I am not quite familiar to proof question about orthonormal basis and hope someone could give some suggestions. Here is the question.
Given the spanning set of vectors $$V=\left\{\vec{v_{1}}, \vec{v_{2}}, ..., \vec{v_{n}}\right\}\in \mathcal{V}=\mathbb{R^{n}}$$ which are a basis for the vector size $\mathcal{V}$, consider the following algorithm,
$$\vec{u_{k}}=\left\{\begin{matrix} \frac{\vec{v_{k}}}{\left\|\vec{v_{k}}\right\|} & for \ k=1\\ \frac{\vec{v_{k}}-\sum^{k-1}_{i=1}\left(\vec{v_{k}}\cdot\vec{u_{i}}\right)\vec{u_{i}}}{\left\|\vec{v_{k}}-\sum^{k-1}_{i=1}\left(\vec{v_{k}}\cdot\vec{u_{i}}\right)\vec{u_{i}}\right\|} & for \ k>1 \end{matrix}\right.$$
Show that the set $$U=\left\{\vec{u_{1}}, \vec{u_{2}}, ..., \vec{u_{n}}\right\}$$ is an orthonormal basis for $\mathcal{V}$.
Thanks very much!
Each element of $U$ has norm $1$. All that remains to be proved is that they're orthogonal. (I'm sorry, but I have no patience for those little arrows).
Note that\begin{align}u_1\perp u_2&\iff\frac{v_1}{\|v_1\|}\perp\left(v_2-\left\langle v_2,\frac{v_1}{\|v_\|}\right\rangle\frac{v_1}{\|v_1\|}\right)\\&\iff\left\langle\frac{v_1}{\|v_1\|},v_2-\left\langle v_2,\frac{v_1}{\|v_1\|}\right\rangle\frac{v_1}{\|v_1\|}\right\rangle=0\\&\iff\frac{\langle v_1,v_2\rangle}{\|v_1\|}-\frac{\langle v_1,v_2\rangle}{\|v_1\|}=0,\end{align}which is true. Therefore, yes, $u_1\perp u_2$.
Now, let $k\in\{1,2\}$. In order to prove that $u_k\perp u_3$, note that $u_3$ is a non-zero scalar times $v_3-\bigl(\langle v_3,u_1\rangle u_1+\langle v_3,u_2\rangle u_2\bigr)$ and therefore that\begin{align}u_k\perp u_3&\iff u_k\perp\left(v_3-\bigl(\langle v_3,u_1\rangle u_1+\langle v_3,u_2\rangle u_2\bigr)\right)\\&\iff \langle u_k,v_3\rangle=\langle v_3,u_1\rangle\langle u_k,u_1\rangle+\langle v_3,u_2\rangle\langle u_k,u_2\rangle,\end{align}which is true, no matter if $k=1$ or $k=2$. And so on.