When is a vector wasted?
A vector is wasted if it is a linear combination of the others — you can already reach it by scaling and adding the vectors you have. Take (1,0), (0,1), and (2,3). The third one is wasted, because (2,3) = 2*(1,0) + 3*(0,1). It adds no new direction; it only repeats what the first two already span.
A set is linearly independent when none of its vectors is wasted — every vector points somewhere the others cannot reach. If even one is wasted, the set is dependent.
The only-zero test
Checking each vector one by one is slow. Here is the standard test: vectors v1, ..., vk are independent exactly when the only way to make c1*v1 + ... + ck*vk = 0 is to take every coefficient c = 0. If some non-zero coefficients also give zero, you can solve for one vector in terms of the rest — so it was wasted.
- Write the combination c1*v1 + ... + ck*vk = 0 as a system A*c = 0, with the vectors as columns of A.
- Solve by elimination.
- If c = 0 is the only solution, the set is independent. If other solutions exist, it is dependent.
A tiny worked example
v1 = (1, 2) v2 = (2, 4) c1*v1 + c2*v2 = 0 c1 + 2 c2 = 0 2 c1 + 4 c2 = 0 -> row 2 is just 2 * row 1 -> c1 = -2 c2 works for ANY c2 -> e.g. c1 = -2, c2 = 1 gives 0 => DEPENDENT (v2 = 2 * v1)