Vector spaces, span and linear independence

Which targets can two moves reach?

Let u=(2,1) and v=(1,−1). Taking one copy of each gives u+v=(3,0); taking two copies of u and subtracting v gives 2u−v=(3,3). A weight may be any real number, including zero or a negative number. The question is which target vectors can be written as au+bv.

For a target (x,y), the equations are 2a+b=x and a−b=y. Adding them gives a=(x+y)/3; substituting gives b=(x−2y)/3. These coefficients work for every real x and y, so the two moves can build every vector in the coordinate plane. The equations also force these coefficients, so each target has just one representation using this pair.

Now replace v by 2u=(4,2). Every combination au+b(2u)=(a+2b)u stays on the same line through the origin. Adding a second listed vector has not added a direction. We will make 'all targets', 'same line' and 'redundant' precise.

Preparation: vector addition and scalar multiplication from Vectors 121, and reversible row operations and free variables from Matrices 122. Read the geometric and subspace sections first if the polynomial examples are unfamiliar. All vector spaces and scalars in this article are real; the operations are the ones explicitly stated. Bases and dimension belong to the next article.

A vector space includes its operations

A real vector space V is a nonempty set with two operations: adding any two elements gives another element of V, and multiplying an element by any real scalar gives an element of V. These operations must obey the laws in the table. The word 'vector' means an element of this structure; it need not mean a drawn arrow.

In the table, u,v,w are arbitrary elements of V and a,b are arbitrary real numbers. The zero vector 0 is the identity for vector addition; the scalar 0 is a real number. Their roles differ even though the same symbol is conventional. The negative vector −v is the additive inverse, and (−1)v=−v follows from the laws.

For example, 0v=(0+0)v=0v+0v. Adding the inverse of 0v to both sides gives 0v=0. The same reasoning with a(0+0) proves a0=0. A definition lists general laws; particular examples must use operations that actually satisfy them.

Ordinary coordinate vectors Rⁿ, real matrices of a fixed shape, and real polynomials of degree at most 2 are vector spaces under their usual entrywise or coefficientwise operations. The last space, called P₂, includes the zero polynomial. The set of all real-valued functions on a fixed domain D is another example: (f+g)(t)=f(t)+g(t) and (af)(t)=a·f(t) for every t∈D. The familiar real-number laws hold at every coordinate, coefficient or input, so these operations satisfy the vector-space laws.

Test whether a subset keeps the same structure

A subspace W of a real vector space V is a subset that is itself a vector space using the addition and scalar multiplication inherited from V. It is enough to check three things: 0 belongs to W; whenever u,v belong to W, u+v belongs to W; and whenever v belongs to W and a∈R, av belongs to W.

Why are there only three checks? The associative, commutative, distributive and scalar-identity laws already hold in V and therefore hold for elements of W. Closure keeps the results in W. The zero-vector check supplies its additive identity, and closure with a=−1 supplies additive inverses. No new operations have been invented.

The line W={(x,2x):x∈R} passes. It contains (0,0); (x,2x)+(z,2z)=(x+z,2(x+z)); and a(x,2x)=(ax,2ax). The shifted line y=2x+1 fails immediately because it excludes (0,0).

Containing zero alone is insufficient. The union of the two coordinate axes contains zero and is closed under scalar multiplication, but (1,0)+(0,1)=(1,1) lies on neither axis. The first quadrant, including its edges, contains zero and is closed under addition, but multiplying (1,0) by −1 leaves it. An empty subset fails the zero-vector check. A set that fails this inherited-operations test is not a subspace under those operations.

A span contains every allowed combination

For a finite indexed list v₁,…,vₖ in V, a linear combination is a₁v₁+⋯+aₖvₖ with real coefficients. Its span is the set of all such combinations. Using a list makes repeated entries meaningful: listing a vector twice gives two coefficient positions even though the two entries have the same value.

The span contains zero by taking every coefficient to be zero. If x=Σaᵢvᵢ and y=Σbᵢvᵢ, then x+y=Σ(aᵢ+bᵢ)vᵢ is in the span. For c∈R, cx=Σ(caᵢ)vᵢ is also in the span. Thus every finite span is a subspace. Here Σ means a finite sum over the listed indices.

