Household reflector or transformation 
Let $A\in\mathbb{R}^{n\times k}$, $n\geq k$, and $rank(A) = k$. Consider the use of Household reflectors, $H_i$, $1\leq i\leq k$, to transform $A$ to upper trapezoidal form, i.e., $$H_{k}H_{k-1}\ldots H_2 H_1 A = \begin{pmatrix}R\\0\end{pmatrix}$$
  where $R\in\mathbb{R}^{k\times k}$ nonsingular upper triangular. We can define $$H_i = I + \alpha_i x_i x_i^{T}$$
  $$\alpha_i = \frac{-2}{\|x_i\|_{2}^{2}}$$
  not sure what $x_i$ is though don't see it in my notes anywhere.

So suppose $$A = \begin{pmatrix}
2 & 1\\
1 & 2\\
1 & 2\\
\end{pmatrix}$$
and I want to get $R$. I tried to apply the formula but my $R$ is not upper trapezoidal. So there must be something I am doing wrong. Could someone kindly show the steps so I know how to do this?
Attempt - $$H_1 = I + \frac{-2}{6}\begin{pmatrix}
2\\
1\\
1\\
\end{pmatrix}\begin{pmatrix} 2 & 1 & 1\end{pmatrix}$$ thus we have $$\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 1\\
0 & 0 & 0\\
\end{pmatrix} - \frac{1}{3}\begin{pmatrix}
4 & 2 & 2\\
2 & 1 & 1\\
2 & 1 & 1\\
\end{pmatrix} = \begin{pmatrix}
-1/3 & -2/3 & -2/3\\
-2/3 & 2/3 & -1/3\\
-2/3 & -1/3 & 2/3\\
\end{pmatrix}$$
Thus $$H_1 A = \begin{pmatrix}
-2 & -3\\
-1 & 0\\
-1 & 0\\
\end{pmatrix}$$
thus I have a problem, please let me know if there is an error in my calculations.

The point of $H_1$ is to make the first column a column with zeros except in the top entry.
To find $x_1$, note that $\|(2,1,1)\| = \sqrt{6}$, so we should aim to get $(2,1,1)$ to $\sqrt{6}(1,0,0)$.  Take
$$
x_1 = (2,1,1) - (\sqrt{6},0,0) = (2-\sqrt{6},1,1)
$$
(why should this work? Try to figure it out geometrically).
In the next step, since the first column is already where we want it, we apply a Householder transformation for the $2 \times 2$ submatrix that comes from deleting the first row and column.  In this way, our upper-triangularization process is recursive.
