Sign in Householder algorithm Our teacher gives us the formula for the first Householder,
$$H_1 = I-(2/||\Bbb v||^2_2)\Bbb v \Bbb v^T$$
where
$$\Bbb v = \Bbb a_1 \pm ||\Bbb a_1||_2 \Bbb e_1.$$
When do we use plus and when do we use minus? All the explanations online are vague and use notation I don't understand.
 A: There is a numerical reason to choose one projection over the other. Let $x$ be the point we want to project and let $H_1$ be the projection generated by $w_1=x-\|x\|e_1$ and $H_2$ be the projection generated by $w_2=x-(-\|x\|e_1)=x+\|x\|e_1$. Note that $H_1x=\|x\|e_1$, whereas $H_2x=-\|x\|e_1$. Now assume that the first component of $x$, $x_1$, is positive. This implies that $H_1x$ is closer to $x$ than $H_2x$. How close? Assume that $\|x-H_1x\|<\varepsilon << 1$. Then $\|x-H_1|=\|x-\|x\|e_1\| = \|w_1\| < \varepsilon$. Recall that the Householder algorithm requires the computation of the normal vector $w_1$. However, since $\|w_1\| < \varepsilon$, and $w_1=x-\|x\|e_1$, it follows that $x\approx \|x\|e_1$, and we could experience catastrophic cancellation when subtracting these two quantities. Consequently, we choose the projection that projects $x$ the farthest from $x$, i.e., $w=x+\text{sign}(x_1)\|x\|e_1$.
A: If I remember correctly, the algorithm (bringing $A$ into upper Hessenberg form) should work regardless; it makes no difference whether you use $+$ or $-$. 
One choice produces a positive upper-left entry after we compute $H_1A$, the other produces a negative entry.  For a positive entry, use $+$.  For a negative entry, use $-$.
