Sketch reconstruction from Gram matrix

Gram-matrix-parameterization.md

@@ -6,9 +6,15 @@ Take a 5d vector space $V$ with a bilinear form $(\_\!\_, \_\!\_)$ of signature
 
 ### Constraints as Gram matrix entries
 
-The vectors $a_1, \ldots, a_n \in V$ representing the elements of our construction can be encoded in a linear map $A \colon \mathbb{R}^n \to V$, whose matrix is
-\[ \left[\begin{array}{cccc} \rule{0.5pt}{16pt} & \rule{0.5pt}{16pt} & & \rule{0.5pt}{16pt} \\ a_1 & a_2 & \cdots & a_n \\ \rule{0.5pt}{16pt} & \rule{0.5pt}{16pt} & & \rule{0.5pt}{16pt} \end{array}\right] \]
-We can then express constraints by fixing elements of the Gram matrix $G = A^\top A$, where $\top$ is the adjoint with respect to the inner product $\langle\_\!\_, \_\!\_\rangle$ on $\mathbb{R}^n$ and the Lorentz form. Since the inner product and the Lorentz form are both non-degenerate, the rank of $G$ matches the dimension of the image of $A$.
+The vectors $a_1, \ldots, a_n \in V$ representing the elements of our construction can be encoded in the linear map $A \colon \mathbb{R}^n \to V$ with $Ae_j = a_j$ for each of the standard basis vectors $e_1, \ldots, e_n$. We can then express constraints by fixing elements of the Gram matrix $G = A^\top A$, where $\top$ is the adjoint with respect to the inner product $\langle\_\!\_, \_\!\_\rangle$ on $\mathbb{R}^n$ and the Lorentz form. Since the inner product and the Lorentz form are both non-degenerate, the rank of $G$ matches the dimension of the span of $a_1, \ldots, a_n$.
 
 The symmetric bilinear form $\langle\_\!\_, G\_\!\_\rangle$ is the pullback of the Lorentz form along $A$:
 \[\begin{align*} \langle\_\!\_, G\_\!\_\rangle & = \langle\_\!\_, A^\top A\_\!\_\rangle \\ & = (A\_\!\_, A\_\!\_). \end{align*}\]
+
+## Reconstruction
+
+### Reconstruction from the Gram matrix
+
+Let's say $a_1, \ldots, a_n$ span $V$. Write $G$ as $LU$, where the matrix $U$ is a row echelon form of $G$, and the the matrix $L$ encodes the sequence of row operations that restores $G$. Since $G$ has rank five, the first five rows of $U$ are non-zero, and the rest are zero. Thus, the first five columns of $L$ are non-zero, and the rest are zero. Since $G$ is symmetric, we should have $L = U^\top D$, where $D$ is a diagonal matrix that has one negative entry and four positive entries at the pivot positions on the diagonal.
+
+The first five rows of $U$ form the matrix of $A$ in some orthogonal basis for $V$. The matrix $D$ describes the Lorentz form in this basis.