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Linear Combinations



Definition
Geometric Interpretation
Span
Spanning Sets
Trivial and Non-Trivial Combinations
Linear Combinations and Systems of Equations
Toward Independence and Basis
Summary: Linear Combinations as the Unifying Construction
Common Pitfalls
Linear Combinations FAQ



Building New Vectors from Old

Every operation in this section — addition, subtraction, scalar multiplication — is a special case of one unifying construction: the linear combination. Scale a collection of vectors by chosen coefficients, then add the results. The output is a single vector assembled from the pieces. This idea is deceptively simple, but it reaches into every corner of linear algebra. Asking which vectors can be built from a given set leads to the concept of span. Asking whether a set contains redundant vectors leads to linear independence. Asking whether a particular vector can be expressed as a combination of others turns out to be equivalent to solving a system of linear equations. The definitions introduced here are developed formally in the vector spaces section, but their computational foundation belongs here, grounded in the concrete algebra of Rn\mathbb{R}^n.



Definition

You already make combinations like this: so much of one thing plus so much of another.
A trail mix comes in two bags. Mix A holds 11 cup of oats and 22 cups of nuts; mix B holds 33 cups of oats and 11 cup of nuts. Buy 22 bags of A and 33 bags of B and you can work out what you have without a formula: scale each bag's contents, then add.
oats (cups)nuts (cups)
one bag of mix A12
one bag of mix B31
2 bags of A24
3 bags of B93
total117
The same calculation turns up in moves on a grid and in columns of numbers.
On a grid. One kind of step, u\mathbf{u}, goes 11 right and 22 up; another, v\mathbf{v}, goes 33 right and 11 up. Take two u\mathbf{u}-steps, then three v\mathbf{v}-steps, and you end up 1111 right and 77 up.
In columns of numbers. Write the bags, or the steps, as vectors u=(1,2)\mathbf{u} = (1, 2) and v=(3,1)\mathbf{v} = (3, 1) and compute 2u+3v2\mathbf{u} + 3\mathbf{v} entry by entry.
entry2 × u3 × vsum
first2 × 1 = 23 × 3 = 911
second2 × 2 = 43 × 1 = 37
Two steps of u = (1, 2) then three steps of v = (3, 1) end at (11, 7)xy123456789101112123456782 steps of u3 steps of v(11, 7)Steps of u and v, laid end to end, land at (11, 7)
Two u\mathbf{u}-steps from the origin reach (2,4)(2, 4); three v\mathbf{v}-steps from there reach (11,7)(11, 7) — the same pair of numbers as the trail mix and the table.
What they share: scale each vector by its own number, then add the results.
Bags, steps, columns: in each case there are a few fixed vectors, each one is multiplied by a number saying how much of it to take, and the scaled copies are added. The objects differ; the structure — 2u+3v=(11,7)2\mathbf{u} + 3\mathbf{v} = (11, 7) — is the same every time.
The definition
Given vectors v1,…,vk\mathbf{v}_1, \ldots, \mathbf{v}_k in Rn\mathbb{R}^n and scalars c1,…,ckc_1, \ldots, c_k, a linear combination of the vectors is the vector
Linear Combination
c1v1+c2v2+⋯+ckvkc_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k
Learn more about this formula: Linear Combination →
A linear combination is a vector, in the same space as the vectors combined.
Fact: Some colours need negative red
In the colour-matching experiments of the late 1920s behind the 1931 CIE colour standard, observers mixed red, green and blue lights to copy pure colours from a prism. Blue-green light could be matched only by adding red to the target itself, so its recipe has a negative coefficient of red.
The expression is a linear combination of those vectors. The scalars cic_i are called the coefficients or weights of the combination. Each term civic_i\mathbf{v}_i is a scaled copy of vi\mathbf{v}_i, and the full expression adds all the scaled copies together.
The definition imposes no restrictions on the coefficients — they can be positive, negative, or zero. Setting all coefficients to 11 reduces the linear combination to ordinary vector addition. Using a single vector with one coefficient gives scalar multiplication. Setting all coefficients to zero produces the zero vector. These familiar operations are not separate concepts; they are boundary cases of the same construction.
RULE
A linear combination always produces a vector in the same Rn\mathbb{R}^n as the inputs.
Both scalar multiplication and addition preserve the dimension, so no matter how many vectors are combined or what coefficients are chosen, the result remains in the original space.
What the name buys: many different questions become one question.
Once "linear combination" has a name, a whole family of questions turns out to be the same question in different words.
questionasked as a linear combination
which vectors can these reach?the span: all their linear combinations
is any vector redundant?independence: only the all-zero combination gives 0
does Ax = b have a solution?is b a combination of the columns of A?
what is a basis?a set that writes every vector as one combination
α·u1×4α·u1α·u2α·u3α·u4+β·v1×4β·v1β·v2β·v3β·v4=w1×4α·u1+β·v1α·u2+β·v2α·u3+β·v3?
αu + βv, third phase frozen

