# All Science Fair Projects

## Science Fair Project Encyclopedia for Schools!

 Search    Browse    Forum  Coach    Links    Editor    Help    Tell-a-Friend    Encyclopedia    Dictionary

# Science Fair Project Encyclopedia

For information on any area of science that interests you,
enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.).
Or else, you can start by choosing any of the categories below.

# Linear independence

In linear algebra, a set of elements of a vector space is linearly independent if none of the vectors in the set can be written as a linear combination of finitely many other vectors in the set. 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

## Definition

Let V be a vector space over a field K. If v1, v2, ..., vn are elements of V, we say that they are linearly dependent over K if there exist elements a1, a2, ..., an in K not all equal to zero such that:

$a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + \cdots + a_n \mathbf{v}_n = \mathbf{0}$

or, more concisely:

$\sum_{i=1}^n a_i \mathbf{v}_i = \mathbf{0} \,$

(Note that the zero on the right is the zero element in V, not the zero element in K.)

If there do not exist such field elements, then we say that v1, v2, ..., vn are linearly independent. An infinite subset of V is said to be linearly independent if all its finite subsets are linearly independent.

To focus the definition on linear independence, we can say that the vectors v1, v2, ..., vn are linearly independent, if and only if the following condition is satisfied:

Whenever a1, a2, ..., an are elements of K such that:

a1v1 + a2v2 + ... + anvn = 0

then ai = 0 for i = 1, 2, ..., n.

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.

## The projective space of linear dependences

A linear dependence among vectors v1, ..., vn is a vector (a1, ..., an) with n scalar components, not all zero, such that

$a_1 v_1 + \cdots + a_n v_n=0. \,$

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 v1, ...., vn is a projective 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:

$a(1, 1) + b(-3, 2) = (0, 0) \,$

Then:

$( a - 3 b , a + 2 b ) = (0, 0) \,$
and
$a - 3b = 0 \,$
and
$a + 2b = 0. \,$

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

## Example II

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

$e_1 = (1,0,0,\ldots,0) \,$
$e_2 = (0,1,0,\ldots,0) \,$
$\cdots \,$
$e_n = (0,0,0,\ldots,1) \,$

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

Proof:

Suppose that a1, a2,...,an are elements of Rn such that

$a_1 e_1 + a_2 e_2 + \cdots + a_n e_n = 0 \,$

Since

$a_1 e_1 + a_2 e_2 + \cdots + a_n e_n = (a_1 ,a_2 ,\ldots, a_n) \,$

then ai = 0 for all i in {1, .., n}.

## Example III: (calculus required)

Let V be the vector space of all functions of a real variable t. Then the functions et and e2t in V are linearly independent.

Proof:

Suppose a and b are two real numbers such that

$a e^t + b e^{2 t} = 0 \,$                 (1)

for all values of t. We need to show that a = 0 and b = 0. In order to do this, we differentiate both sides of (1) to get

$a e^t + 2 b e^{2 t} = 0 \,$               (2)

which also holds for all values of t.

Subtracting the first relation from the second relation, we obtain:

$b e^{2 t} = 0 \,$

and, by plugging in t = 0, we get b = 0.

From the first relation we then get:

$a e^t = 0 \,$

and again for t = 0 we find a = 0.