- Linear combination
In

mathematics ,**linear combinations**are a concept central tolinear algebra and related fields of mathematics.Most of this article deals with linear combinations in the context of avector 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 "v"

_{1},...,"v"_{"n"}are vectors and "a"_{1},...,"a"_{"n"}are scalars, then the "linear combination of those vectors with those scalars as coefficients" is:$a\_1\; v\_1\; +\; a\_2\; v\_2\; +\; a\_3\; v\_3\; +\; cdots\; +\; a\_n\; v\_n\; ,$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 asubset 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 beinfinite ; 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".**Examples and counterexamples**Analytic geometry Let the field "K" be the set

**R**ofreal number s, and let the vector space "V" be theEuclidean 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::$(\; a\_1\; ,\; a\_2\; ,\; a\_3)\; =\; (\; a\_1\; ,0,0)\; +\; (0,\; a\_2\; ,0)\; +\; (0,0,\; a\_3)\; ,$ :::$=\; a\_1\; (1,0,0)\; +\; a\_2\; (0,1,0)\; +\; a\_3\; (0,0,1)\; ,$ :::$=\; a\_1\; e\_1\; +\; a\_2\; e\_2\; +\; a\_3\; e\_3\; ,$Functional analysis Let "K" be the set

**C**of allcomplex number s, and let "V" be the set C_{C}("R") of all continuous functions from thereal line **R**to thecomplex 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 theimaginary unit , a square root of −1.)Some linear combinations of "f" and "g" are:

*.$cosh\; t\; =\; egin\{matrix\}frac12end\{matrix\}\; e^\{i\; t\}\; +\; egin\{matrix\}frac12end\{matrix\}\; e^\{-i\; t\}\; ,$

*$2\; sin\; t\; =\; (-i\; )\; e^\{i\; t\}\; +\; (\; i\; )\; e^\{-i\; t\}\; ,$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.Algebraic geometry Let "K" be any field (

**R**,**C**, or whatever you like best), and let "V" be the set "P" of allpolynomial s 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:$a\_1\; (1)\; +\; a\_2\; (\; x\; +\; 1)\; +\; a\_3\; (\; x^2\; +\; x\; +\; 1)\; =\; x^2\; -\; 1\; ,$Multiplying the polynomials out, this means:$(\; a\_1\; )\; +\; (\; a\_2\; x\; +\; a\_2)\; +\; (\; a\_3\; x^2\; +\; a\_3\; x\; +\; a\_3)\; =\; x^2\; -\; 1\; ,$ and collecting like powers of "x", we get:$a\_3\; x^2\; +\; (\; a\_2\; +\; a\_3\; )\; x\; +\; (\; a\_1\; +\; a\_2\; +\; a\_3\; )\; =\; 1\; x^2\; +\; 0\; x\; +\; (-1)\; ,$ Two polynomials are equalif and only if their corresponding coefficients are equal, so we can conclude:$a\_3\; =\; 1,\; quad\; a\_2\; +\; a\_3\; =\; 0,\; quad\; a\_1\; +\; a\_2\; +\; a\_3\; =\; -1\; ,$ Thissystem of linear equations can easily be solved.First, the first equation simply says that "a"_{3}is 1.Knowing that, we can solve the second equation for "a"_{2}, which comes out to −1.Finally, the last equation tells us that "a"_{1}is also −1.Therefore, the only possible way to get a linear combination is with these coefficients.Indeed,:$x^2\; -\; 1\; =\; -1\; -\; (\; x\; +\; 1)\; +\; (\; x^2\; +\; x\; +\; 1)\; =\; -\; p\_1\; -\; p\_2\; +\; p\_3\; ,$so "x"^{2}− 1 "is" a linear combination of "p"_{1}, "p"_{2}, and "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:$0\; x^3\; +\; a\_3\; x^2\; +\; (\; a\_2\; +\; a\_3\; )\; x\; +\; (\; a\_1\; +\; a\_2\; +\; a\_3\; )\; ,$ :$=\; 1\; x^3\; +\; 0\; x^2\; +\; 0\; x\; +\; (-1)\; ,$ However, when we set corresponding coefficients equal in this case, the equation for "x"^{3}is:$0\; =\; 1\; ,$ which is always false.Therefore, there is no way for this to work, and "x"^{3}− 1 is "not" a linear combination of "p"_{1}, "p"_{2}, and "p"_{3}.**The linear span**"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)::$mathrm\{Sp\}(\; v\_1\; ,ldots,\; v\_n)\; :=\; \{\; a\_1\; v\_1\; +\; cdots\; +\; a\_n\; v\_n\; :\; a\_1\; ,ldots,\; a\_n\; subseteq\; K\; \}.\; ,$**Other related concepts**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 thatthe coefficients "a"_{1},...,"a"_{"n"}sum to unity.**Generalizations**If "V" is a

topological vector space , then there may be a way to make sense of certain "infinite" linear combinations, using the topology of "V".For example, we might be able to speak of "a"_{1}"v"_{1}+ "a"_{2}"v"_{2}+ "a"_{3}"v"_{3}+ ..., 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, "K"_{L}and "K"_{R}.In that case, the most general linear combination looks like:$a\_1\; v\_1\; b\_1\; +\; cdots\; +\; a\_n\; v\_n\; b\_n\; ,$ where "a"_{1},...,"a"_{"n"}belong to "K"_{L}, "b"_{1},...,"b"_{"n"}belong to "K"_{R}, and "v"_{1},...,"v"_{"n"}belong to "V".

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