The two scaled copies have already been formed and are being added slot by slot — the filled entries are settled, the highlighted one is the component being written now, and the pale ones are still to come. Nothing here is a new rule: it is scalar multiplication and addition running in sequence, which is what the definition says a combination is. Choose your own coefficients and step through all three phases on the vector linear combination visualizer.

Scaling and adding are the only two operations at work, which is why every rule that follows on this page is really a rule about those two.
The rest of the page shows what linear combinations look like geometrically, how to tell whether a vector is one, how to find the coefficients, the link to matrix–vector multiplication, and the ideas built on them: span, independence and bases.

Geometric Interpretation

The geometric meaning of a linear combination depends on how many vectors are involved and how they relate to one another.

A single nonzero vector v\mathbf{v} generates all its scalar multiples: cvc\mathbf{v} for every real number cc. These fill a line through the origin in the direction of v\mathbf{v}. Moving cc from −∞-\infty to ∞\infty sweeps the entire line, passing through the origin at c=0c = 0.

Fact: Nearly parallel, huge coefficients
(1,0)(1, 0) and (1,0.001)(1, 0.001) are not parallel, so they span R2\mathbb{R}^2. But reaching (0,1)(0, 1) takes −1000(1,0)+1000(1,0.001)-1000(1, 0) + 1000(1, 0.001): the two vectors differ by only 0.0010.001 sideways, so that difference has to be scaled up a thousand times.

Two vectors u\mathbf{u} and v\mathbf{v} that are not parallel generate all combinations c1u+c2vc_1\mathbf{u} + c_2\mathbf{v}. Varying both coefficients independently fills a plane through the origin — the plane that contains both u\mathbf{u} and v\mathbf{v}. In R2\mathbb{R}^2, two non-parallel vectors already span the entire space. In R3\mathbb{R}^3, they span a flat sheet passing through the origin, leaving one dimension unreached.

Three vectors in R3\mathbb{R}^3 that do not all lie in the same plane span the full space: every point in R3\mathbb{R}^3 is reachable as c1u+c2v+c3wc_1\mathbf{u} + c_2\mathbf{v} + c_3\mathbf{w} for some choice of coefficients. If the three vectors are coplanar, however, the third contributes nothing new — the span remains a plane rather than expanding to fill three dimensions.

The coefficients act as continuous dials. Adjusting one coefficient while holding the others fixed slides the result along the direction of the corresponding vector. Adjusting all of them simultaneously moves the result anywhere within the span.
φabspan = ℝ²φabspan = line
Two vectors spanning a plane, then collapsing to a line

Above, the two arrows point in genuinely different directions and their combinations reach every point of the plane. Below, one is a multiple of the other, so every combination lands back on the same line no matter how the coefficients are dialled — the second vector adds nothing the first did not already reach. Drag the arrows and watch the shaded region fall from a plane to a line on the span and independence explorer.

Whether the coefficients sweep out a line or a plane is decided entirely by the directions of the vectors, not by how many of them there are — the idea the next section names as span.

Span

The span of a set of vectors is the collection of every vector that can be formed as a linear combination of the set. For vectors v1,v2,…,vk\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k:

Span
Span{v1,v2,…,vk}={c1v1+c2v2+⋯+ckvk∣ci∈R}\text{Span}\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\} = \{c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k \mid c_i \in \mathbb{R}\}
Learn more about this formula: Span →


Pitfall: Confusing the set with its span
Saying that (3,−2)(3, -2) is not in Span{(1,0),(0,1)}\text{Span}\{(1, 0), (0, 1)\} because it is neither of the two vectors reads the span as the set itself. Since (3,−2)=3(1,0)−2(0,1)(3, -2) = 3(1, 0) - 2(0, 1), it is in the span.

