Can someone help me understand what's going on in these notes? I don't understand what $e_j$ is supposed to be. How it $\alpha_1$ supposed to be orthogonal to $x_j$? Isn't $\alpha_1$ just a constraint in the primal?
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$e_j$ is the $j$-th standard unit vector. It is a vector with the $j$-th component being $1$ and the other entries being $0$.
The hyperplane $\{v_i=0\}$ is orthogonal to $\alpha_i$ because we can write $v_i=0$ as $0=b_i-\alpha_i^T x$. Let $p$ and $q$ be two points on the hyperplane, then we have
$$0=b_i - \alpha_i^T p$$ $$0=b_i - \alpha_i^T q$$
subtracting the equation, we get $$\alpha_i^T(p-q)=0$$
That is any direction that is parallel to the plane is perpendicular to $\alpha_i$.
Siong Thye Goh
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so how come $\alpha_i$ are the red lines going outwards from the planes? Shouldn't they be the black lines because they are constraints for the feasible region? – Gooby Oct 23 '19 at 05:29
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Perhaps an example might help. Consider the line $x-y=0$, and draw the direction $(1,-1)$, you can see that they are perpendicular. – Siong Thye Goh Oct 23 '19 at 05:52
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Sorry I think I'm just confused to as why $v_i$=0 is the constraint plane and why $\alpha_i$ is perpendicular because the equations representing those planes are the same except that $v_i$ has a constraint and has $v_i$=0 yes? Because $v_i$ is derived from the ith row of the matrix A but setting slack $v_i$=0 and $\alpha_i$ is just the ith row of matrix A without the constraint yes? – Gooby Oct 24 '19 at 21:20
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For example, for the constraint $x-y=1$, $v$ is just $x-y+1$. $x-y+1$ is a linear combination of the variables plus a constant. Equating to zero gives us a hyperplane (a line in the context of $2$ dimensions). $\alpha=(1,-1)$ is the corresponding normal direction. – Siong Thye Goh Oct 25 '19 at 01:07