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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.

- $a\_1\; v\_1\; +\; a\_2\; v\_2\; +\; a\_3\; v\_3\; +\; cdots\; +\; a\_n\; v\_n\; ,$

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 v_{1},...,v_{n}, 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 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.

Let the field K be the set R of real numbers, and let the vector space V be the Euclidean space R^{3}.
Consider the vectors e_{1} := (1,0,0), e_{2} := (0,1,0) and e_{3} = (0,0,1).
Then any vector in R^{3} is a linear combination of e_{1}, e_{2} and e_{3}.

To see that this is so, take an arbitrary vector (a_{1},a_{2},a_{3}) in R^{3}, and write:

- $(a\_1\; ,\; a\_2\; ,\; a\_3)\; =\; (a\_1\; ,0,0)\; +\; (0,\; a\_2\; ,0)\; +\; (0,0,\; a\_3)\; ,$

- $=\; a\_1\; (1,0,0)\; +\; a\_2\; (0,1,0)\; +\; a\_3\; (0,0,1)\; ,$

- $=\; a\_1\; e\_1\; +\; a\_2\; e\_2\; +\; a\_3\; e\_3\; ,$

Let K be the set C of all complex numbers, and let V be the set C_{C}(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) := e^{it} and g(t) := e^{−it}.
(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:

- . $cosh\; t\; =\; begin\{matrix\}frac12end\{matrix\}\; e^\{i\; t\}\; +\; begin\{matrix\}frac12end\{matrix\}\; e^\{-i\; t\}\; ,$
- $2\; sin\; t\; =\; (-i\; )\; e^\{i\; t\}\; +\; (i\; )\; e^\{-i\; t\}\; ,$

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 e^{it} and e^{−it}. This means that there would exist complex scalars a and b such that ae^{it} + be^{−it} = 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.

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) p_{1} := 1, p_{2} := x + 1, and p_{3} := x^{2} + x + 1.

Is the polynomial x^{2} − 1 a linear combination of p_{1}, p_{2}, and p_{3}?
To find out, consider an arbitrary linear combination of these vectors and try to see when it equals the desired vector x^{2} − 1.
Picking arbitrary coefficients a_{1}, a_{2}, and a_{3}, we want

- $a\_1\; (1)\; +\; a\_2\; (x\; +\; 1)\; +\; a\_3\; (x^2\; +\; x\; +\; 1)\; =\; x^2\; -\; 1\; ,$

- $(a\_1\; )\; +\; (a\_2\; x\; +\; a\_2)\; +\; (a\_3\; x^2\; +\; a\_3\; x\; +\; a\_3)\; =\; x^2\; -\; 1\; ,$

- $a\_3\; x^2\; +\; (a\_2\; +\; a\_3\; )\; x\; +\; (a\_1\; +\; a\_2\; +\; a\_3\; )\; =\; 1\; x^2\; +\; 0\; x\; +\; (-1)\; ,$

- $a\_3\; =\; 1,\; quad\; a\_2\; +\; a\_3\; =\; 0,\; quad\; a\_1\; +\; a\_2\; +\; a\_3\; =\; -1\; ,$

- $x^2\; -\; 1\; =\; -1\; -\; (x\; +\; 1)\; +\; (x^2\; +\; x\; +\; 1)\; =\; -\; p\_1\; -\; p\_2\; +\; p\_3\; ,$

On the other hand, what about the polynomial x^{3} − 1?
If we try to make this vector a linear combination of p_{1}, p_{2}, and p_{3}, then following the same process as before, we’ll get the equation

- $0\; x^3\; +\; a\_3\; x^2\; +\; (a\_2\; +\; a\_3\; )\; x\; +\; (a\_1\; +\; a\_2\; +\; a\_3\; )\; ,$

- $=\; 1\; x^3\; +\; 0\; x^2\; +\; 0\; x\; +\; (-1)\; ,$

- $0\; =\; 1\; ,$

Main article: linear span

Take an arbitrary field K, an arbitrary vector space V, and let v_{1},...,v_{n} 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 ={v_{1},...,v_{n}}. We write the span of S as span(S) or sp(S):

- $mathrm\{Sp\}(v\_1\; ,ldots,\; v\_n)\; :=\; \{\; a\_1\; v\_1\; +\; cdots\; +\; a\_n\; v\_n\; :\; a\_1\; ,ldots,\; a\_n\; subseteq\; K\; \}.\; ,$

Sometimes, some single vector can be written in two different ways as a linear combination of v_{1},...,v_{n}.
If that is possible, then v_{1},...,v_{n} 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.

Another related concept is the affine combination, which is a linear combination with the additional constraint that
the coefficients a_{1},...,a_{n} sum to unity.

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, K_{L} and K_{R}.
In that case, the most general linear combination looks like

- $a\_1\; v\_1\; b\_1\; +\; cdots\; +\; a\_n\; v\_n\; b\_n\; ,$

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Last updated on Wednesday October 01, 2008 at 07:39:37 PDT (GMT -0700)

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This article is licensed under the GNU Free Documentation License.

Last updated on Wednesday October 01, 2008 at 07:39:37 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

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