v}_i with i \neq k now carries the combined coefficient a_i + a_k c_i, and \mathbf{v}_k has vanished entirely. So \mathbf{y} is a linear combination of S' alone, giving H \subseteq \operatorname{span}(S'). The two inclusions together give \operatorname{span}(S') = H.
Second claim. If H \neq \{\mathbf{0}\}, then S contains at least one nonzero vector, so among all subsets of S that span H, choose one, call it B, with the fewest vectors possible (such a subset exists because S itself spans H and S is finite, so there are only finitely many subsets to compare). It remains to show B is linearly independent, which makes it a basis for H.
Suppose toward a contradiction that B is linearly dependent. Then some vector in B is a linear combination of the others in B — removing that vector, by the first claim applied to B in place of S, leaves a smaller subset of B (hence of S) that still spans H. This contradicts the minimality of B. So B is linearly independent, and since B already spans H, B is a basis for H.
Sanity check: with S = \{\mathbf{v}_1, \mathbf{v}_2\} where \mathbf{v}_2 = 3\mathbf{v}_1 and \mathbf{v}_1 \neq \mathbf{0}, the second vector is a multiple of the first, so dropping it leaves S' = \{\mathbf{v}_1\} with \operatorname{span}(S') = \operatorname{span}(S) exactly as the theorem predicts — and \{\mathbf{v}_1\} is itself the minimal, linearly independent spanning set, i.e. the basis the second claim guarantees.
Watch dependence collapse a volume to zero
Drag w toward the plane spanned by u and v — the parallelepiped's volume shrinks to zero exactly when w becomes a combination of u and v, and the readout flips to 'dependent.'
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Which candidates are actually subspaces?
Click a candidate set, then click 'Is a subspace' or 'Is not' — a wrong guess names which axiom fails.
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Worked example
A dependent set forces a lower-dimensional span
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1match components — the third is automatically satisfied since all three vectors have zero third entry
Let \mathbf u = (3,0,0), \mathbf v = (0,3,0), \mathbf w = (2,1,0). Decide whether \{\mathbf u, \mathbf v, \mathbf w\} is independent by checking whether \mathbf w is a combination of \mathbf u and \mathbf v. c_1\mathbf u + c_2 \mathbf v = (3c_1,\ 3c_2,\ 0) \overset{?}{=} (2, 1, 0)
match components — the third is automatically satisfied since all three vectors have zero third entry
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2
Solving component-wise: 3c_1 = 2 gives c_1 = \tfrac23, and 3c_2 = 1 gives c_2 = \tfrac13. \tfrac23\mathbf u + \tfrac13 \mathbf v = (2, 1, 0) = \mathbf w
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3
\boxed{\{\mathbf u,\mathbf v,\mathbf w\}\ \text{is dependent} — \operatorname{span}\{\mathbf u,\mathbf v,\mathbf w\} = \operatorname{span}\{\mathbf u,\mathbf v\}, \text{a plane, not all of } \mathbb{R}^3} Three vectors in \mathbb{R}^3$ don't automatically span all of it, and a dependent set never spans more than an independent subset of the same size would.
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Where this shows up: the exchange lemma
Chapter 01 proves something Linear Algebra I only ever used informally: every basis of a vector space has the same number of vectors, no matter which basis you pick. That fact — dimension is well-defined — rests entirely on the interplay between span and independence reviewed here, made precise through a result called the exchange lemma.
Where this goes next
Every later chapter in this course
Every one of this course's central objects — a dual space, a quotient space, a tensor product, an eigenspace — is defined as a vector space first and gets everything else second. A shaky grip on "is this closed under addition and scaling" or "is this set of vectors wasteful" turns every later definition into memorized syntax instead of a checkable claim; a solid one makes chapter 01's generalization to arbitrary fields feel like changing the label on a box you already know how to open.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
For \mathbf{u} = (4, -2) and scalar c = -3 in \mathbb{R}^2, compute the first entry of c\mathbf{u}.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Scalar multiplication scales every entry: the first entry is -3 \times 4 = -12, so c\mathbf{u} = (-12, 6).
Correct answer
Also accepted:
Why
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
Which pair of operations must a set V carry for "vector space" to even be a meaningful question to ask about it?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
The dot and cross products are extra structure some vector spaces carry -- they are not part of the base definition every vector space must have.
