To decide if a set of vectors is linearly independent, we need to check whether or not each one is a linear combination of the others. This process can result in several linear systems to solve: for three vectors in 3-space, for example, there are 3 separate systems to solve. Fortunately, it is possible to reduce the number of systems to solve to just one. Let's look at how to do it for three vectors u, v and w. The basic idea: use a linear dependence relation. Specifically, set up the equation

au + bv + cw = 0

and look at what the coefficients a, b and c can be. There's obviously the trivial solution a = b = c = 0; we're interested in non-trivial solutions (i.e. with at least one of a, b or c not equal to 0).

Suppose there is one, say with a ≠ 0. Then you can solve for u in as a linear combination of the other vectors:

u = –(b/a)v – (c/a)w,

so the vectors are linearly dependent.

On the other hand, suppose the vectors are linearly dependent, say with u a linear combination of the others:

u = pv + qw.

Then (–1)u + pv + qw = 0, so the linear dependence relation has the non-trivial solution a = –1, b = p, c = q.

In other words, linear dependence is equivalent to the existence of a non-trivial solution of the linear dependence relation.

Example: Are the vectors [1, 2, 3], [4, 5, 6] and [7, 8, 9] linearly dependent or linearly independent?

There are three vectors in 3-space, so you can't tell geometrically whether or not they're linearly independent (unless you plot them, perhaps). So set up a linear dependence relation:

a[1, 2, 3] + b[4, 5, 6] + c[7, 8, 9] = [0, 0, 0].

This equation is equivalent to the linear system

  a + 4b + 7c = 0
2a + 5b + 8c = 0
3a + 6b + 9c = 0.

Solve the system (by any means you know). You don't actually have to find all the non-trivial solutions; you just need to find out if the system has any non-trivial solutions. Check the determinant of the coefficient matrix (it turns out to be 0), or find the reduced row echelon form of the coefficient matrix (it's not the identity matrix), or ... . The system does indeed have non-trivial solutions, so the original vectors are linearly dependent.

This works in general: to determine whether or not a set of vectors is linearly independent,

1. Set up the linear dependence equation for those vectors and solve it.
2. If you get only the trivial solution (all coefficients zero), the vectors are linearly independent.
3. If you get any solution other than the trivial solution, the vectors are linearly dependent.

We can even apply this form of the linear independence definition to a set containing just one vector: the set {u} is linearly independent whenever the equation cu = 0 has only the trivial solution c = 0. This happens whenever u ≠ 0, so the set {0} is linearly dependent, while the set {u} is linearly independent for u ≠ 0.