The span is a subset of Rn\mathbb{R}^n that contains all vectors reachable through linear combinations — and no others. It always contains the zero vector, since setting every coefficient to zero produces 0\mathbf{0}.

The geometric shape of the span reflects the structure of the input vectors. A single nonzero vector spans a line. Two non-parallel vectors span a plane. In general, the span of kk vectors that are "sufficiently independent" is a kk-dimensional flat subspace passing through the origin. What "sufficiently independent" means precisely is the subject of linear independence, but the intuition is clear: each new vector expands the span by one dimension only if it points in a direction not already covered.

The span of the standard basis {e1,e2,…,en}\{\mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n\} is all of Rn\mathbb{R}^n, since any vector (v1,v2,…,vn)(v_1, v_2, \ldots, v_n) equals v1e1+v2e2+⋯+vnenv_1\mathbf{e}_1 + v_2\mathbf{e}_2 + \cdots + v_n\mathbf{e}_n. The span of a proper subset of the standard basis is a coordinate subspace — a lower-dimensional slice of Rn\mathbb{R}^n aligned with the coordinate axes.

The formal theory of span, including its relationship to subspaces, is developed in the vector spaces section.

Spanning Sets

A set of vectors spans a space if every vector in that space belongs to the span of the set. Equivalently, the set spans Rn\mathbb{R}^n if for every vector b\mathbf{b} in Rn\mathbb{R}^n, there exist coefficients c1,c2,…,ckc_1, c_2, \ldots, c_k such that c1v1+c2v2+⋯+ckvk=bc_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{b}.

The standard basis {e1,e2,…,en}\{\mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n\} spans Rn\mathbb{R}^n with exactly nn vectors and no redundancy. But spanning sets need not be this efficient. The set {(1,0), (0,1), (1,1)}\{(1, 0),\ (0, 1),\ (1, 1)\} also spans R2\mathbb{R}^2, because any vector (a,b)(a, b) can be written as a(1,0)+b(0,1)+0(1,1)a(1, 0) + b(0, 1) + 0(1, 1). The third vector is unnecessary — it is already a linear combination of the first two — yet its presence does not prevent the set from spanning. A spanning set with redundant vectors is larger than it needs to be but still reaches every point in the space.

The question of which vectors in a spanning set are genuinely needed and which are superfluous is answered by the concept of linear independence. A spanning set with no redundancy is called a basis — the smallest possible spanning set for a given space. Both ideas are formalized in the vector spaces section.
A set spans a space when it reaches every vector, whether or not it carries vectors it does not need.

Trivial and Non-Trivial Combinations

Among all possible linear combinations of a set of vectors, one always exists regardless of the vectors involved: the trivial combination. Setting every coefficient to zero gives:

Trivial and Non-Trivial Combinations
0v1+0v2+⋯+0vk=00\mathbf{v}_1 + 0\mathbf{v}_2 + \cdots + 0\mathbf{v}_k = \mathbf{0}

This produces the zero vector for any choice of v1,…,vk\mathbf{v}_1, \ldots, \mathbf{v}_k. It is called trivial because it carries no information about the vectors themselves — it works automatically, without engaging with the actual components.

A non-trivial combination is one in which at least one coefficient is nonzero. The distinction between trivial and non-trivial combinations is the key to defining linear independence. A set of vectors is linearly independent if the only combination that produces 0\mathbf{0} is the trivial one — there is no way to combine the vectors with nonzero coefficients and arrive back at the origin. If a non-trivial combination does produce 0\mathbf{0}, then at least one vector in the set can be written as a linear combination of the others, revealing redundancy.

Pitfall: Testing one combination only
Trying c1=c2=1c_1 = c_2 = 1 on (1,2)(1, 2) and (2,4)(2, 4) gives (3,6)≠0(3, 6) \neq \mathbf{0}, which does not show independence. Other coefficients reach zero, 2(1,2)−1(2,4)=02(1, 2) - 1(2, 4) = \mathbf{0}, so the pair is dependent.

