LINEAR INDEPENDENCE

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In linear algebra, a family of vectors is 'linearly independent' if none of them can be written as a linear combination of finitely many other vectors in the collection. A family of vectors which is not linearly independent is called 'linearly dependent'. For instance, in the three-dimensional real vector space 'R'³ we have the following example.
:$$
egin{matrix}
mbox{independent}qquad\
underbrace{
overbrace{
egin{bmatrix}0\0\1end{bmatrix},
egin{bmatrix}0\2\-2end{bmatrix},
egin{bmatrix}1\-2\1end{bmatrix}
},
egin{bmatrix}4\2\3end{bmatrix}
}\
mbox{dependent}\
end{matrix}

Here the first three vectors are linearly independent; but the fourth vector equals 9 times the first plus 5 times the second plus 4 times the third, so the four vectors together are linearly dependent. Linear dependence is a property of the family, not of any particular vector; here we could just as well write the first vector as a linear combination of the last three.
:
ight) old{v}_2 + left(-rac{4}{9}
ight) old{v}_3 + rac{1}{9} old{v}_4 .

Contents

Formal definition

Geometric meaning

Example I

Proof

Alternative method using determinants

Example II

Proof

Example III

Proof

Example IV

Proof

The projective space of linear dependences

See also

External links

Formal definition

A subset ''S'' of vector space ''V'' is called ''linearly dependent'' if there exist a finite number of distinct vectors 'v'₁, 'v'₂, ..., 'v'_''n'' in ''S'' and scalars ''a''₁, ''a''₂, ..., ''a''_''n'', not all zero, such that
:$a\_1\; mathbf\{v\}\_1\; +\; a\_2\; mathbf\{v\}\_2\; +\; cdots\; +\; a\_n\; mathbf\{v\}\_n\; =\; mathbf\{0\}.$
Note that the zero on the right is the zero vector, not the number zero.
If such scalars do not exist, then the vectors are said to be ''linearly independent''. This condition can be reformulated as follows: Whenever ''a''₁, ''a''₂, ..., ''a''_''n'' are scalars such that
:$a\_1\; mathbf\{v\}\_1\; +\; a\_2\; mathbf\{v\}\_2\; +\; cdots\; +\; a\_n\; mathbf\{v\}\_n\; =\; mathbf\{0\},$
we have ''a''_''i'' = 0 for ''i'' = 1, 2, ..., ''n'', i.e. ''only'' the trivial solution exists.
A set is linearly independent if and only if the only representations of the zero vector as linear combinations of its elements are trivial solutions.
More generally, let ''V'' be a vector space over a field ''K'', and let {'v'_''i''}_{''i''∈''I''} be a family of elements of ''V''. The family is ''linearly dependent'' over ''K'' if there exists a family {''a''_''j''}_{''j''∈''J''} of elements of ''K'', not all zero, such that
:$sum\_\{j\; in\; J\}\; a\_j\; mathbf\{v\}\_j\; =\; mathbf\{0\}\; ,$
where the index set ''J'' is a nonempty, finite subset of ''I''.
A set ''X'' of elements of ''V'' is ''linearly independent'' if the corresponding family {'x'}_'x'∈''X'' is linearly independent.
Equivalently, a family is dependent if a member is in the linear span of the rest of the family, i.e., a member is a linear combination of the rest of the family.
A set of vectors which is linearly independent and spans some vector space, forms a basis for that vector space.

Geometric meaning

A geographic example may help to clarify the concept of linear independence. A person describing the location of a certain place might say, "It is 5 miles north and 6 miles east of here." This is sufficient information to describe the location, because the geographic coordinate system may be considered a 2-dimensional vector space (ignoring altitude). The person might add, "The place is 7.81 miles northeast of here." Although this last statement is ''true'', it is not necessary.
In this example the "5 miles north" vector and the "6 miles east" vector are linearly independent. That is to say, the north vector cannot be described in terms of the east vector, and vice versa. The third "7.81 miles northeast" vector is a linear combination of the other two vectors, and it makes the set of vectors ''linearly dependent'', that is, one of the three vectors is unnecessary.
Note that in this example, ''any'' of the three vectors may be described as a linear combination of the other two. While it might be inconvenient, one could describe "6 miles east" in terms of north and northeast. (For example, "Go 5 miles south (mathematically, −5 miles north) and then go 7.81 miles northeast.") Similarly, the north vector is a linear combination of the east and northeast vectors.
Also note that if altitude is not ignored, it becomes necessary to add a third vector to the linearly independent set. In general, ''n'' linearly independent vectors are required to describe any location in ''n''-dimensional space.

Example I

The vectors (1, 1) and (−3, 2) in 'R'² are linearly independent.

Proof

Let λ₁ and λ₂ be two real numbers such that
:$(1,\; 1)\; lambda\_1\; +\; (-3,\; 2)\; lambda\_2\; =\; (0,\; 0)\; .\; ,!$
Taking each coordinate alone, this means
:
lambda_1 - 3 lambda_2 &{}= 0 , \
lambda_1 + 2 lambda_2 &{}= 0 .
end{align}
Solving for λ₁ and λ₂, we find that λ₁ = 0 and λ₂ = 0.

