Let $\{ {\bf v}_1,{\bf v}_2 \}$ be two linearly independent vectors. An orthogonal base $\{{\bf u}_1,{\bf u}_2 \}$ of the vector space $\mathrm{span}\{ {\bf v}_1,{\bf v}_2 \}$ can be computed using Gram-Schmidt Orthogonalization process (GSO) which is summarized by the formulas $${\bf u}_1={\bf v}_1 \quad\text{and}\quad {\bf u}_2={\bf v}_2-\mu_{2,1}{\bf u}_1 \quad\text{where}\quad\mu_{2,1}=\frac{{\bf v}_2 \cdot {\bf u}_1}{||{\bf u}_1||^2}$$ Geometrically it is given as
In order to get ${\bf u}_2$ i.e., orthogonal to ${\bf u}_1$ project ${\bf v}_2$ onto ${\bf u}_1$ and the length of the projection on ${\bf u}_1$ is given by $\frac{|{\bf v}_2 \cdot {\bf u}_1|}{||{\bf u}_1||}$ and to get the direction we multiply the length by the unit vector $\frac{{\bf u}_1}{||{\bf u}_1||}$ and the $\mathrm{proj}_{{\bf u}_1} {\bf v}_2= \frac{{\bf v}_2 \cdot {\bf u}_1}{||{\bf u}_1||}\times \frac{{\bf u}_1}{||{\bf u}_1||},$ where the length of the projection is $\frac{|{\bf v}_2 \cdot {\bf u}_1|}{||{\bf u}_1||}$. In GSO process $\mu_{2,1}=\frac{\text{length of the projection of \(v_2\) on \(u_1\)}}{||{\bf u}_1||}$ what does $\mu_{2,1}$ represent geometrically?
If $\{ {\bf v}_1,{\bf v}_2 \}$ is the basis of the lattice to solve the shortest vector problem, our aim is to find the orthogonal basis. But orthogonal vectors computed using GSO need not be a basis because $\mu_{2,1}$ need not be an integer.
Since SVP is a hard problem, the LLL algorithm solves approximate SVP. In defining an LLL reduced basis the first requirement is $\mu_{2,1} \le \frac{1}{2}$ (size reduction condition). What does this imply?
