Gram Schmidt and Inner Product I'm working on this problem:
Show that if $(e_1...e_n)$ is an orthonormal basis constructed using the Gram Schmidt process from $(v_1...v_n)$, then for any $j,$ $\langle e_j, v_j \rangle > 0$. 
Intuitively this makes sense to me, as the Gram Schmidt procedure takes a vector $v_j$ and removes all of its components that are part of $\langle e_1,..., e_ {j-1} \rangle$, but how would I prove this mathematically? 
 A: Let $\{w_1,w_2,\ldots,w_n\}$ be obtained from $\{v_1,v_2,\ldots,v_n\}$
by the Gram-Schmidt process, then $w_1=v_1$ and
$$w_j=v_j-\sum_{k=1}^{j-1}\frac{\langle v_j,w_k\rangle}{\Vert w_k\Vert^2}w_k
  \quad\text{ for }2\leq j\leq n.$$
Thus $\langle e_1,v_1\rangle=\langle e_1,w_1\rangle=\Vert w_1\Vert\langle e_1,e_1\rangle>0$ and
for $2\leq j\leq n,$
\begin{align*}
\langle e_j,v_j\rangle
&=\left\langle e_j,w_j+\sum_{k=1}^{j-1}\frac{\langle v_j,w_k\rangle}{\Vert w_k\Vert^2}w_k\right\rangle
 =\langle e_j,w_j\rangle+\sum_{k=1}^{j-1}\frac{\overline{\langle v_j,w_k\rangle}}{\Vert w_k\Vert^2}\langle e_j,w_k\rangle\\
&=\langle e_j,w_j\rangle+\sum_{k=1}^{j-1}\frac{\overline{\langle v_j,w_k\rangle}\Vert w_k\Vert}{\Vert w_k\Vert^2}\langle e_j,e_k\rangle
 =\langle e_j,w_j\rangle+\sum_{k=1}^{j-1}\frac{\overline{\langle v_j,w_k\rangle}}{\Vert w_k\Vert}\delta_{jk}\\
&= \langle e_j,w_j\rangle=\Vert w_j\Vert\langle e_j,e_j\rangle>0,
\end{align*}
where $\delta_{jk}$ is the Kronecker delta function.
A: By orthonormality of the basis,
$$v_j=\sum_k(e_k\cdot v_j)e^k,$$
and
$$v_j^2=\sum_k(e_k\cdot v_j)^2,$$
(the orthogonal transform preserves the $L_2$ norm) so that
$$v_j^2\ge\sum_{k<j}(e_k\cdot v_j)^2$$ (a vector is longer than its projection).
Then
$$u_j:=\|u_j\|e_j:=v_j-\sum_{k<j}(e_k\cdot v_j)e_k$$
(the vectors $u_j$, then $e_j$ are obtained by projection) and
$$\|u_j\|e_j\cdot v_j=u_j\cdot v_j=v_j^2-\sum_{k<j}(e_k\cdot v_j)^2.$$
