Exercise 6.B.9

Proof. Suppose $v_1,\dots ,v_m$ is a lienarly dependent list in $V$. Let $k$ be the smallest integer such that $v_k\in \mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_{k-1})$. Then $v_1,\dots ,v_{k-1}$ is linearly independent and we can apply the Gram-Schmidt Procedure to produce an orthonormal list $e_1,\dots ,e_{k-1}$ whose span is the same. Therefore $v_k\in \mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(e_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,e_{k-1})$ and, by 6.30,

But the right-hand side is exactly what we subtract from $v_k$ when calculating $e_k$, hence the Gram-Schmidt Procedure cannot continue because we can’t divide by $0$. If, however, you discard $v_k$ (and every other vector to which happens the same thing), you end up producing an orthonormal basis whose span equals $\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_m)$. □