Gram-Schmidt Orthogonalization for lattices in matrix form, how it works?

Question

I'm reading this pdf about lattices and there's this section where it talks about Gram-Schmidt Orthogonalization for a basis of a lattice.

I know what a basis for a lattice is and also know about the Gram-Schmidt process and even done a few by hand for simple vectors in $\mathbb{R}^3$. The idea is that by this process we come up with a new basis set where the vectors are all orthogonal.

In the Gram-Schmidt I studied we used projections. However, I'm not recognizing anything that appears here. I don't even know what $*$ is supposed to mean.

What is happening and why the new basis is orthogonal?

Answer 1

Recall how the the Gram-Schmidt Process is carried out normally:

Here, $Q = \begin{pmatrix} \textbf{e}_1 & \textbf{e}_2 & ... & \textbf{e}_k\end{pmatrix}$, while $\widetilde{\mathbb{B}} = \begin{pmatrix}\textbf{u}_1 & \textbf{u}_2 & ... & \textbf{u}_k\end{pmatrix}$

That is, $Q$ is the orthonormal basis generated via Gram-Schmidt, while $\widetilde{\mathbb{B}}$ is merely orthogonal, generated in much the same way, but without the final step of normalizing all of the vectors. This is because normalizing vectors will change the lattice that they generate.