Take the 2-minute tour ×  Let $v_1, \dots, v_m$ be a linearly
dependent list of vectors. If $v_1 \ne 0$, then there is some
$v_j$ in the span of $v_1, \dots, v_{j-1}$ If we let j be the
smallest integer with this property, and apply the gram-schmidt
procedure to produce an orthonormal list $(e_1, \dots, e_{j-1})$
then $v_j$ is in the span of $(e_1, \dots, e_{j-1})$ and $$v_j =
\langle v_j, e_1\rangle e_1+ \dots + \langle v_j, e_{j-1}\rangle
e_{j-1}$$ Why does this guarantee that length of $v_j$=0? I'm
missing something about linear dependence that should probably
be obvious sorry :\  share|improve this question add comment  2
Answers  up vote 3 down vote accepted  The Gram-Schmidt process
goes as follows: given $v_1,\ldots,v_n$, you define
$$\begin{align*} u_1 &= v_1\\ e_1 &= \frac{1}{||u_1||}u_1;\\ u_2
&= v_2 - \langle v_2,e_1\rangle e_1;\\ e_2 &=
\frac{1}{||u_2||}u_2;\\ &\vdots\\ u_{k+1} &= v_{k+1} -
\left(\langle v_{k+1},e_1\rangle e_1 + \langle
v_{k+1},e_2\rangle e_2 + \cdots + \langle v_{k+1},e_k\rangle