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# Linear independence

### From Intelligent Perception

Vectors are linearly independent if none of them is a linear combination of the rest.

For example, (1,0,1), (1,2,0), (0,1,-1), (3, 2 1/2, 1 1/2) are linearly dependent as the last vector is represented as a linear combination of the rest:

2(1,0,1) + 1(1,2,0) + 1/2(0,1,-1) = (2,0,2) + (1,2,0) + (0,1/2,-1/2) = (2+1+0, 0+2+1/2, 2+0-1/2) = (3, 2 1/2, 1 1/2)

Now, suppose you have three vectors (1,0,1), (1,2,0), (0,1,-1), is it possible to find three (non-zero) numbers x, y, z such that this linear combination is 0? To decide, try to find them algebraically:

x(1,0,1) + y(1,2,0) + z(0,1,-1) = (x,0,x) + (y,2y,0) + (0,z,-z) = (x+y, 2y+z, x-z).

Then we have the following system of linear equations:

x+y=0 2y+z=0 x-z=0

Solving it produces: x=y=z=0. Thus, these vectors are linearly independent.

A more detailed example:

In order to find out if vectors (1,1,1), (2,0,3), (-1,2,-5) are linearly independent, we will find all values of a, b, c such that:

a(1,1,1)+b(2,0,3)+c(-1,2,-5)=(0,0,0)

Then, we obtain a system of linear equations:

(1) a+2b-c=0, (2) a+2c=0, (3) a+3b-5c=0.

Now we solve it. First, we solve (2) for a:

(4) a=-2c,

and substitute it into (1) and (3). Then we have:

-2c+2b-c=0, -2c+3b-5c=0.

Rewrite:

(5) 2b-3c=0, (6) 3b-7c=0.

Now we solve (5) for b:

(7) b=3c/2,

and substitute into (6):

3(3c/2)-7c=0.

Solving (6) for c we discover that

c=0.

Substitute c=0 into (7) we discover that

b=0.

Substituting c=0 into (4) we discover that a=0. Thus,

a=0,b=0,c=0.

Therefore the three vectors are linearly independent.

Linear independence is important in homology theory.