# Linear combination

In mathematics, linear combinations are a concept central to linear algebra and related fields of mathematics. Most of this article deals with linear combinations in the context of a vector space over a field, with some generalisations given at the end of the article.

Definition   Suppose that K is a field and V is a vector space over K. As usual, we call elements of V vectors and call elements of K scalars. If v1,...,vn are vectors and a1,...,an are scalars, then the linear combination of those vectors with those scalars as coefficients is
[itex]a_1 v_1 + a_2 v_2 + a_3 v_3 + \cdots + a_n v_n \,[itex]

In a given situation, K and V may be specified explicitly, or they may be obvious from context. In that case, we often speak of a linear combination of the vectors v1,...,vn, with the coefficients unspecified (except that they must belong to K). Or, if S is a subset of V, we may speak of a linear combination of vectors in S, where both the coefficients and the vectors are unspecified, except that the vectors must belong to the set S (and the coefficients must belong to K). Finally, we may speak simply of a linear combination, where nothing is specified (except that the vectors must belong to V and the coefficients must belong to K).

Note that by definition, a linear combination involves only of finitely many vectors (except as described in Generalisations below). However, the set S that the vectors are taken from (if one is mentioned) can still be infinite; each individual linear combination will only involve finitely many vectors. Also, there is no reason that n cannot be zero; in that case, we declare by convention that the result of the linear combination is the zero vector in V.

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## Examples and counterexamples

### Analytic geometry

Let the field K be the set R of real numbers, and let the vector space V be the Euclidean space R3. Consider the vectors e1 := (1,0,0), e2 := (0,1,0) and e3 = (0,0,1). Then any vector in R3 is a linear combination of e1, e2 and e3.

To see that this is so, take an arbitrary vector (a1,a2,a3) in R3, and write:

[itex] ( a_1 , a_2 , a_3) = ( a_1 ,0,0) + (0, a_2 ,0) + (0,0, a_3) \,[itex]
[itex] = a_1 (1,0,0) + a_2 (0,1,0) + a_3 (0,0,1) \,[itex]
[itex] = a_1 e_1 + a_2 e_2 + a_3 e_3 \,[itex]

### Functional analysis

Let K be the set C of all complex numbers, and let V be the set CC(R) of all continuous functions from the real line R to the complex plane C. Consider the vectors (functions) f and g defined by f(t) := eit and g(t) := eit. (Here, e is the base of the natural logarithm, about 2.71828..., and i is the imaginary unit, a square root of −1.) Some linear combinations of f and g are:

• .
[itex] \cos t = \begin{matrix}\frac12\end{matrix} e^{i t} + \begin{matrix}\frac12\end{matrix} e^{-i t} \,[itex]
• [itex] 2 \sin t = (-i ) e^{i t} + ( i ) e^{-i t} \,[itex]

On the other hand, the constant function 3 is not a linear combination of f and g. To see this, suppose that 3 could be written as a linear combination of eit and eit. This means that there would exist complex scalars a and b such that aeit + beit = 3 for all real numbers t. Setting t = 0 and t = π gives the equations a + b = 3 and a + b = −3, and clearly this cannot happen.

### Algebraic geometry

Let K be any field (R, C, or whatever you like best), and let V be the set P of all polynomials with coefficients taken from the field K. Consider the vectors (polynomials) p1 := 1, p2 := x + 1, and p3 := x2 + x + 1.

Is the polynomial x2 − 1 a linear combination of p1, p2, and p3? To find out, consider an arbitrary linear combination of these vectors and try to see when it equals the desired vector x2 − 1. Picking arbitrary coefficients a1, a2, and a3, we want

[itex] a_1 (1) + a_2 ( x + 1) + a_3 ( x^2 + x + 1) = x^2 - 1 \,[itex]

Multiplying the polynomials out, this means

[itex] ( a_1 ) + ( a_2 x + a_2) + ( a_3 x^2 + a_3 x + a_3) = x^2 - 1 \,[itex]

and collecting like powers of x, we get

[itex] a_3 x^2 + ( a_2 + a_3 ) x + ( a_1 + a_2 + a_3 ) = 1 x^2 + 0 x + (-1) \,[itex]

Two polynomials are equal if and only if their corresponding coefficients are equal, so we can conclude

[itex] a_3 = 1, \quad a_2 + a_3 = 0, \quad a_1 + a_2 + a_3 = -1 \,[itex]

This system of linear equations can easily be solved. First, the first equation simply says that a3 is 1. Knowing that, we can solve the second equation for a2, which comes out to −1. Finally, the last equation tells us that a1 is also −1. Therefore, the only possible way to get a linear combination is with these coefficients. Indeed,

[itex] x^2 - 1 = -1 - ( x + 1) + ( x^2 + x + 1) = - p_1 - p_2 + p_3 \,[itex]

so x2 − 1 is a linear combination of p1, p2, and p3.

On the other hand, what about the polynomial x3 − 1? If we try to make this vector a linear combination of p1, p2, and p3, then following the same process as before, we’ll get the equation

[itex] 0 x^3 + a_3 x^2 + ( a_2 + a_3 ) x + ( a_1 + a_2 + a_3 ) \,[itex]
[itex] = 1 x^3 + 0 x^2 + 0 x + (-1) \,[itex]

However, when we set corresponding coefficients equal in this case, the equation for x3 is

[itex] 0 = 1 \,[itex]

which is always false. Therefore, there is no way for this to work, and x3 − 1 is not a linear combination of p1, p2, and p3.

## The linear span

Main article: linear span

Take an arbitrary field K, an arbitrary vector space V, and let v1,...,vn be vectors (in V). It’s interesting to consider the set of all linear combinations of these vectors. This set is called the linear span (or just span) of the vectors, say S ={v1,...,vn}. We write the span of S as span(S) or sp(S):

[itex] \mathrm{Sp}( v_1 ,\ldots, v_n) := \{ a_1 v_1 + \cdots + a_n v_n : a_1 ,\ldots, a_n \subseteq K \} \,[itex]

## Other related concepts

Sometimes, some single vector can be written in two different ways as a linear combination of v1,...,vn. If that is possible, then v1,...,vn are called linearly dependent; otherwise, they are linearly independent. Similarly, we can speak of linear dependence or independence of an arbitrary set S of vectors.

If S is linearly independent and the span of S equals V, then S is a basis for V.

We can think of linear combinations as the most general sort of operation on a vector space. The basic operations of addition and scalar multiplication, together with the existence of an additive identity and additive inverses, cannot be combined in any more complicated way than the generic linear combination. Ultimately, this fact lies at the heart of the usefulness of linear combinations in the study of vector spaces.

## Generalisations

If V is a topological vector space, then there may be a way to make sense of certain infinite linear combination, using the topology of V. For example, we might be able to speak of a1v1 + a2v2 + a3v3 + ..., going on forever. Such infinite linear combinations do not always make sense; we call them convergent when they do. Allowing more linear combinations in this case can also lead to a different concept of span, linear independence, and basis. The articles on the various flavours of topological vector spaces go into more detail about these.

If K is a commutative ring instead of a field, then everything that has been said above about linear combinations generalises to this case without change. The only difference is that we call spaces like V modules instead of vector spaces. If K is a noncommutative ring, then the concept still generalises, with one caveat: Since modules over noncommutative rings come in left and right versions, our linear combinations may also come in either of these versions, whatever is appropriate for the given module. This is simply a matter of doing scalar multiplication on the correct side.

A more complicated twist comes when V is a bimodule over two rings, KL and KR. In that case, the most general linear combination looks like

[itex] a_1 v_1 b_1 + \cdots + a_n v_n b_n \, [itex]

where a1,...,an belong to KL, b1,...,bn belong to KR, and v1,...,vn belong to V.de:Linearkombination es:Combinación lineal ko:선형결합 he:צירוף לינארי ja:線型結合 pl:Kombinacja liniowa wektorów

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