You can find the description of the inertial partition algorithm either in [1] or [2].
I think, I understood the main idea behind this algorithm:
We have the spatial coordinates of points that we want to bisect. We find a line (or, hyperplane, in general) $L$ through these points, such that it is a total least squares fit of the nodes. We project the points onto this line. We find the "median projection", which then is used to split the points into two partitions: that is, the perpendicular hyperplane to $L$ that goes through the "median projection" divides the points into two partitions.
Nonetheless, I have a doubt regarding the theory behind the inertial partitioning algorithm.
Now, it turns out that the idea above can be framed as an eigenvalue-eigenvector problem (as for spectral partitioning algorithm). More specifically, it turns out, after a derivation (see [2]), we want to minimise the quantity $$\vec{u}^T M \vec{u} = \lambda,$$ i.e. find $\vec{u}$ such that $\lambda$ is minimised.
According to [2], $u=[a, b]$ is the unit eigenvector of $M$ corresponding to the smallest eigenvalue.
The fact that $\vec{u}$ is a unit eigenvector, I suppose, is related to the fact that the author of [2] assumes (w.l.o.g.) that $a^2 + b^2 = 1$. Why does the author assume $a^2 + b^2 = 1$? Why would that be convenient? Why is the assumption w.l.o.g.? Note $-\frac{a}{b}$ is the slow of the line $L$.
Given the assumption $a^2 + b^2 = 1$, I understand that $\vec{u} = [a, b]$ is a unit eigenvector, because $\|\vec{u}\| = \sqrt{a^2 + b^2} = 1 \iff a^2 + b^2 = 1$.
Why is $\vec{u}$ the eigenvector corresponding to the smallest eigenvalue of $M$? Why do we care about the smallest eigenvalue of $M$?
Once we have found $\vec{u}$, we have the slow of the line. We can also retrieve the center of mass, that is $(\bar{x},\bar{y})$, which is used to formulate the equation of the line of $L$: it is basically an average of the points.