It is the smallest subspace containing the listed vectors: any subspace containing them must contain all their scalar multiples and sums. Conversely, the span itself is such a subspace. 'Smallest' refers to inclusion of sets, not to a length or area.

For an empty list there are no terms; the empty sum is defined as the zero vector. Its span is therefore {0}, not an empty set. A list spans V only when its span equals V. A target can belong to a smaller span without the list spanning the whole ambient space.

Keep the ambient space and the reachable set distinct

For a nonzero u in R², span(u) is the entire line {au:a∈R}. For u=0 it is only {0}. The pair u=(2,1), v=(1,−1) spans R² because the opening equations construct coefficients for every target. The dependent pair u,2u spans just span(u). These are conclusions about all real coefficients, not just the finite sample shown in a diagram.

A vector such as (3,6) belongs to span((1,2)) because (3,6)=3(1,2). The target (3,5) does not: the first component would force the coefficient to be 3, giving second component 6. A membership question fixes the target; a spanning question asks about every vector in a stated space.

The same distinction matters in larger spaces. The vectors (1,0,0) and (0,1,0) span the plane z=0 inside R³; they do not span R³. A drawing of that plane on a flat screen is a representation of a subset of R³, not a change of the ambient space.

Independence asks whether zero has another recipe

The finite list v₁,…,vₖ is linearly independent when a₁v₁+⋯+aₖvₖ=0 forces every coefficient aᵢ to be zero. Otherwise it is linearly dependent: at least one combination with some nonzero coefficient gives zero. Such an equation is a dependence relation. The condition concerns the whole list.

A list containing the zero vector is dependent: put coefficient 1 at that position and zero elsewhere. Repeated equal entries give a relation with coefficients 1 and −1. A list with one nonzero vector is independent because av=0 forces a=0. The empty list is independent: its only coefficient list is empty, so there is no nontrivial zero combination.

Two nonzero vectors that are not scalar multiples are independent. But the analogous pairwise test fails for three or more vectors. In R² take e₁=(1,0), e₂=(0,1) and w=(1,1). No pair consists of scalar multiples, yet e₁+e₂−w=0. Checking pairs has missed a relation involving all three.

Independence and spanning answer different questions. The one-vector list ((1,0)) is independent in R² but does not span R². The triple e₁,e₂,e₁+e₂ spans R² but is dependent. The opening pair u,v does both.

A dependence relation exposes redundancy

If a list has a dependence relation Σaᵢvᵢ=0 with aⱼ≠0, rearrange it to vⱼ=−Σᵢ≠ⱼ(aᵢ/aⱼ)vᵢ. Thus that member is a combination of the others and can be removed without changing the span. Conversely, expressing one member as a combination of the rest produces a nontrivial zero combination by moving it to the other side.

At least one member can be removed from a dependent list; it need not be every member. The list (1,0),(2,0),(0,1) is dependent because twice the first minus the second is zero. Removing (0,1), however, reduces its span from R² to the x-axis.

An independent list gives unique coefficients for every target in its span. If Σaᵢvᵢ=Σbᵢvᵢ, subtraction gives Σ(aᵢ−bᵢ)vᵢ=0, so independence forces aᵢ=bᵢ for each position. Existence of the coefficients comes from membership in the span; independence supplies uniqueness.

The converse holds too. If every vector in the span has unique coefficients, zero has only its all-zero representation, so the list is independent. A nontrivial dependence relation instead gives multiple representations of every target in the span: add any real multiple of that relation to one coefficient list. Targets outside the span have no representation.

Use columns and homogeneous systems to test a list

For coordinate vectors v₁,…,vₖ in Rⁿ with k>0, put them in the columns of A, an n×k matrix. A target b is in their span exactly when Ac=b is consistent. The list is independent exactly when Ac=0 has only the zero solution. The entries of c are coefficients of the listed vectors. They generally differ from the target’s coordinates in the original ambient axes; the standard coordinate list makes the two lists of numbers agree.

The homogeneous system is always consistent because c=0 works. The RREF result from Matrices 122 therefore applies without a consistency exception: every column having a pivot means no free coefficients and an independent list; a missing pivot column gives a free coefficient and a nontrivial dependence relation. There can be at most n pivot columns in n rows, so any list with more than n vectors in Rⁿ is dependent.