This characterization is developed formally in the linear independence section, but the computational meaning is accessible now: if the equation c1v1+⋯+ckvk=0c_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k = \mathbf{0} has only the solution c1=c2=⋯=ck=0c_1 = c_2 = \cdots = c_k = 0, the vectors are independent. If it has other solutions, they are dependent.
Combination Coefficient pattern What it produces What it tells you about the vectors
Trivial every ci = 0 always the zero vector 0, regardless of the vi nothing — it exists for every set and reveals no structure
Non-trivial at least one ci ≠ 0 depends on the vectors and the specific coefficients chosen if a non-trivial combination produces 0, the vectors are linearly dependent — at least one is a combination of the others
Any vectors reach zero with all-zero coefficients; only a second way of reaching it reveals a redundant vector.

Linear Combinations and Systems of Equations

The question "Is b\mathbf{b} a linear combination of v1,v2,…,vk\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k?" is an algebraic question about the existence of coefficients. Writing it out component by component produces a system of linear equations:

Linear Combinations and Systems of Equations
c1v1+c2v2+⋯+ckvk=bc_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{b}

Arranging the vectors v1,…,vk\mathbf{v}_1, \ldots, \mathbf{v}_k as columns of a matrix AA and the coefficients as an unknown vector x\mathbf{x}, the equation becomes:

Linear System Matrix Form
Ax=bA\mathbf{x} = \mathbf{b}
Learn more about this formula: Linear System Matrix Form →


Pitfall: Writing the vectors as rows
To reach b=(3,7)\mathbf{b} = (3, 7) from (1,2)(1, 2) and (0,1)(0, 1), rows give c1+2c2=3c_1 + 2c_2 = 3 and c2=7c_2 = 7, so c1=−11c_1 = -11, but −11(1,2)+7(0,1)=(−11,−15)≠b-11(1, 2) + 7(0, 1) = (-11, -15) \neq \mathbf{b}. As columns, c1=3c_1 = 3 and c2=1c_2 = 1.

The linear combination question and the system-of-equations question are identical — they are two descriptions of the same problem. The combination exists if and only if the system has a solution. Solving the system finds the coefficients; the coefficients reconstruct the combination.

This connection transforms a geometric question (does b\mathbf{b} lie in the span of the columns?) into a computational one (does the augmented matrix [A∣b][A \mid \mathbf{b}] reduce to a consistent system?). The tools for answering the computational question — row reduction, echelon forms, pivot analysis — belong to the linear systems section, but the conceptual link originates here, in the definition of a linear combination.
510-22468v1c1v2c2v3c3Av
Av assembled from the columns of A

Each column of A has been scaled by its entry of v and laid tail to head, and the last arrow lands on the answer. That walk is the picture behind the question this section asks: b is a linear combination of the columns exactly when some choice of weights ends the walk on b, and no choice of weights can leave the plane the columns span. Set your own weights and watch the walk change on the column picture visualizer.

Row reduction, echelon forms and pivots are the machinery for deciding whether such a walk exists; the question they answer is the one stated here.

Toward Independence and Basis

Linear combinations open two questions that this page has raised but not resolved. The first: given a set of vectors, can any of them be removed without shrinking the span? A vector that is a linear combination of the others in the set contributes nothing new — every point it helps reach is already reachable without it. A set where no vector is redundant in this sense is called linearly independent. Independence is a property of the set as a whole, not of any individual vector, and it is characterized by the trivial-combination criterion: the only way to combine the vectors into 0\mathbf{0} is with all-zero coefficients.

The second question: what is the smallest set of vectors whose span equals the entire space? Such a set must span Rn\mathbb{R}^n (reaching every vector) and be linearly independent (containing no excess). A set with both properties is a basis for Rn\mathbb{R}^n, and every basis of Rn\mathbb{R}^n contains exactly nn vectors — a fact that defines the dimension of the space.

Both concepts — linear independence and basis — are developed rigorously in the vector spaces section. The machinery of linear combinations built on this page provides the raw material from which those definitions are constructed.

Summary: Linear Combinations as the Unifying Construction

Every operation introduced earlier in the section can be written as a linear combination with a particular pattern of coefficients. Scalar multiplication is a one-vector combination; ordinary addition is a two-vector combination with both coefficients equal to 1; the trivial combination uses all-zero coefficients. The table below recovers each familiar operation as a special case of the general formula c₁v₁ + ⋯ + cₖvₖ — showing that the construction on this page is not a new tool, but a unifying lens through which every operation earlier in the section can be viewed.
The entries below are one expression with the coefficients set differently. Pin them to specific numbers and each of the basic operations appears; leave them free and the span appears; ask whether a particular choice reaches 0\mathbf{0} and independence appears. The grouping is that distinction — fixed coefficients above, free coefficients below.
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Vectors · the unifying construction

One construction, chosen coefficients

Every vector operation is c1v1+⋯+ckvkc_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k with the coefficients fixed in advance. What separates the entries below is only which coefficients are allowed — pinned to specific values, or left free.

