Symmetric preconditioning as energy geometry ($\Vert x-x_{\ast }{\Vert }_A=\Vert M^{1/2}(x-x_{\ast }){\Vert }_B$)

For a symmetric positive-definite matrix $A$, solving $A{x}_{\ast }=b$ is equivalent to minimizing the quadratic form with linear termWriting the error vector $e=x-x_{\ast }$ and expanding gives the exact energy norm identityThe cross term vanishes because $A{x}_{\ast }=b$. With $\eta =Se$ and $B=S^{-1}A{S}^{-1}$, the same quantity is ${\eta }^{T}B\eta /2$. Symmetric positive-definite preconditioning changes coordinates without changing the energy assigned to a corresponding error vector.

In an orthonormal eigenbasis of $A$, the level set $e^{T}Ae=t>0$ has semiaxis lengths $\sqrt{t/{\lambda }_i}$. Its longest-to-shortest axis ratio is $\sqrt{{\kappa }_2(A)}$. A large spectral condition number of a positive-definite matrix therefore appears geometrically as a thin ellipsoid. Good scaling makes the transformed ellipsoid rounder, allowing a simple search direction to improve all coordinates together.

The negative gradient of $F$ is the linear-system residual $r$. Relative to the norm $\Vert v{\Vert }_M=\sqrt{v^{T}Mv}$, a steepest direction is $z=M^{-1}r$: the Cauchy-Schwarz inequality applied to $r^{T}v=(S^{-1}r{)}^T(Sv)$ shows that the greatest decrease per unit $M$-length occurs along this direction. Equivalently, ordinary gradient descent in $y=Sx$ becomes preconditioned steepest descent in $x$.^[ref]