Row operations preserve the solutions for the coefficient vector, but generally change the column vectors themselves. For A=[[1,1],[1,1]], the RREF is [[1,1],[0,0]]. Both homogeneous systems say c₁+c₂=0. Yet the original columns span the diagonal line through (1,1), while the reduced columns span the x-axis through (1,0). Do not replace the original column space by the reduced column space.

To remove redundant columns, locate the pivot-column positions in RREF and select those columns from the original A. In RREF, the pivot columns are independent and every nonpivot column is their combination, using the entries in its pivot rows as coefficients. Apply the inverse row operations to these equations: they give the same combinations of the original columns, and preserve their independence. This justifies using the original pivot columns; a discussion of bases and dimension follows next.

Separate a homogeneous solution space from its translate

For a real m×n matrix A, the null space N(A)={x∈Rⁿ:Ax=0} is a subspace. It contains zero. If Ax=0 and Ay=0, then A(x+y)=Ax+Ay=0; for a∈R, A(ax)=aAx=0. The linearity identities follow by expanding the matrix-vector product.

Consider A=[[1,2,−1],[2,4,−2]]. Its homogeneous equations reduce to x+2y−z=0. Taking y=s and z=t gives every solution (−2s+t,s,t)=s(−2,1,0)+t(1,0,1), with s,t∈R. Every combination satisfies both original rows, and every solution has this form because its second and third components determine s and t.

Change the right-hand side to b=(3,6). A particular solution is x₀=(3,0,0); all solutions are x₀+N(A). Indeed, if Ax=b, then A(x−x₀)=0. Conversely, adding any element of N(A) to x₀ gives another solution. This solution set is an affine translate of the null space. It excludes zero, so it is not a subspace.

For any consistent Ax=b this translated description holds; if b=0 the translate is the null space itself. If the system is inconsistent its solution set is empty and is not a vector space. A diagram resembling a plane does not decide whether it contains the required zero vector.

Polynomials are vectors with a different meaning

In P₂, a vector is a polynomial a+bt+ct² with real coefficients; t is an input variable, not one of the coefficients being chosen. Addition and scalar multiplication combine coefficients, so recording (a,b,c) preserves both operations. The zero vector is the polynomial that is zero for every real t.

Take p=1+t, q=t+t² and r=1−t². Their coefficient columns are (1,1,0), (0,1,1) and (1,0,−1). The identity p−q−r=0 proves that this three-member list is dependent. The pair p,q is independent: in ap+bq=0, the constant coefficient forces a=0 and the t² coefficient forces b=0.

The target f=2+3t+t² equals 2p+q. Using all three polynomials, every real s gives f=(2−s)p+(1+s)q+sr, because r=p−q. Conversely, equating constant and t² coefficients forces exactly those first two coefficients once the r coefficient is named s. The nonunique coefficients describe the same polynomial, not different function values.

The span of 1 and t is the subspace of P₂ with zero t² coefficient, so it does not include t². For function spaces on R, zero means zero at every real input, not just at one sampled point: f(t)=t has f(0)=0 but is not the zero function. A finite set of sample values is generally insufficient to prove equality of arbitrary functions.

Intersections preserve a subspace; unions may not

If U and W are subspaces of the same V, then U∩W is a subspace. Zero belongs to both. The sum of two vectors belonging to both stays in both, and so does every real scalar multiple. These are exactly the three subspace checks.

In R³, the planes y=0 and z=0 are subspaces. Their intersection consists of (x,0,0), the x-axis. Adding both constraints narrows the set while preserving the linear structure.

A union accepts a vector that belongs to either set, so it need not be closed under sums taken from different sets. The two-axis example earlier supplies the failure. Do not confuse a union with span(U∪W), which also allows sums and scalar multiples and is a subspace. Here span(S), for a possibly infinite subset S, means all finite linear combinations of elements of S, including the empty sum. No infinite sums are being taken.

Ask the three questions separately

First identify the ambient space, the scalars and the operations. Then ask whether a proposed subset is a subspace. Next ask what a particular list spans, and whether that list is independent. Finally, if a target is specified, ask whether it belongs to that span and whether its coefficient representation is unique.

