To show that vectors **a, b, c** are linearly independent, show that it is not possible to find real numbers m, n such that **c** = m**a** + n**b**. This will generally involve finding a set of simultaneous equations in m and n, and showing that there is no simultaneous solution.

I'm in the middle of exams right now, so I'll keep this brief, but this actually is

**not** true and is a common misconception among Specialist Maths students. I'll be writing a concise guide on linear dependence/independence in the near future.

In a nutshell though, the

**correct** statement relating linear combinations of 3 vectors \((\mathbf{a},\mathbf{b},\mathbf{c})\) and linear independence/dependence is:

\[\exists \,m,n\in\mathbb{R}\setminus\{0\}\ \ \text{s.t.}\ \ \mathbf{c}=m\mathbf{a}+n\mathbf{b}\ \implies\ \mathbf{a},\mathbf{b},\mathbf{c}\ \text{are linearly dependent vectors}\] Simply negating both sides of the statement doesn't make it equivalent. The following is an

**incorrect** statement:

\[\forall\,m,n\in\mathbb{R}\setminus\{0\},\ \ \mathbf{c}\neq m\mathbf{a}+n\mathbf{b}\ \implies\ \mathbf{a},\mathbf{b},\mathbf{c}\ \text{are linearly independent vectors}\]A simple counterexample is shown by the vectors \[\mathbf{a}=2\mathbf{i}-\mathbf{j}+\mathbf{k},\ \ \mathbf{b}=-4\mathbf{i}+2\mathbf{j}-2\mathbf{k},\ \ \mathbf{c}=3\mathbf{i}+2\mathbf{j}+\mathbf{k}\] It can very easily be shown that \(\mathbf{c}\neq m\mathbf{a}+n\mathbf{b}\ \ \forall\,m,n\in\mathbb{R}\) yet its obvious that \(\mathbf{a},\mathbf{b},\mathbf{c}\) are linearly dependent since \[2\mathbf{a}+\mathbf{b}+0\mathbf{c}=\mathbf{0}.\]