Alternative method using determinants

An alternative method uses the fact that ''n'' vectors in 'R'^''n'' are linearly dependent if and only if the determinant of the matrix formed by taking the vectors as its columns is zero.
In this case, the matrix formed by the vectors is
:
We may write a linear combination of the columns as
:
We are interested in whether ''A''Λ = '0' for some nonzero vector Λ. This depends on the determinant of ''A'', which is
:$det\; A\; =\; 1cdot2\; -\; 1cdot(-3)\; =\; 5$
e 0 . ,!
Since the determinant is non-zero, the vectors (1, 1) and (−3, 2) are linearly independent.
When the number of vectors equals the dimension of the vectors, the matrix is square and hence the determinant is defined.
Otherwise, suppose we have ''m'' vectors of ''n'' coordinates, with ''m'' < ''n''. Then ''A'' is an ''n''×''m'' matrix and Λ is a column vector with ''m'' entries, and we are again interested in ''A''Λ = '0'. As we saw previously, this is equivalent to a list of ''n'' equations. Consider the first ''m'' rows of ''A'', the first ''m'' equations; any solution of the full list of equations must also be true of the reduced list. In fact, if ⟨''i''₁,…,''i''_''m''⟩ is any list of ''m'' rows, then the equation must be true for those rows.
:$A\_\{\{lang\; i\_1,dots,i\_m\}$
ang} Lambda = old{0} . ,!
Furthermore, the reverse is true. That is, we can test whether the ''m'' vectors are linearly dependent by testing whether
:$det\; A\_\{\{lang\; i\_1,dots,i\_m\}$
ang} = 0 ,!
for all possible lists of ''m'' rows. (In case ''m'' = ''n'', this requires only one determinant, as above. If ''m'' > ''n'', then it is a theorem that the vectors must be linearly dependent.) This fact is valuable for theory; in practical calculations more efficient methods are available.

Example II

Let ''V'' = 'R'^''n'' and consider the following elements in ''V'':
:
mathbf{e}_1 & = & (1,0,0,ldots,0) \
mathbf{e}_2 & = & (0,1,0,ldots,0) \
& dots \
mathbf{e}_n & = & (0,0,0,ldots,1).end{matrix}
Then 'e'₁, 'e'₂, ..., 'e_n' are linearly independent.

Proof

Suppose that ''a''₁, ''a''₂, ..., ''a_n'' are elements of 'R' such that
:$a\_1\; mathbf\{e\}\_1\; +\; a\_2\; mathbf\{e\}\_2\; +\; cdots\; +\; a\_n\; mathbf\{e\}\_n\; =\; 0\; .\; ,!$
Since
:$a\_1\; mathbf\{e\}\_1\; +\; a\_2\; mathbf\{e\}\_2\; +\; cdots\; +\; a\_n\; mathbf\{e\}\_n\; =\; (a\_1\; ,a\_2\; ,ldots,\; a\_n)\; ,\; ,!$
then ''a_i'' = 0 for all ''i'' in {1, ..., ''n''}.

Example III

Let ''V'' be the vector space of all functions of a real variable ''t''. Then the functions ''e^t'' and ''e''^2''t'' in ''V'' are linearly independent.

Proof

Suppose ''a'' and ''b'' are two real numbers such that
:''ae^t'' + ''be''^2''t'' = 0
for ''all'' values of ''t''. We need to show that ''a'' = 0 and ''b'' = 0. In order to do this, we divide through by ''e''^''t'' (which is never zero) and subtract to obtain
:''be^t'' = −''a''
In other words, the function ''be''^''t'' must be independent of ''t'', which only occurs when ''b'' = 0. It follows that ''a'' is also zero.

Example IV

The following vectors in 'R'⁴ are linearly dependent.
:$$
egin{matrix}
\
egin{bmatrix}1\4\2\-3end{bmatrix},
egin{bmatrix}7\10\-4\-1end{bmatrix} mathrm{and},
egin{bmatrix}-2\1\5\-4end{bmatrix}
\
\
end{matrix}

Proof

We need to find scalars $lambda\_1$, $lambda\_2$ and $lambda\_3$ such that
:$$
egin{matrix}
\
lambda_1 egin{bmatrix}1\4\2\-3end{bmatrix}+
lambda_2 egin{bmatrix}7\10\-4\-1end{bmatrix}+
lambda_3 egin{bmatrix}-2\1\5\-4end{bmatrix}=
egin{bmatrix}0\0\0\0end{bmatrix}
end{matrix}

Forming the simultaneous equations:
:$$
egin{align}
lambda_1& ;+ 7lambda_2& &- 2lambda_3& = 0\
4lambda_1& ;+ 10lambda_2& &+ lambda_3& = 0\
2lambda_1& ;- 4lambda_2& &+ 5lambda_3& = 0\
-3lambda_1& ;- lambda_2& &- 4lambda_3& = 0\
end{align}

we can solve (using for example Gaussian elimination) to obtain:
:$$
egin{align}
lambda_1 &= -3/2 \
lambda_2 &= 1/2 \
lambda_3 &= 1 \
end{align}

Since these are nontrivial results, the vectors are linearly dependent.

The projective space of linear dependences

A 'linear dependence' among vectors 'v'₁, ..., 'v'_''n'' is a tuple (''a''₁, ..., ''a''_''n'') with ''n'' scalar components, not all zero, such that
:$a\_1\; mathbf\{v\}\_1\; +\; cdots\; +\; a\_n\; mathbf\{v\}\_n=0.\; ,$
If such a linear dependence exists, then the ''n'' vectors are linearly dependent. It makes sense to identify two linear dependences if one arises as a non-zero multiple of the other, because in this case the two describe the same linear relationship among the vectors. Under this identification, the set of all linear dependences among 'v'₁, ...., 'v'_''n'' is a projective space.

See also

★ orthogonality

★ matroid (generalization of the concept)

★ Wronskian

External links

★ MIT Linear Algebra Lecture on Linear Independence at Google Video, from MIT OpenCourseWare

★ Linearly Dependent Functions at WolframMathWorld