# LINEAR DEPENDENCE AND INDEPENDENCE OF VECTORS

## Linear Dependence

The set {v1, v2, v3, ............, vn} is linearly dependent if the equation

b1v+ b2v+ b3v+ ............ + bnvn = 0

such that

b1 or b2 or b3 ............ or bn  0

That is, all the weights b1, b2, b3 ............ bn are not zero.

## Linear Independence

A collection of vectors {v1, v2, v3, ............, vn} is linearly independent if the equation

a1v+ a2v+ a3v+ ............ + anvn = 0

implies that

a1 = a2 = a3 = ............ = an = 0

It has only trivial solution.

Example 1 :

Show the following set of vectors is linearly dependent.

{(1, 2, 1), (0, 1, 0), (-2, 0, -2)}

Solution :

Let a, b and c be the scalars.

We can write

2b - (½)c = 2(0, 1, 0) - (½)(-2, 0, -2)

2b - (½)c = (0, 2, 0) - (-1, 0, -1)

2b - (½)c = (0, 2, 0) + (1, 0, 1)

2b - (½)c = (1, 2, 1)

2b - (½)c = a

Since a can be written as a linear combination of b, c, the set is linearly dependent.

Example 2 :

Show the following set of vectors is linearly independent.

{(1, 0, 1), (1, 1, -1), (0, 1, 0)}

Solution :

Let a, b and c be the scalars. Then, we have

a(1, 0, 1) + b(1, 1, -1) + c(0, 1, 0) = (a + b, b + c, a - b)

The zero vector is (0, 0, 0).

If the given set of vetcors is linearly independent,

a(1, 0, 1) + b(1, 1, -1) + c(0, 1, 0) = 0

Then, we have

a + b = 0 ----(1)

b + c = 0 ----(2)

a - b = 0 ----(3)

From (3), we see that a = b.

Substitute b = a into (1).

a + a = 0

2a = 0

a = 0

Since, a = b, we have

b = 0

Substitute b = 0 into (2).

0 + c = 0

c = 0

Since all of the scalars a, b and c are zero, the set is linearly independent.

We can show that a set of vectors is linearly independent by arranging them in a matrix form. Then row reduce the matrix; if each row has a nonzero pivot, then the vectors are linearly independent.

Example 3 :

Determine if the following set of vectors is linearly independent.

{(1, 3, 5) , (4,−1, 2) , (0,−1, 2)}

Solution :

We can arrange the given vectors as a matric, using each vector as a column. Then, the matrix is

Now we row reduce the matrix.

Since all the columns in the reduced matrix contain a pivot entry, no vector can be written as a linear combination of the other vectors; therefore, the set is linearly independent.

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