Linear combination

In Mathematical, the linear combinations are a central concept of the Linear algebra and other fields of related mathematics. The major part of this article treats linear combinations in the context of vector Space on a commutative body, and indicates some generalizations at the end of the article.

Definitions

Let us suppose that $K$ is a commutative body and $E$ a vector Space on $K$. As usual we call the elements of $E$ the vectors and the scalar elements of $K$ the . If $v\_1,\; \backslash \; ldots,\; v\_n$ are vectors of $E$ and $a\_1,\; \backslash \; ldots,\; a\_n$ of the scalars, then the linear combination of these vectors having as coefficients these scalars is:

$a\_1\; v\_1+a\_2\; v\_2+\backslash cdots+a\_n\; v\_n$

By convention, a linear combination relating to no vector is declared null.

One can wish to speak about linear combination on an infinity of terms; it is agreed whereas all the scalars intervening are null except a finished number: $(x\_i)\; \_\; \{I\; \backslash \; in\; I\}$ being an unspecified family of vectors of $E$ and $(\backslash \; lambda\_i)\; \_\; \{I\; \backslash \; in\; I\}$ a family of null scalars almost all (i.e. all null except possibly a finished number), the linear combination of the family $(x\_i)\; \_\; \{I\; \backslash \; in\; I\}$ of coefficients $(xi)$ is the following sum:

$\backslash \; sum\_\; \{I\; \backslash \; in\; I\}\; \backslash \; lambda\_ix\_i$

A linear relation of dependence is a linear combination equal to the null vector. The relation of commonplace dependence linaire is that given by a family of null coefficients all.

Examples and counterexamples

analytical Geometry

Either $K$ the body $\backslash \; mathbb\; \{R\}$ of the real numbers, and or $E$ the Euclidean vector Space $\backslash \; mathbb\; \{R\}\; ^3$.

Let us consider the vectors $e\_1=\; (1,0,0)$, $e\_2=\; (0,1,0)$ and $e\_3=\; (0,0,1)$.

Then any vector of $\backslash \; mathbb\; \{R\}\; ^3$ is a linear combination of $e\_1$, $e\_2$ and $e\_3$.

To show it, let us consider an arbitrary vector $(a\_1,\; a\_2,\; a\_3)$ of $\backslash \; mathbb\; \{R\}\; ^3$, and let us write:

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

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

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

functional Analysis

Either $K$ the unit $\backslash \; mathbb\; \{C\}$ of all the complex numbers, and or $E$ the unit $\backslash \; mathcal\; \{C\}\; (\backslash \; mathbb\; \{R\},\; \backslash \; mathbb\; \{C\})$ of the continuous functions of the real Right $\backslash \; mathbb\; \{R\}$ in the Plane complex $\backslash \; mathbb\; \{C\}$.

Let us consider the vectors (functions) $f$ and $g$définies by $f\; (T)\; =e^\; \{it\}$ and $g\; (T)\; =e^\; \{-\; it\}$.

(Here, $e$ indicates the bases Napierian logarithm, roughly equalizes with 2,71828, and $i$ the Imaginary number, a square root of $-1$.

Linear combinations of $f$ and $g$ are:

• $\backslash \; cos\; =\; \backslash \; begin\; \{matrix\}\; \backslash \; frac12\; \backslash \; end\; \{matrix\}\; F\; +\; \backslash \; begin\; \{matrix\}\; \backslash \; frac12\; \backslash \; end\; \{matrix\}\; G\; \backslash ,$
• $2\; \backslash \; sin\; =\; (-\; I)\; F\; +\; (I)\; G\; \backslash ,$

On the other hand, the constant function 3 is not not a linear combination of $f$ and $g$. To see it, let us suppose by the absurdity that 3 can be written like linear combination of the functions $t\; \backslash \; mapsto\; e^\; \{it\}$ and $t\; \backslash \; mapsto\; e^\; \{-\; it\}$. That would mean that there would exist complex scalars $a$ and $b$ such as for any reality $t$, $ae^\; \{it\}\; +be^\; \{-\; it\}\; =3$. By posing $t=0$ and $t=\; \backslash \; pi$, that would give the relations $a+b=3$ and $a+b=-3$, which could not clearly occur.

algebraic Geometry

That is to say $K$ an unspecified commutative body ($\backslash \; mathbb\; \{R\}$ or $\backslash \; mathbb\; \{C\}$), and $E$ the $K$ unit of the polynomials to coefficients in the body $K$.

Let us consider the vectors (polynomials) $p\_1=1$, $p\_2=X+1$ and $p\_3=X^2+X+1$.

Is the polynomial $X^2-1$ combination linear of $p\_1$, $p\_2$ and $p\_3$?

To see it, let us consider an arbitrary linear combination of these vectors and try to see when it is equal to this vector $X^2-1$.

Let us take, arbitrary coefficients $a\_1,\; a\_2$ and $a\_3$. We want:

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

By distributing the coefficients on the polynomials we obtain

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

and according to the power of $X$, we gather obtain

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

Two polynomials are equal if and only if their coefficients correpondants are equal, thus we can deduce from it that

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

This Système of linear equations can easily be solved.