Determinants and inverses are matrix-specific tools, not part of the vector space axioms, which apply to any set at all.
Length and angle come from an inner product, which is optional extra structure -- plenty of vector spaces (like polynomials with no dot product defined) never mention either.
Why
A vector space is a set V equipped with exactly two operations -- vector addition and scalar multiplication -- that together satisfy ten axioms. Nothing about length, angle, or determinants is required.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
Write, in terms of x, y, and z, the left-hand side of the single linear equation (set equal to 0) whose solution set is the plane through the origin containing all points where twice the x-coordinate plus the z-coordinate equals the y-coordinate.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
"Twice x plus z equals y" is 2x+z=y, i.e. 2x - y + z = 0. Any equivalent rearrangement like 2x+z-y is accepted.
Correct answer
Also accepted:
Why
Translate 'twice x plus z equals y' directly into an equation and move everything to one side, set equal to 0.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
For which value of k does the line y = 6x + k in \mathbb{R}^2 pass through the origin (and therefore have a chance of being a subspace)?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Plugging (0,0) into y = 6x + k gives 0 = 6(0) + k, so k = 0. Only k=0 puts the origin on the line.
Correct answer
Also accepted:
Why
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
Write (9,5) as c_1(1,1) + c_2(2,-1). What is c_1?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Solving c_1+2c_2=9 and c_1-c_2=5: subtract to get 3c_2 = 4, so c_2 = 4/3. Then c_1 = 5+c_2 = 5+4/3 = 19/3 \approx 6.33.
Correct answer
Also accepted:
Why
Set up the two equations from matching coordinates, then eliminate one variable by subtracting.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
Is (6,2,5) in \operatorname{span}((1,0,2),(0,1,1))?
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Check the third coordinate: 6(2)+2(1) = 14 \ne 5. Matching the first two coordinates alone doesn't guarantee the third also matches.
Try the first coordinate: c_1(1,0,2) contributes c_1 to the first slot, so c_1 must equal 6, not 2.
This is fully determined -- solve the system from the first two coordinates, then check the third as a consistency test.
Why
The first two coordinates force c_1=6, c_2=2 (since c_1=6 directly and c_2=2 directly). Checking the third coordinate: 6(2)+2(1)=14\ne 5, so no combination reaches (6,2,5) -- it lies outside the span, which is a plane in \mathbb{R}^3.
Use the first two coordinates to pin down c1 and c2, then check whether that same combination also matches the third coordinate.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
Four vectors are given in \mathbb{R}^3. Without computing anything, what can you conclude about their independence?
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Four vectors can never be independent in \mathbb{R}^3 -- independence caps out at the space's dimension, and that hasn't been covered yet, but the pattern is forced by having more vectors than coordinates to pin them down.
It's not a coin flip here -- any set of more vectors than the ambient space's coordinate count is forced to be dependent, no computation needed.
Dependence doesn't say anything about spanning either -- four vectors in \mathbb{R}^3 could span all of \mathbb{R}^3 or only a smaller subspace, independent of this fact.
Why
Any homogeneous system with more unknowns (c_1,\dots,c_4) than equations (3, one per coordinate) always has a free variable, hence a nontrivial solution -- so more than n vectors in an n-coordinate space are automatically dependent. This becomes the dimension bound formalized later in the chapter.
Compare the number of vectors to the number of coordinates each one has, and think about how many equations that gives the homogeneous system.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
A student claims: "If no vector in the set is a scalar multiple of another, the whole set must be linearly independent." For a set of 3 or more vectors, this claim is:
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
That's the definition for exactly TWO vectors. With three or more, no single vector needs to be a multiple of just one other -- it can instead equal a combination of several others, which pairwise checks completely miss.
The dimension of the ambient space doesn't change why this claim fails -- the counterexample (three vectors, no two parallel, but one is a combination of the other two) exists in \mathbb{R}^2 and higher alike.
Pairwise non-parallel vectors can be either independent or dependent -- it isn't a guarantee of dependence either.
Why
Example: (1,0,0), (0,1,0), and (1,1,0) -- no two of these are scalar multiples of each other, yet the third equals the sum of the first two, a genuine dependence relation. Independence must be checked against ALL combinations at once, not pair by pair.
Try to construct three vectors where no two are parallel but a combination of all three still reaches zero.