Here is a general proof for a Vector Space that shows when \(\displaystyle k+1\) will be lin. ind. and lin. dep.

Let \(\displaystyle \mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k\) be lin. ind. vectors in V. If we add a vector \(\displaystyle \mathbf{x}_{k+1}\), do we still have a set of lin. ind. vectors?

(i) Assume \(\displaystyle \mathbf{x}_{k+1}\in\) \(\displaystyle Span (\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k)\)

\(\displaystyle \mathbf{x}_{k+1}=c_1\mathbf{x}_{1}+...+c_k\mathbf{x}_{k}\)

\(\displaystyle c_1\mathbf{x}_{1}+...+c_k\mathbf{x}_{k}+c_{k+1}\mathbf{x}_{k+1}=0\)

\(\displaystyle c_{k+1}=-1\)

\(\displaystyle -1\neq 0\); therefore, \(\displaystyle (\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k, \mathbf{x}_{k+1})\) are lin. dep.

(ii) Assume \(\displaystyle \mathbf{x}_{k+1}\notin\) \(\displaystyle Span (\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k)\)

\(\displaystyle c_1\mathbf{x}_{1}+...+c_k\mathbf{x}_{k}+c_{k+1}\mathbf{x}_{k+1}=0\)

\(\displaystyle c_{k+1}=0\) otherwise \(\displaystyle \mathbf{x}_{k+1}=\frac{-c_1}{c_{k+1}}\mathbf{x}+....+\frac{-c_k}{c_{k+1}}\mathbf{x}_k\) which is a contradiction.

\(\displaystyle c_1\mathbf{x}_{1}+...+c_k\mathbf{x}_{k}+c_{k+1}\mathbf{x}_{k+1}=0\) since \(\displaystyle (\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k, \mathbf{x}_{k+1})\) are lin ind., \(\displaystyle c_1=...c_{k+1}=0\).