Exercise 1.18

Proof. Let $v_1,\dots ,v_n$ be a finite collection of vectors.

• Suppose that $v_1,\dots ,v_n$ are linearly independent. Suppose for the sake of contradiction that there is a vector $v_j$, $1\le j\le n$, which can be expressed as a linear combination of other vectors in the family, i.e, there exists a collection of scalars $a_1,\dots ,a_{j-1},a_{j+1},\dots ,a_n\in \Re$ such that

  $$v_j=\sum _{i\ne j}{a}_i{v}_i.$$

  Setting $a_j=-1$ we obtain

  $$\sum _{i=1}^n{a}_i{v}_i=\sum _{i\ne j}{a}_i{v}_i-v_j=0$$

  meaning that we have found a non-trivial linear combination of $v_1,\dots ,v_n$ which sums up to a zero vector. This is a contradiction to the assumption that these vectors are linearly independent.

• Now assume that none of the vectors $v_1,\dots ,v_n$ can be expressed as a linear combination of the others. Suppose for the sake of contradiction that $v_1,\dots ,v_n$ are not linearly independent, i.e., we can find scalars $a_1,\dots ,a_n\in \Re$ such that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{a}_i{v}_i=0$. Solving for $v_1$, we arrive at a contradiction:

  $$v_1=\sum _{i=2}^n\frac{a_i}{a_1}{v}_i.$$

  Thus, $v_1,\dots ,v_n$ must be linearly independent.