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. For instance, in three-dimensional Euclidean space R³, the three vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) are linearly independent, while (2, −1, 1), (1, 0, 1) and (3, −1, 2) are not (since the third vector is the sum of the first two). Vectors which are not linearly independent are called linearly dependent.

Contents 1 Definition 2 Geometric meaning 3 Example I 3.1 Proof 3.2 Alternative method using determinants 4 Example II 4.1 Proof 5 Example III 5.1 Proof 6 The projective space of linear dependences 7 See also

Definition

Let v₁, v₂, ..., v_n be vectors. We say that they are linearly dependent if there exist numbers a₁, a₂, ..., a_n, not all equal to zero, such that:

<math> a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + \cdots + a_n \mathbf{v}_n = \mathbf{0}. </math>

Note that the zero on the right is the zero vector, not the number zero.

If such numbers do not exist, then the vectors are said to be linearly independent. This condition can be reformulated as follows: Whenever a₁, a₂, ..., a_n are numbers such that

<math> a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + \cdots + a_n \mathbf{v}_n = \mathbf{0}, </math>

we have a_i = 0 for i = 1, 2, ..., n.

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 nonzero elements of K such that

<math> \sum_{j \in J} a_j \mathbf{v}_j = \mathbf{0} \,</math>

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.

The concept of linear independence is important because 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 a location in n-dimensional space.

Example I

The vectors (1, 1) and (−3, 2) in R² are linearly independent.

Proof

Let a, b be two real numbers such that

<math>a \left( 1, 1 \right) + b \left( -3, 2 \right) = \left( 0, 0 \right)</math>

Then

<math>\left( a - 3b, a + 2b\right) = \left(0, 0\right)</math> and

<math>\ a - 3b = 0</math> and <math>\ a + 2b = 0</math>.

Solving for a and b, we find that a = 0 and b = 0.

Alternative method using determinants

An alternative method uses the fact that n vectors in Rⁿ are linearly dependent if and only if the determinant of the matrix formed by the vectors is zero.

In this case, the matrix formed by the vectors is

<math>A = \begin{bmatrix}1&-3\\1&2\end{bmatrix}. \,</math>

The determinant of this matrix is

<math>\det(A) = 1\cdot2 - 1\cdot(-3) = 5 \ne 0.</math>

Since the determinant is non-zero, the vectors (1, 1) and (−3, 2) are linearly independent.

This method can only be applied when the number of vectors equals the length of the vectors.

Example II

Let V = Rⁿ and consider the following elements in V:

<math>\begin{matrix}

\mathbf{e}_1 & = & (1,0,0,\ldots,0) \\ \mathbf{e}_2 & = & (0,1,0,\ldots,0) \\ & \vdots \\ \mathbf{e}_n & = & (0,0,0,\ldots,1).\end{matrix}</math>

Then e₁, e₂, ..., e_n are linearly independent.

Proof

Suppose that a₁, a₂, ..., a_n are elements of R such that

<math> a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 + \cdots + a_n \mathbf{e}_n = 0 \,</math>

Since

<math> a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 + \cdots + a_n \mathbf{e}_n = (a_1 ,a_2 ,\ldots, a_n) \,</math>

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^2t in V are linearly independent.

Proof

Suppose a and b are two real numbers such that

ae^t + be^2t = 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.

The projective space of linear dependences

A linear dependence among vectors v₁, ..., v_n is a tuple (a₁, ..., a_n) with n scalar components, not all zero, such that

<math>a_1 \mathbf{v}_1 + \cdots + a_n \mathbf{v}_n=0. \,</math>

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 v₁, ...., v_n is a projective space.

See also

• orthogonality
• matroid (generalization of the concept)de:Lineare Unabhängigkeit

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