7forms
Coefficients pinned — the basic operations4
1
Scalar multiplication§ 1
one vector, one coefficient
cvc\mathbf{v}
The shortest combination there is. Its span is the line through v\mathbf{v}, which is the k=1k = 1 case of everything below — see basic operations.
2
Addition§ 1
both coefficients 11
1⋅u+1⋅v1 \cdot \mathbf{u} + 1 \cdot \mathbf{v}
Addition is not a separate operation that combinations generalise — it is the combination with both coefficients set to one. The tip-to-tail picture is the geometric reading of exactly this.
3
Subtraction§ 1
coefficients 11 and −1-1
1⋅u+(−1)⋅v1 \cdot \mathbf{u} + (-1) \cdot \mathbf{v}
Likewise, with one coefficient negated. This is why subtraction needs no rules of its own: everything true of combinations applies to it directly.
4
Trivial combination§ 5
always available, always gives 0\mathbf{0}
0v1+⋯+0vk=00\mathbf{v}_1 + \cdots + 0\mathbf{v}_k = \mathbf{0}
Available for any set whatsoever, which is why finding a combination equal to 0\mathbf{0} proves nothing. Independence asks whether it is the only one — the whole subject turns on that word.
Coefficients free — what the construction generates3
5
General combination§ 1
arbitrary scalars
c1v1+⋯+ckvkc_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k
The definition the rest are instances of. Note it is a finite sum with arbitrary real coefficients — no convergence question arises, which is what separates this from an infinite series.
6
Span§ 3
the set of every reachable vector
Span⁡{vi}={\operatorname{Span}\{\mathbf{v}_i\} = \{ all combinations }\}
Let the coefficients range over everything and collect the results. The span is automatically a subspace — a combination of combinations is a combination — which is the cheapest subspace proof available.
7
Membership as a system§ 6
columns of AA are the vi\mathbf{v}_i
b∈Span⁡  ⟺  Ac=b\mathbf{b} \in \operatorname{Span} \iff A\mathbf{c} = \mathbf{b} solvable
The bridge to computation. Asking whether b\mathbf{b} is a combination is asking whether a linear system is consistent, and the coefficients are its solution — which is why row reduction answers questions about spans.
The progression is one of freedom. Fix every coefficient and you get a single operation; free them and you get the span; ask whether a particular combination can reach 0\mathbf{0} and you get independence. Nothing new is introduced at any stage — only the question asked about the same expression.
One construction, chosen coefficients·/linear-algebra/vectors/linear-combinationsLearn Math Class
Reading it this way changes what the basic operations are. Addition is not an operation that linear combinations later generalise — it is the combination with both coefficients equal to one, and subtraction is the same with one of them negated. Nothing is added to the vocabulary between the first row and the last; only the freedom in the coefficients changes.
The trivial combination is the row that matters most later, and it is easy to pass over. Every set of vectors admits it, so exhibiting a combination equal to 0\mathbf{0} establishes nothing at all. Independence is the statement that it is the only one, and that single word — only — is what the rest of the subject is built on.

Common Pitfalls

Three mistakes account for most wrong answers about linear combinations. Each one is marked where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.

Confusing the set with its span

The span of a set is treated as if it were the set itself, a short list of the given vectors, rather than the collection of every combination of them. A vector that is not on the list is then wrongly left out.

Wrong working: (3,−2)(3, -2) is neither (1,0)(1, 0) nor (0,1)(0, 1), so (3,−2)∉Span{(1,0),(0,1)}(3, -2) \notin \text{Span}\{(1, 0), (0, 1)\}.

Right working: look for coefficients.
c1(1,0)+c2(0,1)=(c1,c2)=(3,−2)c_1(1, 0) + c_2(0, 1) = (c_1, c_2) = (3, -2)

gives c1=3c_1 = 3 and c2=−2c_2 = -2, so (3,−2)=3(1,0)−2(0,1)(3, -2) = 3(1, 0) - 2(0, 1) is in the span. In fact every vector of R2\mathbb{R}^2 is.

