# Geometry of complementary slackness

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?

$$e_j$$ 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 $$\alpha_i$$.

• 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? – Dylan Y Oct 23 '19 at 5:29
• 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 5:52
• 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? – Dylan Y Oct 24 '19 at 21:20
• 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 1:07