A nearly dependent finite basis has an ill-conditioned overlap matrix

Both vectors have norm one and $v_1^T{v}_2=1-\epsilon$, so their Gram matrix is the displayed $S_{\epsilon }$. Its eigenvectors $(1,1)$ and $(1,-1)$ have eigenvalues $2-\epsilon$ and $\epsilon$, both positive. Thus $S_{\epsilon }$ is positive definite and its spectral condition number is $(2-\epsilon )\mathrm{/}\epsilon$. Also $\mathrm{\parallel }{v}_1-v_2{\mathrm{\parallel }}^2=2\epsilon$, which tends to zero.

For $H=I$, diagonalizing $S_{\epsilon }$ in the same two eigenvectors shows that $Hc=\lambda S_{\epsilon }c$ has eigenvalues $\lambda =(2-\epsilon {)}^{-1}$ and $\lambda ={\epsilon }^{-1}$. The Cholesky factor is $L=\left(\begin{array}{cc}1& 0\\ 1-\epsilon & \sqrt{2\epsilon -{\epsilon }^2}\end{array}\right),$ so $S_{\epsilon }=L{L}^{\ast }$ and its positive diagonal makes it the factor in [F2]. By [F3], the whitened Hermitian problem has the same two eigenvalues. For example, at $\epsilon =1\mathrm{/}N$ with integer $N\ge 2$, the condition number is $2N-1$, which grows without bound although whitening is an exact change of coordinates.