These questions connect geometry to computation without making them interchangeable. Elimination tests coefficient relations exactly for the finite coordinate examples here. The diagrams show their geometric consequences; the polynomial examples show why the definitions do not depend on drawing arrows.

The next lesson combines independence and spanning in the notion of a basis, defines dimension, and studies linear transformations. Those ideas build on this article's definitions rather than being assumed in its tests. Numerical near-dependence and floating-point conditioning require later numerical analysis; an almost-zero computed entry is not an exact dependence proof.

Worked examples

Try it yourself

1. For u=(2,1), v=(1,−1), find 2u−v.

   Show the answer and reasoning

   (3,3).

   Compute (4,2)−(1,−1) component by component.

2. Express (5,1) as au+bv for u=(2,1), v=(1,−1).

   Show the answer and reasoning

   a=2, b=1.

   The component equations give 2a+b=5 and a−b=1; solving gives the unique coefficients.

3. Does (4,7) belong to span((1,2))?

   Show the answer and reasoning

   No.

   The first component forces coefficient 4, which would give second component 8.

4. Which of y=2x and y=2x+1 defines a subspace of R² under the usual operations?

   Show the answer and reasoning

   Only y=2x.

   It passes zero, addition and scalar closure. The shifted line excludes zero.

5. Why is the closed first quadrant not a subspace of R²?

   Show the answer and reasoning

   It is not closed under all real scalar multiplication.

   It contains (1,0), but not its multiple −(1,0)=(−1,0).

6. Are (1,0),(0,1),(1,1) independent merely because no pair is a pair of scalar multiples?

   Show the answer and reasoning

   No; the list is dependent.

   The first plus the second minus the third is zero, with nonzero coefficients.

7. The list (1,0),(2,0),(0,1) is dependent. Can you remove any member and keep its span?

   Show the answer and reasoning

   No. Removing (0,1) loses every vector with nonzero second component.

   Dependence guarantees at least one removable member. Either of the first two is redundant, but the third is not.

8. What is the span of an empty list in V, and is the list independent?

   Show the answer and reasoning

   Its span is {0}; the empty list is independent.

   Its only combination is the empty sum, defined as zero; there is no nontrivial coefficient list.

9. Is a list containing the zero vector linearly independent?

   Show the answer and reasoning

   No.

   Put coefficient 1 at a zero entry and zero at every other position to obtain a nontrivial zero combination.

10. Write every solution of x+2y−z=0 as a combination of two fixed vectors.

    Show the answer and reasoning

    s(−2,1,0)+t(1,0,1), with s,t∈R.

    Choose y=s and z=t freely; then x=−2s+t. Substitution verifies every choice and the construction includes every solution.

11. For p=1+t, q=t+t² and r=1−t², give a nontrivial zero combination.

    Show the answer and reasoning

    p−q−r=0.

    The constant, t and t² coefficients each cancel. This is an identity for every input t.

12. Do row operations generally preserve the set spanned by a matrix's columns as vectors in the original ambient coordinates?

    Show the answer and reasoning

    No.

    [[1,1],[1,1]] and its RREF [[1,1],[0,0]] have the same homogeneous coefficient relations, but their columns span different lines. Use pivot positions to select original columns.

13. If a function satisfies f(0)=0, must it be the zero vector in a space of functions on R?

    Show the answer and reasoning

    No.

    The function f(t)=t is zero at 0 but nonzero at 1. The zero function must vanish at every input in its domain.

What to carry forward

• A real vector space consists of a set and operations satisfying the vector-space laws; its vectors need not be arrows.
• A subspace uses inherited operations and must contain zero and be closed under addition and every real scalar multiple.
• A finite span contains all real linear combinations and is the smallest subspace containing its listed vectors.
• Membership of one target and spanning an entire ambient space are different questions.
• Independence means that only the all-zero coefficient list builds zero; a dependence relation identifies a removable member.
• Independent lists give unique representations within their span; dependent lists give nonunique representations of every member of their span.
• Homogeneous systems test coordinate dependence; row reduction preserves coefficient relations, so choose original pivot columns.
• Polynomial and function examples obey the same laws, with equality and the zero vector interpreted for the whole polynomial or function.