In linear algebra, a set of elements of a vector space is **linearly independent** if none of the vectorss 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 **R**^{3}, 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**.

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2 Example I 3 Example II 4 Example III: (Calculus required) |

If there do not exist such field elements, then we say that v_{1},v_{2},...,v_{n} are *linearly independent*.
An infinite subset of V is said to linearly independent if all its finite subsets are linearly independent.

To focus the definition on linear independence, we can say that the vectors v_{1},v_{2},..,v_{n} are *linearly independent*, if and only if the following condition is satisfied:

- Whenever a
_{1},a_{2},...,a_{n}are elements of K such that: - a
_{1}v_{1}+ a_{2}v_{2}+ ... + a_{n}v_{n}= 0 - then a
_{i}= 0 for all i=1,2,...,n.

Proof:

Let a, b be two real numbers such that:

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

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

- e
_{1}= (1,0,0,...,0) - e
_{2}= (0,1,0,...,0) - ...
- e
_{n}= (0,0,0,...,1)

Proof:

Suppose that a_{1}, a_{2}, ,a_{n} are elements of **R**^{n} such that

- a
_{1}e_{1}+ a_{2}e_{2}+ ... + a_{n}e_{n}= 0

- a
_{1}e_{1}+ a_{2}e_{2}+ .. + a_{n}e_{n}= (a_{1},a_{2},..,a_{n})

Proof:

Suppose a and b are two real numbers such that

- ae
^{t}+ be^{2t}= 0**(1)**

we differentiate equation **(1)** to get

- ae
^{t}+ 2be^{2t}= 0**(2)**

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

- be
^{2t}= 0

From the first relation we then get:

- ae
^{t}= 0

A **linear dependence** among vectors *v*_{1},...,*v*_{n} is a vector (*a*_{1},...,*a*_{n}) with *n* scalar components, not all zero, such that *a*_{1}*v*_{1}+...+*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 *v*_{1}, ...., *v*_{n} is a projective space.

See also: