Editing Gram matrix (section)

==Constructing an orthonormal basis==

Given a set of linearly independent vectors <math>\{v_i\}</math> with Gram matrix <math>G</math> defined by <math>G_{ij}:= \langle v_i,v_j\rangle</math>, one can construct an orthonormal basis
:<math>u_i := \sum_j \bigl(G^{-1/2}\bigr)_{ji} v_j.</math>
In matrix notation, <math>U = V G^{-1/2} </math>, where <math>U</math> has orthonormal basis vectors <math>\{u_i\}</math> and the matrix <math>V</math> is composed of the given column vectors <math>\{v_i\}</math>.

The matrix <math>G^{-1/2}</math> is guaranteed to exist. Indeed, <math>G</math> is Hermitian, and so can be decomposed as <math>G=UDU^\dagger</math> with <math>U</math> a unitary matrix and <math>D</math> a real diagonal matrix. Additionally, the <math>v_i</math> are linearly independent if and only if <math>G</math> is positive definite, which implies that the diagonal entries of <math>D</math> are positive. <math>G^{-1/2}</math> is therefore uniquely defined by <math>G^{-1/2}:=UD^{-1/2}U^\dagger</math>. One can check that these new vectors are orthonormal:
:<math>\begin{align}
\langle u_i,u_j \rangle
&= \sum_{i'} \sum_{j'} \Bigl\langle \bigl(G^{-1/2}\bigr)_{i'i} v_{i'},\bigl(G^{-1/2}\bigr)_{j'j} v_{j'} \Bigr\rangle \\[10mu]
&= \sum_{i'} \sum_{j'} \bigl(G^{-1/2}\bigr)_{ii'} G_{i'j'} \bigl(G^{-1/2}\bigr)_{j'j} \\[8mu]
&=  \bigl(G^{-1/2} G G^{-1/2}\bigr)_{ij} = \delta_{ij}
\end{align}</math>
where we used <math>\bigl(G^{-1/2}\bigr)^\dagger=G^{-1/2} </math>.