Explanation The easiest way (I think!) is to look at the first two vectors (u₁ and u₂) in R⁵. Their respective coordinates all differ by 2, so that u₂–u₁ is the vector which has all 2's as entries. But add two-fifths (that is, 2/5) of u₃ to that difference, and the result will be 0. So here u₂–u₁+(2/5)u₃=0, and we have a non-trivial linear combination of the elements of the set {u₁, u₂, u₃} equal to 0. This set is not linearly independent.