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In the solution of an exercise about separation of the two sets of ellipsoids I found a thought like this: "We first note that the problem is homogeneous in a and b, so we can replace the strict inequalities $a^Tx + b > 0$ and $a^Tx + b < 0$ with $a^Tx + b \ge 1$ and $a^Tx + b \le −1$, respectively." (http://egrcc.github.io/docs/math/cvxbook-solutions.pdf, page $96$, ex. $4.25$)

Now I wonder should one shift the inequalities by $-1$ and $1$ respectively? How does that not change the solution sets?

user13
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1 Answers1

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The main idea is that a hyperplane $H_{a,b}(x) = a^\top x + b$ defined by parameters $a,b$ gives a halfspace $X = \{x : H_{a,b}(x) \geq 0\}$ that is positive homogenous. That is, for $\hat{X} = \{\hat{x}: H_{a/t,b/t}(\hat{x})\geq0/t\}$ and $t>0$ (to preserve the direction of the inequality), we have $X=\hat{X}$.

To see this, take some $x\in X$. Fix $t>0$. Then $a^\top x + b \geq 0$. This means that $$(1/t) \times (a^\top x + b) = (a/t)^\top x + (b/t) \geq 0,$$ so that $x\in\hat{X}$. For the other containment, take some $\hat{x}\in\hat{X}$. Then $$(a/t)^\top\hat{x} + (b/t) = (1/t) \times (a^\top \hat{x} + b) \geq0$$ so that $\hat{x} \in X$. Thus $X=\hat{X}$.

For strict inequalities, we can do a $0<\epsilon\ll1$ argument and send $\epsilon\to0$. That is, we know $$a^\top x + b > 0 \iff a^\top x + b \geq \epsilon$$ for $\epsilon>0$. But sending $\epsilon\to0$ will give us a "$\geq$" inequality in the limit. This is why the "$\geq$" appears, and by the previous argument involving positive homogeneity, we can scale the RHS to 1 (and move the $b$ term around).

The same sort of argument should work for the other half of the hyperplane, i.e., the "$\leq$" part.

So in the solutions, the $a$ and $b$ are probably "different" but still characterize the same set. And we could take $b=0$ after making the RHS $\pm1$.

jjjjjj
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