Consider set of arbitrary non-zero vectors $\{v_1,...,v_n\}$ in an inner product space $V$. Consider set $S = \{p_1,...,p_n\}$, where $p_1 = v_1$ and $$p_j = v_j - \biggr[\sum_{x=1}^{j-1}\frac{\langle v_j,p_x \rangle}{||p_x||^2}p_x\biggr]$$ where $j > 1$
Show that $S$ is pairwise orthogonal.
My attempt:
By induction.
Base case: $j = 2$
We have $$ \langle p_1,p_2 \rangle = \big\langle p_1, v_2 - \frac{\langle v_2, p_1 \rangle}{||p_1||^2}p_1 \big\rangle = 0$$
Induction step:
Suppose for arbitrary $j < n$, it is true that for all $i < j$, $\langle p_i,p_j \rangle = 0$
Take $p_{j+1}$, and take arbitrary $k < j + 1$.
We have
$$\begin{align}\big \langle p_k, p_{j+1} \big \rangle & = \bigl \langle p_k,v_{j+1} - \biggr[\sum_{x=1}^{j}\frac{\langle v_{j+1},p_x \rangle}{||p_x||^2}p_x\biggr] \bigr \rangle \\ & = \bigl \langle p_k,v_{j+1} \bigl \rangle - \underbrace{\bigl \langle p_k, \sum_{x=1}^{j}\frac{\langle v_{j+1},p_x \rangle}{||p_x||^2}p_x\bigl \rangle}_{= \langle p_k,v_{j+1} \rangle} \\ & =0 \end{align}$$
Hence $S$ is pairwise orthogonal. $\blacksquare$
Is it correct?