Why it happens: the notation Span{…}\text{Span}\{\ldots\} puts braces around the same short list that names the set, and the infinite collection it stands for is never written out.

How to check: the span of a set that contains a nonzero vector is never finite; it holds every multiple of that vector, and 0\mathbf{0} as well. An answer that treats the span as a few listed vectors has read the braces as the set.

Testing one combination only

To decide whether vectors are independent, one choice of nonzero coefficients is tried. When it does not give 0\mathbf{0}, the vectors are declared independent.

Wrong working: for (1,2)(1, 2) and (2,4)(2, 4),
1(1,2)+1(2,4)=(3,6)≠0,so the vectors are independent1(1, 2) + 1(2, 4) = (3, 6) \neq \mathbf{0}, \quad \text{so the vectors are independent}


Right working: solve c1(1,2)+c2(2,4)=0c_1(1, 2) + c_2(2, 4) = \mathbf{0} for all coefficients at once. The components give c1+2c2=0c_1 + 2c_2 = 0 and 2c1+4c2=02c_1 + 4c_2 = 0; the second equation is twice the first, so every c1=−2c2c_1 = -2c_2 works. With c2=−1c_2 = -1 and c1=2c_1 = 2:
2(1,2)−1(2,4)=(0,0)2(1, 2) - 1(2, 4) = (0, 0)

A non-trivial combination reaches 0\mathbf{0}, so the pair is dependent.

Why it happens: one failed attempt feels like evidence, but independence is a statement about every possible choice of coefficients, and a single trial cannot cover them all.

How to check: for two vectors, look for a scalar multiple. Here (2,4)=2(1,2)(2, 4) = 2(1, 2), so the vectors are parallel and therefore dependent. For more vectors, solve the system instead of sampling it.

Writing the vectors as rows

When the combination question is turned into Ax=bA\mathbf{x} = \mathbf{b}, the given vectors are placed as the rows of AA instead of its columns. The resulting system may still have a solution, but its numbers are not the coefficients of the combination.

Wrong working: to write b=(3,7)\mathbf{b} = (3, 7) as a combination of (1,2)(1, 2) and (0,1)(0, 1), the rows (1,2)(1, 2) and (0,1)(0, 1) give the system
c1+2c2=3,c2=7c_1 + 2c_2 = 3, \qquad c_2 = 7

so c2=7c_2 = 7 and c1=−11c_1 = -11.

Right working: each equation collects one component from every vector. The first components give c1⋅1+c2⋅0=3c_1 \cdot 1 + c_2 \cdot 0 = 3, and the second components give c1⋅2+c2⋅1=7c_1 \cdot 2 + c_2 \cdot 1 = 7, so
c1=3,c2=7−6=1c_1 = 3, \qquad c_2 = 7 - 6 = 1

and 3(1,2)+1(0,1)=(3,7)=b3(1, 2) + 1(0, 1) = (3, 7) = \mathbf{b}.

Why it happens: vectors are usually written across the page, as (1,2)(1, 2), so they are copied into the matrix the way they are written, as rows.

How to check: substitute the coefficients back into the combination; the result must be b\mathbf{b}. The wrong answer gives −11(1,2)+7(0,1)=(−11,−15)-11(1, 2) + 7(0, 1) = (-11, -15), which is not (3,7)(3, 7).

Linear Combinations FAQ

Does the span of a set always contain the zero vector?

+
Always, whatever the vectors are. Setting every coefficient to zero is a legal linear combination and it produces 0\mathbf{0}, so the zero vector belongs to every span without exception. This is one reason a span is always a flat subspace passing through the origin rather than an arbitrary region of the space.Read more →

Is the span of two vectors always a plane?

+
Only if they point in genuinely different directions. Two non-parallel vectors span a plane, but two parallel ones span merely a line, since the second contributes no direction the first did not already cover. A new vector enlarges the span by a dimension only when it reaches outside what the existing vectors can already build.Read more →

Can a spanning set contain unnecessary vectors?

+
Yes, spanning and efficiency are separate questions. The set {(1,0),(0,1),(1,1)}\{(1,0), (0,1), (1,1)\} spans R2\mathbb{R}^2 even though the third vector is already a combination of the first two: every point is still reachable, just with a redundant option. A spanning set carrying no redundancy at all is what gets called a basis.Read more →