First of all, the first equation shows that $a\_3=1$.

Knowing that, we can solve the second equation which gives $a\_2=-1$.

Finally, the last equation indicates to us that $a\_3$ is worth also $-1$.

Reciprocally, the equality $X^2\; -\; 1\; =\; -1\; -\; (X\; +\; 1)\; +\; (X^2\; +\; X\; +\; 1)\; =\; -\; p\_1\; -\; p\_2\; +\; p\_3\; \backslash ,$ is well checked. Thus $X^2-1$ is linear combination of $p\_1,\; p\_2$ and $p\_3$.

Consequently, $X^2-1$ is written in a only one possible way in the form of a linear combination of $p\_1,\; p\_2$ and $p\_3$.

In addition, that is it polynomial $X^3-1$?

If we try to write this vector like a linear combination of $p\_1,\; p\_2$ and $p\_3$, then while following the same reasoning that front, we obtain the equation:

$0\; X^3\; +\; a\_3\; X^2\; +\; (a\_2\; +\; a\_3)\; X\; +\; (a\_1\; +\; a\_2\; +\; a\_3)\; \backslash ,$

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

However, when we translate that the corresponding coefficients must be equal in this case, the relation obtained by considering $X^3$ becomes

$0\; =\; 1\; \backslash ,$ which is contradictory.

Consequently, there is no manner so that this is true, thus $X^3-1$ is not linear combination of $p\_1,\; p\_2$ and $p\_3$.

Generated vectorial subspace

Let us consider a commutative body $K$ and a vector space $E$ arbitrary, and is $v\_1,\; \backslash \; ldots,\; v\_n$ vectors of $E$. It is interesting to consider the whole of all the linear combinations of these vectors. This unit is called the “vectorial Sous-espace generated” (or right “generated subspace”) by these vectors, say by the $A\; unit\; =\; \backslash \; \{v\_1,\; \backslash \; ldots,\; v\_n\; \backslash \}$. Let us note $\{\backslash \; rm\; Vect\}\; (v\_1,\; \backslash \; ldots,\; v\_n)$ or $$ the unit

$\backslash \; mathrm\; \{Vect\}\; (v\_1,\; \backslash \; ldots,\; v\_n)\; =\; \backslash \; \{a\_1\; v\_1\; +\; \backslash \; cdots\; +\; a\_n\; v\_n/a\_1,\; \backslash \; ldots,\; a\_n\; \backslash \; in\; K\; \backslash \}\; \backslash ,$

Other relative concepts

Sometimes, a certain vector can be written in two manners different like linear combination from $v\_1,\; \backslash \; ldots,\; v\_n$. If that occurs the vectors $v\_1\; then,\; \backslash \; ldots,\; v\_n$ is linearly dependant, and in the contrary case, when any writing of a vector like linear combination of $v\_1,\; \backslash \; ldots,\; v\_n$ is single, then the vectors are linearly independent.

In the same way, we can speak about the dependence or the linear independence of the vectors of an arbitrary unit $A$. If the vectors of $A$ are linearly independent then the $A$ part is known as free and so moreover the vectorial subspace generated by $A$ to $E$ then $A$ is equal is a basic left $E$.

We can compare the linear combinations to the most general possible operation on a vector space. The basic operations of addition and multiplication by a scalar, as well as the existence of a neutral element and of opposite, cannot be combined in a way more complicated than in a linear combination. Finally, this fact is in the middle of the utility of the linear combinations in the study of spaces of vector.

Generalizations

If $E$ is a topological vector Space, then it is possible to give a direction to a linear combination infinite , by using the topology of $E$. For example, we could speak about the infinite sum $a\_1\; v\_1+a\_2\; v\_2+a\_3\; v\_3+\; \backslash \; cdots$.

Such infinite linear combinations always do not have a direction; we qualify them convergent when they have one of them. The fact of being able to consider more linear combinations in this case can also lead to broader concepts of generated vectorial subspace, linear independence, and bases.

If $K$ is a commutative Anneau instead of being a body, then all that was known as above on the linear combinations spreads without any change. The only difference is that we call these spaces $E$ modules instead of vector spaces.

If $K$ is a noncommutative ring, then the concept of combination linear still spreads, however with a restriction: Since the modules on the noncommutative rings can be modules on the right or on the left, our linear combinations can also be written on the right or on the left, i.e. with scalars placed on the right or on the left, according to the nature of the module. It is simply a question of multiplication by a scalar of the good side.

A more complicated adaptation occurs when $E$ is a Bimodule on two two rings, $K\_G$ and $K\_D$.

In this case, the linear combination most general resembles:

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

where $a\_1,\; \backslash \; ldots,\; a\_n$ belong to $K\_G$, $b\_1,\; \backslash \; ldots,\; b\_n$ belong to $K\_D$, and $v\_1,\; \backslash \; ldots,\; v\_n$ belong to $E$.

Random links:

Sixth IV | Direct object | Line of bus the RATP 189 | Dendrobates histrionicus | Record of France of swimming ladies of the 200 meters back | Le_yard_de_la_bouilloire

© 2007-2008 speedlook.com; article text available under the terms of GFDL, from fr.wikipedia.org